Instructional Materials for Fundamentals of Mixed-Signals Sensors (PDF)

Table of Contents LESSON 1: Operational Amplifiers........................................................................................................ 6 Introduction to Operational Amplifiers................................................................................................. 6 The Inverting Amplifier......................................................................................................................... 12 The Noninverting Amplifier................................................................................................................. 16 Two Op-Amp Applications.................................................................................................................. 18 Linear ICs.............................................................................................................................................. 21 Activities/Assessments........................................................................................................................ 22 LESSON 2: Differential Amplifiers........................................................................................................ 23 Differential Amplifier............................................................................................................................. 23 DC Analysis of a Diff Amp................................................................................................................... 26 AC Analysis of a Diff Amp................................................................................................................... 27 Input Characteristics of an Op Amp.................................................................................................. 31 Common-Mode Gain........................................................................................................................... 35 Activities/Assessments........................................................................................................................ 37 LESSON 3: Linear Op-Amp Applications............................................................................................ 38 Inverting-Amplifier Circuits.................................................................................................................. 38 Noninverting-Amplifier Circuits........................................................................................................... 38 Inverter/Non-inverter Circuits.............................................................................................................. 40 Differential Amplifiers........................................................................................................................... 42 Instrumentation Amplifiers.................................................................................................................. 43 Current Boosters.................................................................................................................................. 45 Voltage-Controlled Current Sources................................................................................................. 47 Single-Supply Operation..................................................................................................................... 48 Activities/Assessments........................................................................................................................ 50 LESSON 4: Non-Linear Op-Amp Applications.................................................................................... 51 Comparators with Zero Reference.................................................................................................... 51 Comparators with Nonzero References............................................................................................ 55 Comparators with Hysteresis............................................................................................................. 58 The Integrator....................................................................................................................................... 62 Active-Diode Circuits........................................................................................................................... 64 The Differentiator.................................................................................................................................. 66 Class-D Amplifier.................................................................................................................................. 67 Activities/Assessments........................................................................................................................ 67 LESSON 5: Analog to Digital Conversion............................................................................................ 68 Conversion Parameters....................................................................................................................... 68 Sampling................................................................................................................................................ 70 Quantization Noise............................................................................................................................... 70 Methods of Analog-to-Digital Conversion......................................................................................... 71 Activities/Assessments........................................................................................................................ 74 LESSON 6: Digital to Analog Conversion............................................................................................ 75 Simple DAC........................................................................................................................................... 75 Improved DAC with Output Op-Amp................................................................................................. 76 DAC Specification Jargon................................................................................................................... 77 DAC Designs......................................................................................................................................... 77 Activities/Assessments........................................................................................................................ 80 LESSON 7: Basic Measurement Theory.............................................................................................. 81 Temperature Measurement................................................................................................................ 81 Strain, Pressure, and Flow Rate Measurements............................................................................ 90 Motion Measurement........................................................................................................................... 95 Level Measurement............................................................................................................................. 97 Neutron Flux Measurement.............................................................................................................. 100 Activities/Assessments...................................................................................................................... 103 LESSON 8: Sensors and Transducers............................................................................................... 104 Transducer.......................................................................................................................................... 104 Types of Transducers........................................................................................................................ 104 Sensors................................................................................................................................................ 106 Types of Sensors............................................................................................................................... 107 Other Types of Sensors.................................................................................................................... 109 Activities/Assessments...................................................................................................................... 109 LESSON 9: Basic Control Devices...................................................................................................... 110 Switches.............................................................................................................................................. 110 Diodes.................................................................................................................................................. 111 Capacitors........................................................................................................................................... 112 Transistors........................................................................................................................................... 112 Circuit Protection Devices................................................................................................................. 114 Electromagnetic Devices................................................................................................................... 117 Motors.................................................................................................................................................. 118 Activities/Assessments...................................................................................................................... 118 APPENDICES......................................................................................................................................... 119 REFERENCES....................................................................................................................................... 123 LESSON 1: Operational Amplifiers Introduction to Operational Amplifiers The term operational amplifier (op amp) refers to an amplifier that performs a mathematical operation. Historically, the first op amps were used in analog computers, where they did addition, subtraction, multiplication, and so on. At one time, op amps were built as discrete circuits. Now, most op amps are integrated circuits (ICs). The typical op amp is a dc amplifier with very high voltage gain, very high input impedance, and very low output impedance. The unity-gain frequency is from 1 to more than 20 MHz, depending on the part number. An IC op amp is a complete functional block with external pins. By connecting these pins to supply voltages and a few components, we can quickly build all kinds of useful circuits. The figure below shows a block diagram of an op amp. The input stage is a diff amp, followed by more stages of gain, and a Class-B push-pull emitter follower. Because a diff amp is the first stage, it determines the input characteristics of the op amp. In most op amps, the output is single- ended, as shown. With positive and negative supplies, the single-ended output is designed to have a quiescent value of zero. This way, zero input voltage ideally results in zero output voltage. Not all op amps are designed like this. For instance, some do not use a Class-B push-pull output, and others may have a double-ended output. The schematic symbol of an op amp has noninverting and inverting inputs and a single-ended output, illustrated in the left figure below. Ideally, this symbol means that the amplifier has infinite voltage gain, infinite input impedance, and zero output impedance. The ideal op amp represents a perfect voltage amplifier and is often referred to as a voltage-controlled voltage source (VCVS), as shown in the figure on the right below. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 6 The following table summarizes the characteristics of an ideal op amp. Quantity Symbol Ideal LM741C LF157A Open-loop voltage gain 𝐴𝑉𝑂𝐿 Infinite 100,000 200,000 Unity-gain frequency 𝑓𝑢𝑛𝑖𝑡𝑦 Infinite 1 MHz 20 MHz Input resistance 𝑅𝑖𝑛 Infinite 2 MΩ 1012 Ω Output resistance 𝑅𝑜𝑢𝑡 Zero 75 Ω 100 Ω Input bias current 𝐼𝑖𝑛(𝑏𝑖𝑎𝑠) Zero 80 nA 30 pA Input off set current 𝐼𝑖𝑛(𝑜𝑓𝑓) Zero 20 nA 3 pA Input off set voltage 𝑉𝑖𝑛(𝑜𝑓𝑓) Zero 2 mV 1 mV Common-mode rejection ratio 𝐶𝑀𝑅𝑅 Infinite 90 dB 100 dB Operation of 741 Op Amp The following figure is a simplified schematic diagram of the 741. The input stage is a diff amp (Q1 and Q2). In the 741, Q14 is a current source that replaces the tail resistor. R2, Q13, and Q14 are a current mirror that produces the tail current for Q1 and Q2. Instead of using an ordinary resistor as the collector resistor of the diff amp, the 741 uses an active-load Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 7 resistor. This active-load Q4 acts like a current source with an extremely high impedance. Because of this, the voltage gain of the diff amp is much higher than with a passive-load resistor. The amplified signal from the diff amp drives the base of Q5, an emitter follower. This stage steps up the impedance level to avoid loading down the diff amp. The signal out of Q5 goes to Q6. Diodes Q7 and Q8 are part of the biasing for the final stage. Q11 is an active-load resistor for Q6. Therefore, Q6 and Q11 are like a CE driver stage with a very high voltage gain. The amplified signal out of the CE driver stage (Q6) goes to the final stage, which is a Class-B push-pull emitter follower (Q9 and Q10). Because of the split supply (equal positive VCC and negative VEE voltages), the quiescent output is ideally 0 V when the input voltage is zero. Any deviation from 0 V is called the output error voltage. When v1 is greater than v2, the input voltage vin produces a positive output voltage vout. When v2 is greater than v1, the input voltage vin produces a negative output voltage vout. Ideally, vout can be as positive as +VCC and as negative as -VEE before clipping occurs. The output swing is normally within 1 to 2 V of each supply voltage because of voltage drops inside the 741. Active Loading In the previous figure, we have two examples of active loading (using transistors instead of resistors for loads): 1. Active-load Q4 on the input diff amp 2. Active-load Q11 in the CE driver stage Because current sources have high output impedances, active loads produce much higher voltage gain than is possible with resistors. These active loads produce a typical voltage gain of 100,000 for the 741C. Active loading is very popular in integrated circuits (ICs) because it is easier and less expensive to fabricate transistors on a chip than it is to fabricate resistors. Frequency Compensation Referring to the previous figure, CC is a compensating capacitor. Because of the Miller effect, this small capacitor (typically 30 pF) is multiplied by the voltage gain of Q5 and Q6 to get a much larger equivalent capacitance of: 𝐶𝑖𝑛(𝑀) = (𝐴𝑣 + 1)𝐶𝐶 where Av is the voltage gain of the Q5 and Q6 stages. The resistance facing this Miller capacitance is the output impedance of the diff amp. Therefore, we have a lag circuit, which produces a cutoff frequency of 10 Hz in a 741C. The open-loop gain of the op amp is down 3 dB at this cutoff frequency. Then, 𝐴𝑉𝑂𝐿 decreases approximately 20 dB per decade until reaching the unity-gain frequency. The 741C has an open-loop voltage gain of 100,000, equivalent to 100 dB. Since the open-loop cutoff frequency is 10 Hz, the voltage gain breaks at 10 Hz and then rolls off at a rate of 20 dB per decade until it is 0 dB at 1 MHz. An op amp that is internally compensated, such as the 741C, has a first-order response. Incidentally, not all op amps are internally compensated. Some require the user to connect an external compensating capacitor to prevent oscillations. The advantage of using external compensation is that a designer has more control over the high-frequency performance. Although Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 8 an external capacitor is the simplest way to compensate, more complicated circuits can be used that not only provide compensation but also produce a higher f unity than is possible with internal compensation. Bias and Offsets A diff amp has input bias and offsets that produce an output error when there is no input signal. In many applications, the output error is small enough to ignore. But when the output error cannot be ignored, a designer can reduce it by using equal base resistors. This eliminates the problem of bias current, but not the offset current or offset voltage. This is why it is best to eliminate output error by using the nulling circuit given on the data sheet. This nulling circuit works with the internal circuitry to eliminate the output error and also to minimize thermal drift, a slow change in output voltage caused by the effect of changing temperature on op-amp parameters. In case the data sheet of an op amp does not include a nulling circuit, a small input voltage is applied to null the output. The figure below shows the nulling method from the data sheet of a 741C. The ac source driving the inverting input has a Thevenin resistance of RB. To neutralize the effect of input bias current (80 nA) flowing through this source resistance, a discrete resistor of equal value is added to the noninverting input, as shown. To eliminate the effect of an input offset current of 20 nA and an input offset voltage of 2 mV, the data sheet of a 741C recommends using a 10-kΩ potentiometer between pins 1 and 5. By adjusting this potentiometer with no input signal, we can null or zero the output voltage. Common-Mode Rejection Ratio (CMRR) For a 741C, CMRR is 90 dB at low frequencies. Given equal signals, one a desired signal and the other a common-mode signal, the desired signal will be 90 dB larger at the output than the common-mode signal. In ordinary numbers, this means that the desired signal will be approximately 30,000 times larger than the common-mode signal. At higher frequencies, reactive effects degrade CMRR. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 9 Maximum Peak-to-Peak (MPP) Output The MPP value of an amplifier is the maximum peak-to-peak output that the amplifier can produce without clipping. Since the quiescent output of an op amp is ideally zero, the ac output voltage can swing positively or negatively. For load resistances that are much larger than 𝑅𝑜𝑢𝑡 , the output voltage can swing almost to the supply voltages. For instance, if 𝑉𝐶𝐶 = +15 V and 𝑉𝐸𝐸 = -15 V, the MPP value with a load resistance of 10 kΩ is ideally 30 V. With a nonideal op amp, the output cannot swing all the way to the value of the supply voltages because there are small voltage drops in the final stage of the op amp. Furthermore, when the load resistance is not large compared to 𝑅𝑜𝑢𝑡 , some of the amplified voltage is dropped across 𝑅𝑜𝑢𝑡 , which means that the final output voltage is smaller. When the load resistance decreases, MPP decreases. For instance, if the load resistance is only 275 Ω, MPP decreases to 16 V, which means that the output saturates positively at +8 V and negatively at -8 V. Short-Circuit Current In some applications, an op amp may drive a load resistance of approximately zero. In this case, you need to know the value of the short-circuit output current. The data sheet of a 741C lists a short-circuit output current of 25 mA. This is the maximum output current the op amp can produce. If you are using small load resistors (less than 75 Ω), don’t expect to get a large output voltage because the voltage cannot be greater than the 25 mA times the load resistance. Frequency Response The unity-gain frequency is the frequency at which the voltage gain equals 1. Data sheets usually specify the value of 𝑓𝑢𝑛𝑖𝑡𝑦 because it represents the upper limit on the useful gain of an op amp. For instance, the data sheet of a 741C lists an 𝑓𝑢𝑛𝑖𝑡𝑦 of 1 MHz. This means that the 741C can amplify signals up to 1 MHz. Beyond 1 MHz, the voltage gain is less than 1 and the 741C is useless. If a designer needs a higher 𝑓𝑢𝑛𝑖𝑡𝑦 , better op amps are available. For instance, the LM318 has an 𝑓𝑢𝑛𝑖𝑡𝑦 of 15 MHz, which means that it can produce usable voltage gain all the way to 15 MHz Slew Rate The compensating capacitor inside a 741C performs a very important function: It prevents oscillations that would interfere with the desired signal. But there is a disadvantage: the compensating capacitor needs to be charged and discharged. This creates a speed limit on how fast the output of the op amp can change. Here is the basic idea: Suppose the input voltage to an op amp is a positive voltage step, a sudden transition in voltage from one dc level to a higher dc level. If the op amp were perfect, we would get the ideal response shown in the first graph below. Instead, the output is the positive exponential waveform shown. This occurs because the compensating capacitor must be charged before the output voltage can change to the higher level. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 10 In the first graph, the initial slope of the exponential waveform is called the slew rate, symbolized 𝑆𝑅. The definition of slew rate is: 𝛥𝑣𝑜𝑢𝑡 𝑆𝑅 = 𝛥𝑡 where the Greek letter ∆ (delta) stands for “the change in.” In words, the equation says that slew rate equals the change in output voltage divided by the change in time. The second graph above illustrates the meaning of slew rate. The initial slope equals the vertical change divided by the horizontal change between two points on the early part of the exponential wave. For instance, if the exponential wave increases 0.5 V during the first microsecond, as shown in the third graph, the slew rate is: 0.5 𝑉 𝑆𝑅 = = 0.5 𝑉/𝜇𝑠 1𝜇𝑠 The slew rate represents the fastest response that an op amp can have. For instance, the slew rate of a 741C is 0.5 V/s. This means that the output of a 741C can change no faster than 0.5 V in a microsecond. In other words, if a 741C is driven by a large step in input voltage, we do not get a sudden step in output voltage. Instead, we get an exponential output wave. The initial part of this output waveform will look like the third graph. The data sheet of an op amp always specifies the slew rate because this quantity limits the large- signal response of an op amp. If the output sine wave is very small or the frequency is very low, slew rate is no problem. But when the signal is large and the frequency is high, slew rate will distort the output signal. With calculus, it is possible to derive this equation: 𝑆𝑆 = 2𝜋𝑓𝑉𝑃 where 𝑆𝑆 is the initial slope of the sine wave, f is its frequency, and 𝑉𝑃 is its peak value. To avoid slew-rate distortion of a sine wave, 𝑆𝑆 has to be less than or equal to 𝑆𝑅. When the two are equal, we are at the limit, on the verge of slew-rate distortion. In this case: 𝑆𝑅 = 𝑆𝑆 = 2𝜋𝑓𝑉𝑃 Solving for 𝑓 gives: 𝑆𝑅 𝑓𝑚𝑎𝑥 = 2𝜋𝑉𝑃 where 𝑓𝑚𝑎𝑥 is the highest frequency that can be amplified without slew-rate distortion. Given the slew rate of an op amp and the peak output voltage desired, we can use this equation to calculate the maximum undistorted frequency. Above this frequency, we will see slew-rate distortion on an oscilloscope. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 11 The frequency 𝑓𝑚𝑎𝑥 is sometimes called the power bandwidth or large-signal bandwidth of the op amp. For instance, suppose we are using a 741C. To get an undistorted output peak voltage of 8 V, the frequency can be no higher than 10 kHz. One way to increase the 𝑓𝑚𝑎𝑥 is to accept less output voltage. By trading off peak value for frequency, we can improve the power bandwidth. As an example, if our application can accept a peak output voltage of 1 V, 𝑓𝑚𝑎𝑥 increases to 80 kHz. There are two bandwidths to consider when analyzing the operation of an op-amp circuit: 1. the small-signal bandwidth determined by the first-order response of the op amp 2. the large-signal or power bandwidth determined by the slew rate. The Inverting Amplifier The inverting amplifier is the most basic op-amp circuit. It uses negative feedback to stabilize the overall voltage gain. The reason we need to stabilize the overall voltage gain is because 𝐴𝑉𝑂𝐿 is too high and unstable to be of any use without some form of feedback. For instance, the 741C has a minimum 𝐴𝑉𝑂𝐿 of 20,000 and a maximum 𝐴𝑉𝑂𝐿 of more than 200,000. An unpredictable voltage gain of this magnitude and variation is useless without feedback. Inverting Negative Feedback The following figure shows the ac-equivalent of an inverting amplifier. Input voltage vin drives the inverting input through resistor 𝑅1. This results in an inverting input voltage of 𝑣2. The input voltage is amplified by the open-loop voltage gain to produce an inverted output voltage. The output voltage is fed back to the input through feedback resistor 𝑅𝑓. This results in negative feedback because the output is 180° out of phase with the input. In other words, any changes in 𝑣2 produced by the input voltage are opposed by the output signal Here is how the negative feedback stabilizes the overall voltage gain: If the open-loop voltage gain 𝐴𝑉𝑂𝐿 increases for any reason, the output voltage will increase and feedback more voltage to the inverting input. This opposing feedback voltage reduces 𝑣2. Therefore, even though 𝐴𝑉𝑂𝐿 has increased, 𝑣2 has decreased, and the final output increases much less than it would without the negative feedback. The overall result is a very slight increase in output voltage, so small that it is hardly noticeable. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 12 Virtual Ground When we connect a piece of wire between some point in a circuit and ground, the voltage of the point becomes zero. Furthermore, the wire provides a path for current to flow to ground. A mechanical ground (a wire between a point and ground) is ground to both voltage and current. A virtual ground is different. This type of ground is a widely used shortcut for analyzing an inverting amplifier. With a virtual ground, the analysis of an inverting amplifier and related circuits becomes incredibly easy. The concept of a virtual ground is based on an ideal op amp. When an op amp is ideal, it has infinite open-loop voltage gain and infinite input resistance. Because of this, we can deduce the following ideal properties for the inverting amplifier (refer to the figure below): 1. Since 𝑅𝑖𝑛 is infinite𝑖2 , is zero. 2. Since 𝐴𝑉𝑂𝐿 is infinite, 𝑣2 is zero Since 𝑖2 is zero in the figure, the current through 𝑅𝑓 must equal the input current through 𝑅1 , as shown. Furthermore, since 𝑣2 is zero, the virtual ground shown in the figure means that the inverting input acts like a ground for voltage but an open for current! Virtual ground is very unusual. It is like half of a ground because it is a short for voltage but an open for current. To remind us of this half-ground quality, the figure above uses a dashed line between the inverting input and ground. The dashed line means that no current can flow to ground. Although virtual ground is an ideal approximation, it gives very accurate answers when used with heavy negative feedback. Voltage Gain In the figure, visualize a virtual ground on the inverting input. Then, the right end of 𝑅1 is a voltage ground, so we can write: 𝑣𝑖𝑛 = 𝑖𝑖𝑛 𝑅1 Similarly, the left end of 𝑅𝑓 is a voltage ground, so the magnitude of output voltage is: 𝑣𝑜𝑢𝑡 = 𝑖𝑖𝑛 𝑅𝑓 Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 13 𝑣𝑜𝑢𝑡 −𝑖𝑖𝑛 𝑅𝑓 Divide 𝑣𝑜𝑢𝑡 by 𝑣𝑖𝑛 to get the voltage gain where 𝐴𝑣 = 𝑣𝑖𝑛 = 𝑖𝑖𝑛 𝑅1. Therefore: −𝑅𝑓 𝐴𝑣(𝐶𝐿) = 𝑅𝑓 where 𝐴𝑣(𝐶𝐿) is the closed-loop voltage gain. This is called the closed-loop voltage gain because it is the voltage when there is a feedback path between the output and the input. Because of the negative feedback, the closed-loop voltage gain 𝐴𝑣(𝐶𝐿) is always smaller than the open-loop voltage gain 𝐴𝑉𝑂𝐿. The closed-loop voltage gain equals the ratio of the feedback resistance to the input resistance. For instance, if 𝑅1 = 1 kΩ and 𝑅𝑓 = 50 kΩ, the closed-loop voltage gain is 50. Because of the heavy negative feedback, this closed-loop voltage gain is very stable. If 𝐴𝑉𝑂𝐿 varies because of temperature change, supply voltage variations, or op-amp replacement, 𝐴𝑣(𝐶𝐿) will still be very close to 50. The negative sign in the voltage gain equation indicates a 180° phase shift. Input Impedance In some applications, a designer may want a specific input impedance. This is one of the advantages of an inverting amplifier; it is easy to set up a desired input impedance. Here is why: Since the right end of 𝑅1 is virtually grounded, the closed-loop input impedance is: 𝑧𝑖𝑛(𝐶𝐿) = 𝑅1 This is the impedance looking into the left end of𝑅1. For instance, if an input impedance of 2 kΩ and a closed-loop voltage gain of 50 is needed, a designer can use 𝑅1 = 2 kΩ and 𝑅𝑓 = 100 kΩ. Bandwidth The open-loop bandwidth or cutoff frequency of an op amp is very low because of the internal compensating capacitor. For a 741C: 𝑓2(𝑂𝐿) = 10 𝐻𝑧 At this frequency, the open-loop voltage gain breaks and rolls off in a first-order response. When negative feedback is used, the overall bandwidth increases. Here is the reason: When the input frequency is greater than 𝑓2(𝑂𝐿) , 𝐴𝑉𝑂𝐿 decreases 20 dB per decade. When 𝑣𝑜𝑢𝑡 tries to decrease, less opposing voltage is fed back to the inverting input. Therefore, 𝑣2 increases and compensates for the decrease in 𝐴𝑉𝑂𝐿. Because of this, 𝐴𝑣(𝐶𝐿) breaks at a higher frequency than 𝑓2(𝑂𝐿). The greater the negative feedback, the higher the closed-loop cutoff frequency. Stated another way: The smaller 𝐴𝑣(𝐶𝐿) is, the higher 𝑓2(𝐶𝐿) is. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 14 The graph above illustrates how the closed-loop bandwidth increases with negative feedback. As you can see, the heavier the negative feedback (smaller 𝐴𝑣(𝐶𝐿) ), the greater the closed-loop bandwidth. Here is the equation for an inverting amplifier closed-loop bandwidth: 𝑓𝑢𝑛𝑖𝑡𝑦 𝑓2(𝐶𝐿) = 𝐴𝑣(𝐶𝐿) + 1 In most applications, 𝐴𝑣(𝐶𝐿) is greater than 10 and the equation simplifies to: 𝑓𝑢𝑛𝑖𝑡𝑦 𝑓2(𝐶𝐿) = 𝐴𝑣(𝐶𝐿) For instance, when 𝐴𝑣(𝐶𝐿) is 10: 1 𝑀𝐻𝑧 𝑓2(𝐶𝐿) = = 100 𝑘𝐻𝑧 10 If 𝐴𝑣(𝐶𝐿) is 100: 1 𝑀𝐻𝑧 𝑓2(𝐶𝐿) = = 10 𝑘𝐻𝑧 100 The equation can be rearranged into: 𝑓𝑢𝑛𝑖𝑡𝑦 = 𝐴𝑣(𝐶𝐿) 𝑓2(𝐶𝐿) Notice that the unity-gain frequency equals the product of gain and bandwidth. For this reason, many data sheets refer to the unity-gain frequency as the gain-bandwidth product (GBW). Bias and Off sets Negative feedback reduces the output error caused by input bias current, input offset current, and input offset voltage. The three input error voltages and the equation for total output error voltage is: 𝑉𝑒𝑟𝑟𝑜𝑟 = 𝐴𝑉𝑂𝐿 (𝑉1𝑒𝑟𝑟 + 𝑉2𝑒𝑟𝑟 + 𝑉3𝑒𝑟𝑟 ) When negative feedback is used, this equation may be written as: 𝑉𝑒𝑟𝑟𝑜𝑟 ≅ ±𝐴𝑉𝑂𝐿 (±𝑉1𝑒𝑟𝑟 ± 𝑉2𝑒𝑟𝑟 ± 𝑉3𝑒𝑟𝑟 ) Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 15 where 𝑉𝑒𝑟𝑟𝑜𝑟 is the total output error voltage. In mass production, the input errors may add up in the worst possible way. The equations of input errors are: 𝑉1𝑒𝑟𝑟 = (𝑅𝐵1 − 𝑅𝐵2 )𝐼𝑖𝑛(𝑏𝑖𝑎𝑠) 𝐼𝑖𝑛(𝑜𝑓𝑓) 𝑉2𝑒𝑟𝑟 = (𝑅𝐵1 + 𝑅𝐵2 ) 2 𝑉3𝑒𝑟𝑟 = 𝑉𝑖𝑛(𝑜𝑓𝑓) When 𝐴𝑣(𝐶𝐿) is small, the total output error given by the equation of 𝑉𝑒𝑟𝑟𝑜𝑟 may be small enough to ignore. If not, resistor compensation and offset nulling will be necessary. In an inverting amplifier, 𝑅𝐵2 is the Thevenin resistance seen when looking back from the inverting input toward the source. This resistance is given by: 𝑅𝐵2 = 𝑅1 || 𝑅𝑓 If it is necessary to compensate for input bias current, an equal resistance 𝑅𝐵1 should be connected to the noninverting input. This resistance has no effect on the virtual-ground approximation because no ac signal current flows through it. The Noninverting Amplifier The noninverting amplifier is another basic op-amp circuit. It uses negative feedback to stabilize the overall voltage gain. With this type of amplifier, the negative feedback also increases the input impedance and decreases the output impedance. Basic Circuit The figure below shows the ac-equivalent circuit of a noninverting amplifier. An input voltage 𝑣𝑖𝑛 drives the noninverting input. This input voltage is amplified to produce the in-phase output voltage shown. Part of output voltage is fed back to the input through a voltage divider. The voltage across 𝑅1 is the feedback voltage applied to the inverting input. This feedback voltage is almost equal to the input voltage. Because of the high open-loop voltage gain, the difference between 𝑣1 and 𝑣2 is very small. Since the feedback voltage opposes the input voltage, we have negative feedback. Here is how the negative feedback stabilizes the overall voltage gain: If the open-loop voltage gain 𝐴𝑉𝑂𝐿 increases for any reason, the output voltage will increase and feed back more voltage to the inverting input. This opposing feedback voltage reduces the net input voltage 𝑣1 – 𝑣2. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 16 Therefore, even though 𝐴𝑉𝑂𝐿 increases, 𝑣1 – 𝑣2 decreases, and the final output increases much less than it would without the negative feedback. The overall result is only a very slight increase in output voltage. Virtual Short When we connect a piece of wire between two points in a circuit, the voltage of both points with respect to ground is equal. Furthermore, the wire provides a path for current to flow between the two points. A mechanical short (a wire between two points) is a short for both voltage and current. A virtual short is different. This type of short can be used for analyzing noninverting amplifiers. With a virtual short, we can quickly and easily analyze noninverting amplifiers and related circuits. The virtual short uses these two properties of an ideal op amp: 1. Since 𝑅𝑖𝑛 is infinite, both input currents are zero. 2. Since 𝐴𝑉𝑂𝐿 is infinite, 𝑣1 – 𝑣2 is zero. The figure above shows a virtual short between the input terminals of the op amp. The virtual short is a short for voltage but an open for current. As a reminder, the dashed line means that no current can flow through it. Although the virtual short is an ideal approximation, it gives very accurate answers when used with heavy negative feedback. Here is how we will use the virtual short: Whenever we analyze a noninverting amplifier or a similar circuit, we can visualize a virtual short between the input terminals of the op amp. As long as the op amp is operating in the linear region (not positively or negatively saturated), the open- loop voltage gain approaches infinity and a virtual short exists between the two input terminals. One more point: Because of the virtual short, the inverting input voltage follows the noninverting input voltage. If the noninverting input voltage increases or decreases, the inverting input voltage immediately increases or decreases to the same value. This follow-the-leader action is called bootstrapping (as in “pulling yourself up by your bootstraps”). The noninverting input pulls the inverting input up or down to an equal value. Described another way, the inverting input is bootstrapped to the noninverting input. Voltage Gain In the following figure, visualize a virtual short between the input terminals of the op amp. Then, the virtual short means that the input voltage appears across 𝑅1 , as shown. So, we can write: 𝑣𝑖𝑛 = 𝑖1 𝑅1 Since no current can flow through a virtual short, the same 𝑖1 current must flow through 𝑅𝑓 , which means that the output voltage is given by: Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 17 𝑣𝑜𝑢𝑡 = 𝑖1 (𝑅𝑓 + 𝑅1 ) Divide 𝑣𝑜𝑢𝑡 by 𝑣𝑖𝑛 to get the voltage gain: 𝑅𝑓 + 𝑅1 𝐴𝑣(𝐶𝐿) = 𝑅1 or 𝑅𝑓 𝐴𝑣(𝐶𝐿) = +1 𝑅1 This is easy to remember because it is the same as the equation for an inverting amplifier, except that we add 1 to the ratio of resistances. Also, note that the output is in phase with the input. Therefore, no (-) sign is used in the voltage gain equation. Other Quantities The closed-loop input impedance approaches infinity. Since the open-loop input impedance is already very high (2 MΩ for a 741C), the closed-loop input impedance will be even higher. The effect of negative feedback on bandwidth is the same as with an inverting amplifier: 𝑓𝑢𝑛𝑖𝑡𝑦 𝑓2(𝐶𝐿) = 𝐴𝑣(𝐶𝐿) Again, we can trade off voltage gain for bandwidth. The smaller the closed-loop voltage gain, the greater the bandwidth. The input error voltages caused by input bias current, input offset current, and input offset voltage are analyzed the same way as with an inverting amplifier. After calculating each input error, we can multiply by the closed-loop voltage gain to get the total output error. 𝑅𝐵2 is the Thevenin resistance seen when looking from the inverting input toward the voltage divider. This resistance is the same as for an inverting amplifier: 𝑅𝐵2 = 𝑅1 ||𝑅𝑓 If it is necessary to compensate for input bias current, an equal resistance 𝑅𝐵1 should be connected to the noninverting input. This resistance has no effect on the virtual-short approximation because no ac signal current flows through it. Output Error Voltage Reduces MPP If we are amplifying ac signals, we can capacitively couple the output signal to the load. In this case, we can ignore the output error voltage unless it is excessively large. If the output error voltage is large, it will significantly reduce the MPP, the maximum unclipped peak-to-peak output. In other words, the greater the output error voltage, the smaller the MPP value. Two Op-Amp Applications The following are two practical circuits of an op-amp. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 18 The Summing Amplifier Whenever we need to combine two or more analog signals into a single output, the summing amplifier of the figure below is a natural choice. For simplicity, the circuit shows only two inputs, but we can have as many inputs as needed for the application. A circuit like this amplifies each input signal. The gain for each channel or input is given by the ratio of the feedback resistance to the appropriate input resistance. For instance, the closed-loop voltage gains of the figure above are: −𝑅𝑓 −𝑅𝑓 𝐴𝑣1(𝐶𝐿) = and 𝐴𝑣2(𝐶𝐿) = 𝑅1 𝑅2 The summing circuit combines all the amplified input signals into a single output, given by: 𝑣𝑜𝑢𝑡 = 𝐴𝑣1(𝐶𝐿) 𝑣1 + 𝐴𝑣2(𝐶𝐿) 𝑣2 It is easy to prove the equation above. Since the inverting input is a virtual ground, the total input current is: 𝑣1 𝑣2 𝑖𝑖𝑛 = 𝑖1 + 𝑖2 = + 𝑅1 𝑅2 Because of the virtual ground, all this current flows through the feedback resistor, producing an output voltage with a magnitude of: 𝑅𝑓 𝑅𝑓 𝑣𝑜𝑢𝑡 = (𝑖1 + 𝑖2 ) 𝑅𝑓 = − ( 𝑣1 + 𝑣 ) 𝑅1 𝑅2 2 Here you see that each input voltage is multiplied by its channel gain and added to produce the total output. The same result applies to any number of inputs. In some applications, all resistances are equal, as shown in the figure above. In this case, each channel has a closed-loop voltage gain of unity (1) and the output is given by: 𝑣𝑜𝑢𝑡 = −(𝑣1 + 𝑣2 + … + 𝑣𝑛 ) This is a convenient way of combining input signals and maintaining their relative sizes. The combined output signal can then be processed by more circuits. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 19 A final point: If a summing circuit needs to be compensated by adding an equal resistance to the noninverting input, the resistance to use is the Thevenin resistance looking from the inverting input back to the sources. This resistance is given by the parallel equivalent of all resistances connected to the virtual ground: 𝑅𝐵2 = 𝑅1 ||𝑅2 ||𝑅𝑓 || … ||𝑅𝑛 Voltage Follower The figure above shows the ac-equivalent circuit for a voltage follower. Although it appears deceptively simple, the circuit is very close to ideal because the negative feedback is maximum. As you can see, the feedback resistance is zero. Therefore, all the output voltage is fed back to the inverting input. Because of the virtual short between the op-amp inputs, the output voltage equals the input voltage: 𝑣𝑜𝑢𝑡 = 𝑣𝑖𝑛 which means that the closed-loop voltage gain is: 𝐴𝑣(𝐶𝐿) = 1 We can get the same result by calculating the closed-loop voltage gain using the equation 𝑅𝑓 𝐴𝑣(𝐶𝐿) = 𝑅 + 1. Since 𝑅𝑓 = 0 and 𝑅1 = ∞: 1 𝑅𝑓 𝐴𝑣(𝐶𝐿) = +1=1 𝑅1 Therefore, the voltage follower is a perfect follower circuit because it produces an output voltage that is exactly equal to the input voltage (or close enough to satisfy almost any application). Furthermore, the maximum negative feedback produces a closed-loop input impedance that is much higher than the open-loop input impedance (2 MΩ for a 741C). Also, a maximum negative feedback produces a closed-loop output impedance that is much lower than the open-loop output impedance (75 Ω for a 741C). Therefore, we have an almost perfect method for converting a high- impedance source to a low-impedance source. The following figure illustrates the idea. The input ac source has a high output impedance 𝑅ℎ𝑖𝑔ℎ. The load has a low impedance 𝑅𝑙𝑜𝑤. Because of the maximum negative feedback in a voltage follower, the closed-loop input impedance 𝑧𝑖𝑛(𝐶𝐿) is incredibly high and the closed-loop output impedance 𝑧𝑜𝑢𝑡(𝐶𝐿) is incredibly low. As a result, all the input source voltage appears across the load resistor. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 20 The crucial point to understand is this: The voltage follower is the ideal interface to use between a high-impedance source and a low-impedance load. Basically, it transforms the high-impedance voltage source into a low-impedance voltage source. You will see the voltage follower used a great deal in practice. Since 𝐴𝑣(𝐶𝐿) = 1 in a voltage follower, the closed-loop bandwidth is maximum and equal to: 𝑓2(𝐶𝐿) = 𝑓𝑢𝑛𝑖𝑡𝑦 Another advantage is the low output offset error because the input errors are not amplified. Since 𝐴𝑣(𝐶𝐿) = 1, the total output error voltage equals the worst-case sum of the input errors. Linear ICs Op amps represent about a third of all linear ICs. With op amps, we can build a wide variety of useful circuits. Although the op amp is the most important linear IC, other linear ICs, such as audio amplifiers, video amplifiers, and voltage regulators, are also widely used. There are two more characteristics that is essential in the discussion of op amps. First, there is the power supply rejection ratio (PSRR). This quantity is defined as: ∆𝑉𝑖𝑛(𝑜𝑓𝑓) 𝐴𝑣(𝐶𝐿) = ∆𝑉𝑆 In words, the equation says that the power-supply rejection ratio equals the change in the input offset voltage divided by the change in the supply voltages. In making this measurement, the manufacturer varies both supplies simultaneously and symmetrically. If 𝑉𝐶𝐶 = +15 𝑉 , 𝑉𝐶𝐶 = −15 𝑉, and ∆𝑉𝑆 = +1 𝑉, then 𝑉𝐶𝐶 becomes +16 𝑉 and 𝑉𝐸𝐸 becomes −16 𝑉. Here is what this equation means: Because of the imbalance in the input diff amp plus other internal effects, a change in the supply voltage will produce an output error voltage. Dividing this output error voltage by the closed-loop voltage gain gives the change in the input offset voltage. For instance, the LF353 has a PSRR in decibels of 276 dB. When we convert this to an ordinary number, we get: −76 𝑑𝐵 𝑃𝑆𝑅𝑅 = 𝑎𝑛𝑡𝑖𝑙𝑜𝑔 = 0.000158 20 or, as it is sometimes written: 𝑃𝑆𝑅𝑅 = 158 𝜇𝑉/𝑉 This tells us that a change of 1 𝑉 in the supply voltage will produce a change in the input offset voltage of 158 𝜇𝑉. Therefore, we have one more source of input error that joins the three input errors discussed earlier. The last parameter shown for the LF353 is the drift of 10 μV/°C. This is defined as the temperature coefficient of the input offset voltage. It tells us how much the input offset voltage increases with Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 21 temperature. A drift of 10 μV/°C means that the input offset voltage increases 10 V for each degree increase in degrees Celsius. If the internal temperature of the op amp increases by 50°C, the input offset voltage of an LF353 increases by 500 μV. Not all op amps need two supply voltages. For instance, the LM324 has four internally compensated op amps. Although it can operate with two supplies like most op amps, it was specifically designed for a single power supply, a definite advantage in many applications. Internal compensation is convenient and safe because an internally compensated op amp will not break into oscillations under any condition. The price paid for this safety is a loss of design control. This is why some op amps offer external compensation. Overcompensation can improve low- frequency operation, whereas under compensation can increase the bandwidth and slew rate. All op amps have imperfections, as we have seen. Precision op amps try to minimize these imperfections. Precision op amps are necessary for stringent applications such as measurement and control. Activities/Assessments 1. Objective type quiz and problem solving a. Use the answer sheet below for the objective type quiz. b. Answers must be handwritten on the space provided. c. Submission is on the next lecture period. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 22 LESSON 2: Differential Amplifiers The input circuit used in most op amps is the differential amplifier. This amplifier configuration establishes many of the IC’s input characteristics. The differential amplifier may also be configured in a discrete form to be used in communications, instrumentation, and industrial control circuits. Differential Amplifier Transistors, diodes, and resistors are the only practical components in typical ICs. Capacitors may also be used, but they are small, usually less than 50 pF. For this reason, IC designers cannot use coupling and bypass capacitors the way a discrete circuit designer can. Instead, the IC designer has to use direct coupling between stages and also needs to eliminate the emitter bypass capacitor without losing too much voltage gain. The differential amplifier (diff amp) is the key. The design of this circuit is extremely clever because it eliminates the need for an emitter bypass capacitor. For this and other reasons, the diff amp is used as the input stage of almost every IC op amp. Differential Input and Output The figure below shows a diff amp. It is two CE stages in parallel with a common emitter resistor. Although it has two input voltages (𝑣1 and 𝑣2 ) and two collector voltages (𝑣𝑐1 and 𝑣𝑐2 ), the overall circuit is considered to be one stage. Because there are no coupling or bypass capacitors, there is no lower cutoff frequency. The ac output voltage 𝑣𝑜𝑢𝑡 is defined as the voltage between the collectors with the polarity: 𝑣𝑜𝑢𝑡 = 𝑣𝑐2 − 𝑣𝑐1 This voltage is called a differential output because it combines the two ac collector voltages into one voltage that equals the difference of the collector voltages. Note: We will use lowercase letters for 𝑣𝑜𝑢𝑡 , 𝑣𝑐1, and 𝑣𝑐2 because they are ac voltages that include zero Hertz (0 Hz) as a special case. Ideally, the circuit has identical transistors and equal collector resistors. With perfect symmetry, 𝑣𝑜𝑢𝑡 is zero when the two input voltages are equal. When 𝑣1 is greater than 𝑣2 , the output voltage has the polarity shown in the figure above. When 𝑣2 is greater than 𝑣1 , the output voltage is inverted and has the opposite polarity. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 23 The diff amp of the previous figure has two separate inputs. Input 𝑣1 is called the noninverting input because 𝑣𝑜𝑢𝑡 is in phase with 𝑣1. On the other hand, 𝑣2 is called the inverting input because 𝑣𝑜𝑢𝑡 is 180° out of phase with 𝑣2. In some applications, only the noninverting input is used and the inverting input is grounded. In other applications, only the inverting input is active and the noninverting input is grounded. When both the noninverting and inverting input voltages are present, the total input is called a differential input because the output voltage equals the voltage gain times the difference of the two input voltages. The equation for the output voltage is: 𝑣𝑜𝑢𝑡 = 𝐴𝑣 (𝑣1 − 𝑣2 ) where 𝐴𝑣 is the voltage gain. Single-Ended Output A differential output like that of the previous figure requires a floating load because neither end of the load can be grounded. This is inconvenient in many applications since loads are often single- ended; that is, one end is grounded. The figure below shows a widely used form of the diff amp. This has many applications because it can drive single-ended loads like CE stages, emitter followers, and other circuits. As you can see, the ac output signal is taken from the collector on the right side. The collector resistor on the left has been removed because it serves no useful purpose. Because the input is differential, the ac output voltage is still given by 𝐴𝑣 (𝑣1 − 𝑣2 ). With a single- ended output, however, the voltage gain is half as much as with a differential output. We get half as much voltage gain with a single-ended output because the output is coming from only one of the collectors. Incidentally, the following figure shows the block-diagram symbol for a diff amp with a differential input and a single-ended output. The same symbol is used for an op amp. The plus sign (1) represents the noninverting input, and the minus sign (2) is the inverting input. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 24 Noninverting-Input Configurations Often, only one of the inputs is active and the other is grounded, as shown in the left figure below. This configuration has a noninverting input and a differential output. Since 𝑣2 = 0: 𝑣𝑜𝑢𝑡 = 𝐴𝑣 (𝑣1 ) The right figure above shows another configuration for the diff amp. This one has a noninverting input and a single-ended output. Since 𝑣𝑜𝑢𝑡 is the ac output voltage, the equation above is still valid, but the voltage gain 𝐴𝑣 will be half as much because the output is taken from only one side of the diff amp. Inverting-Input Configurations In some applications, 𝑣2 is the active input and 𝑣1 is the grounded input, as shown in the left figure below. In this case, the equation simplifies to: 𝑣𝑜𝑢𝑡 = −𝐴𝑣 (𝑣2 ) The minus sign indicates phase inversion. The right figure above shows the inverting input with a single-ended output. In this case, the ac output voltage is still the same previously given equation. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 25 Conclusion The table below summarizes the four basic configurations of a diff amp. The general case has a differential input and differential output. The remaining cases are subsets of the general case. When the input is single-ended, either the noninverting input 𝑣1 or the inverting input 𝑣2 may be used. Input Output 𝑣𝑖𝑛 𝑣𝑜𝑢𝑡 Differential Differential 𝑣1 − 𝑣2 𝑣𝑐2 − 𝑣𝑐1 Differential Single-ended 𝑣1 − 𝑣2 𝑣𝑐2 Single-ended Differential 𝑣1 𝑜𝑟 𝑣2 𝑣𝑐2 − 𝑣𝑐1 Single-ended Single-ended 𝑣1 𝑜𝑟 𝑣2 𝑣𝑐2 DC Analysis of a Diff Amp The figure below shows the dc-equivalent circuit for a diff amp. Throughout this unit, we will assume identical transistors and equal collector resistors. Also, both bases are grounded in this preliminary analysis. Ideal Analysis A diff amp is sometimes called a long-tail pair because the two transistors share a common resistor 𝑅𝐸. The current through this common resistor is called the tail current. If we ignore the 𝑉𝐵𝐸 drops across the emitter diodes of the figure above, then the top of the emitter resistor is ideally a dc ground point. In this case, all of 𝑉𝐸𝐸 appears across 𝑅𝐸 and the tail current is: 𝑉𝐸𝐸 𝐼𝑇 = 𝑅𝐸 This equation is fine for troubleshooting and preliminary analysis because it quickly gets to the point, which is that almost all the emitter supply voltage appears across the emitter resistor. When the two halves of the figure above are perfectly matched, the tail current will split equally. Therefore, each transistor has an emitter current of: 𝐼𝑇 𝐼𝐸 = 2 The dc voltage on either collector is given by this familiar equation: 𝑉𝐶 = 𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶 Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 26 Second Approximation We can improve the dc analysis by including the 𝑉𝐵𝐸 drop across each emitter diode. In the figure below, the voltage at the top of the emitter resistor is one VBE drop below ground. Therefore, the tail current is: 𝑉𝐸𝐸 − 𝑉𝐵𝐸 𝐼𝑇 = 𝑅𝐸 where 𝑉𝐵𝐸 = 0.7 V for silicon transistors. Effect of Base Resistors on Tail Current In the figure above, both bases are grounded for simplicity. When base resistors are used, they have a negligible effect on the tail current in a well-designed diff amp. Here is the reason: When base resistors are included in the analysis, the equation for tail current becomes: 𝑉𝐸𝐸 − 𝑉𝐵𝐸 𝐼𝑇 = 𝑅𝐸 + 𝑅𝐵 /2𝛽𝑑𝑐 In any practical design, 𝑅𝐵 /2𝛽𝑑𝑐 is less than 1 percent of 𝑅𝐸. This is why we prefer using either the previous equations of 𝐼𝑇 to calculate tail current. Although base resistors have a negligible effect on the tail current, they can produce input error voltages when the two halves of the diff amp are not perfectly symmetrical. AC Analysis of a Diff Amp Theory of Operation The following figure shows a noninverting input and single-ended output. With a large 𝑅𝐸 , the tail current is almost constant when a small ac signal is present. Because of this, the two halves of a diff amp respond in a complementary manner to the noninverting input. In other words, an increase in the emitter current of Q1 produces a decrease in the emitter current of Q2. Conversely, a decrease in the emitter current of Q1 produces an increase in the emitter current of Q2. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 27 The left transistor Q1 acts like an emitter follower that produces an ac voltage across the emitter resistor. This ac voltage is half of the input voltage 𝑣1. On the positive half-cycle of input voltage, the Q1 emitter current increases, the Q2 emitter current decreases, and the Q2 collector voltage increases. Similarly, on the negative half-cycle of input voltage, the Q1 emitter current decreases, the Q2 emitter current increases, and the Q2 collector voltage decreases. This is why the amplified output sine wave is in phase with the noninverting input. Single-Ended Output Gain The figure on the left below shows the ac-equivalent circuit. Notice that each transistor has an 𝑟′𝑒. Also, the biasing resistor 𝑅𝐸 is in parallel with the 𝑟′𝑒 of the right transistor because the base of Q2 is grounded. In any practical design, 𝑅𝐸 is much greater than 𝑟′𝑒. Because of this, we can ignore 𝑅𝐸 in a preliminary analysis. The figure on the right above shows the simplified equivalent circuit. Notice that the input voltage 𝑣1 is across the first 𝑟′𝑒 in series with the second 𝑟′𝑒. Since the two resistances are equal, the voltage across each 𝑟′𝑒 is half of the input voltage. Its ac output voltage is: 𝑣𝑜𝑢𝑡 = 𝑖𝑐 𝑅𝐶 and the ac input voltage is: 𝑣𝑖𝑛 = 𝑖𝑒 𝑟′𝑒 + 𝑖𝑒 𝑟′𝑒 = 2𝑖𝑒 𝑟′𝑒 Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 28 Dividing 𝑣𝑜𝑢𝑡 by vin gives the voltage gain: 𝑅𝐶 𝐒𝐢𝐧𝐠𝐥𝐞 − 𝐞𝐧𝐝𝐞𝐝 𝐨𝐮𝐭𝐩𝐮𝐭: 𝐴𝑣 = 2r′𝑒 A final point: a quiescent dc voltage 𝑉𝐶 exists at the output terminal. This voltage is not part of the ac signal. The ac voltage 𝑣𝑜𝑢𝑡 is any change from the quiescent voltage. In an op amp, the quiescent dc voltage is removed in a later stage because it is unimportant. Differential-Output Gain This figure shows the ac-equivalent circuit for a noninverting input and differential output. The analysis is almost identical to the previous example, except that the output voltage is twice as much since there are two collector resistors: 𝑣𝑜𝑢𝑡 = 𝑣𝑐2 − 𝑣𝑐1 = 𝑖𝑐 𝑅𝐶 − (−𝑖𝑐 𝑅𝐶 ) = 2𝑖𝑐 𝑅𝐶 (Note: The second minus sign appears because the 𝑣𝑐1 signal is 180° out of phase with 𝑣𝑐2 , as shown in the figure.) The ac input voltage is still equal to: 𝑣𝑖𝑛 = 2𝑖𝑒 𝑟′𝑒 Dividing the output voltage by the input voltage gives the voltage gain: 𝑅𝐶 𝐃𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐥 𝐨𝐮𝐭𝐩𝐮𝐭: 𝐴𝑣 = r′𝑒 This is easy to remember because it is the same as the voltage gain for a CE stage. Inverting-Input Configurations The left figure below shows an inverting input and single-ended output. The ac analysis is almost identical to the noninverting analysis. In this circuit, the inverting input v2 produces an amplified and inverted ac voltage at the final output. The r′𝑒 of each transistor is still part of a voltage divider in the ac-equivalent circuit. This is why the ac voltage across R 𝐸 is half of the inverting input voltage. If a differential output is used, the voltage gain is twice as much as previously discussed. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 29 The diff amp in the right figure above is an upside-down pnp version of the left figure. Frequently, pnp transistors are used in transistor circuits using positive power supplies. These pnp transistors are drawn in an upside-down configuration. As with the npn versions, the inputs and outputs may be either differential or single-ended. Differential-Input Configurations The differential-input configurations have both inputs active at the same time. The ac analysis can be simplified by using the superposition theorem as follows: Since we know how a diff amp behaves with noninverting and inverting inputs, we can combine the two results to get the equations for differential-input configurations. The output voltage for a noninverting input is: A𝑣 (v1 ) and the output voltage for an inverting input is: v𝑜𝑢𝑡 = −A𝑣 (v2 ) By combining the two results, we get the equation for a differential input: v𝑜𝑢𝑡 = A𝑣 (v1 − v2 ) Table of Voltage Gains The table below summarize the voltage gains for the diff-amp configurations. As you can see, the voltage gain is maximum with a differential output. The voltage gain is cut in half when a single- ended output is used. Also, when a single-ended output is used, the input may be noninverting or inverting. Input Output vin vout Differential Differential RC/r'e Av (v1 – v2) Differential Single-ended RC/2r'e Av (v1 – v2) Single-ended Differential RC/r'e Avv1 or – Avv2 Single-ended Single-ended RC/2r'e Avv1 or – Avv2 Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 30 Input Impedance In a CE stage, the input impedance of the base is: z𝑖𝑛 = 𝛽r′𝑒 In a diff amp, the input impedance of either base is twice as high: z𝑖𝑛 = 2𝛽r′𝑒 The input impedance of a diff amp is twice as high because there are two ac emitter resistances r′𝑒 in the ac-equivalent circuit instead of one. The equation of z𝑖𝑛 in a diff-amp is valid for all configurations because any ac input signal sees two ac emitter resistances in the path between the base and ground. Input Characteristics of an Op Amp Assuming perfect symmetry in a diff amp is a good approximation for many applications. But in precision applications, we can no longer treat the two halves of a diff amp as identical. There are three characteristics on the data sheet of every op amp that a designer uses when more accurate answers are needed. They are the input bias current, the input offset current, and the input offset voltage. Input Bias Current In an integrated op amp, the dc of each transistor in the first stage is slightly different, which means that the base currents in the figure above are slightly different. The input bias current is defined as the average of the dc base currents: 𝐼𝐵1 + 𝐼𝐵2 𝐼𝑖𝑛(𝑏𝑖𝑎𝑠) = 2 For instance, if 𝐼𝐵1 = 90 nA and 𝐼𝐵2 = 70 nA, the input bias current is: 90 𝑛𝐴 + 70 𝑛𝐴 Iin(bias) = = 80 𝑛𝐴 2 Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 31 With bipolar op amps, the input bias current is typically in nanoamperes. When op amps use JFETs in the input diff amp, the input bias current is in picoamperes. The input bias current will flow through the resistances between the bases and ground. These resistances may be discrete resistances, or they may be the Thevenin resistances of the input sources. Input Off set Current The input offset current is defined as the difference of the dc base currents: 𝐼𝑖𝑛(𝑜𝑓𝑓) = 𝐼𝐵1 − 𝐼𝐵2 This difference in the base currents indicates how closely the transistors are matched. If the transistors are identical, the input offset current is zero because both base currents will be equal. But almost always, the two transistors are slightly different, and the two base currents are not equal. The Q1 transistor has 20 nA more base current than the Q2 transistor. This can cause a problem when large base resistances are used. Base Currents and Off sets By rearranging the previous 2 equations, we can derive these two equations for the base currents: 𝐼𝑖𝑛(𝑜𝑓𝑓) 𝐼𝐵1 = 𝐼𝑖𝑛(𝑏𝑖𝑎𝑠) + 2 𝐼𝑖𝑛(𝑜𝑓𝑓) 𝐼𝐵2 = 𝐼𝑖𝑛(𝑏𝑖𝑎𝑠) − 2 Data sheets always list 𝐼𝑖𝑛(𝑏𝑖𝑎𝑠) and 𝐼𝑖𝑛(𝑜𝑓𝑓) , but not 𝐼𝐵1 and 𝐼𝐵2. With these equations, we can calculate the base currents. These equations assume that 𝐼𝐵1 is greater than 𝐼𝐵2. If 𝐼𝐵2 is greater than 𝐼𝐵1 , transpose the equations. Effect of Base Current Some diff amps operate with a base resistance on only one side, as shown in the figure above. Because of the base current direction, the base current through 𝑅𝐵 produces a noninverting dc input voltage of: 𝑉1 = −𝐼𝐵1 𝑅𝐵 Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 32 (Note: Capital letters are used here and elsewhere for dc error voltages like 𝑉1. For simplicity, we will treat 𝑉1 as an absolute value. This voltage has the same effect as a genuine input signal. When this false signal is amplified, an unwanted dc voltage 𝑉𝑒𝑟𝑟𝑜𝑟 appears across the output, as shown in the figure above.) Effect of Input Off set Current One way to reduce the output error voltage is by using an equal base resistance on the other side of the diff amp, as shown in the figure below. In this case, we have a differential dc input of: 𝑉𝑖𝑛 = 𝐼𝐵1 𝑅𝐵 − 𝐼𝐵2 𝑅𝐵 = (𝐼𝐵1 − 𝐼𝐵2 )𝑅𝐵 or 𝑉𝑖𝑛 = 𝐼𝑖𝑛(𝑜𝑓𝑓) 𝑅𝐵 Since 𝐼𝑖𝑛(𝑜𝑓𝑓) is usually less than 25 percent of 𝐼𝑖𝑛(𝑏𝑖𝑎𝑠) , the input error voltage is much less when equal base resistors are used. For this reason, designers often include an equal base resistance on the opposite side of a diff amp, as shown above. Input Offset Voltage When a diff amp is integrated as the first stage of an op amp, the two halves are almost, but not quite, identical. To begin with, the two collector resistances may be different, as shown in the left figure below. Because of this, an error voltage appears across the output. Another source of error is the different 𝑉𝐵𝐸 curves for each transistor. For instance, suppose that the two base-emitter curves have the same current. Because the curves are slightly different, there is a difference between the two 𝑉𝐵𝐸 values. This difference adds to the error voltage. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 33 Besides 𝑅𝐶 and 𝑉𝐵𝐸 , other transistor parameters may differ slightly on each half of the diff amp, as shown in the chart above. The input offset voltage is defined as the input voltage that would produce the same output error voltage in a perfect diff amp. As an equation: 𝑉𝑒𝑟𝑟𝑜𝑟 𝑉𝑖𝑛(𝑜𝑓𝑓) = 𝐴𝑣 In this equation, 𝑉𝑒𝑟𝑟𝑜𝑟 does not include the effects of input bias and offset current because both bases are grounded when Verror is measured. Combined Effects In the figure above, the output voltage is the superposition of all input effects. To begin with, there is the ideal ac input: 𝑣𝑖𝑛 = 𝑣1 − 𝑣2 This is what we want. It is the voltage coming from the two input sources. It is amplified to produce the desired ac output: 𝑣𝑜𝑢𝑡 = 𝐴(𝑣1 − 𝑣2 ) Then, there are the three unwanted dc error inputs. With the previously given equations, we can derive these formulas: V1err = (R B1 − R B2 )Iin(bias) Iin(off) V2err = (R B1 + R B2 ) 2 V3err = Vin(off) The advantage of these formulas is that they use Iin(bias) and Iin(off), quantities on the data sheet. The three dc errors are amplified to produce the output error voltage: Verror = Av (V1err + V2err + V3err ) In many cases, Verror can be ignored. This will depend on the application. For instance, if we are building an ac amplifier, Verror may not be important. It is only when we are building some kind of precision dc amplifier that Verror needs to be taken into account. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 34 Equal Base Resistances When the bias and offset errors cannot be ignored, here are the remedies. As already mentioned, one of the first things a designer can do is to use equal base resistances: R B1 = R B2 = R BB. This brings the two halves of the diff amp into a closer alignment because the previously given equations of the errors become: V1err = 0 V2err = R B Iin(off) V3err = Vin(off) If further compensation is necessary, the best approach is to use the nulling circuits suggested on the data sheets. Manufacturers optimize the design of these nulling circuits, which should be used if output error voltage is a problem. Conclusion The following table summarizes the sources of output error voltage. In many applications, the output error voltage is either small enough to ignore or not important in the particular application. In precision applications in which the dc output is important, some form of nulling is used to eliminate the effects of input bias and offset. Designers usually null the output with methods suggested on the manufacturer’s data sheet. Description Cause Solution Input bias current Voltage across a single Use equal RB on RB other side Input offset current Unequal current gains Data sheet nulling methods Input offset voltage Unequal RC and VBE Data sheet nulling methods Common-Mode Gain The figure below shows a differential input and single-ended output. The same input voltage 𝑣𝑖𝑛(𝐶𝑀) is being applied to each base. This voltage is called a common-mode signal. If the diff amp is perfectly symmetrical, there is no ac output voltage with a common-mode input signal because 𝑣1 = 𝑣2. When a diff amp is not perfectly symmetrical, there will be a small ac output voltage. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 35 In this figure, equal voltages are applied to the noninverting and inverting inputs. Nobody would deliberately use a diff amp this way because the output voltage is ideally zero. The reason for discussing this type of input is because most static, interference, and other kinds of undesirable pickup are common-mode signals. Here is how a common-mode signal appears: The connecting wires on the input bases act like small antennas. If the diff amp is operating in an environment with a lot of electromagnetic interference, each base acts like a small antenna that picks up an unwanted signal voltage. One of the reasons the diff amp is so popular is because it discriminates against these common-mode signals. In other words, a diff amp does not amplify common-mode signals. Here is an easy way to find the voltage gain for a common-mode signal: We can redraw the circuit, as shown in the left figure below. Since equal voltages 𝑣𝑖𝑛(𝐶𝑀) drive both inputs simultaneously, there is almost no current through the wire between the emitters. Therefore, we can remove the connecting wire, as shown in the right figure below. With a common-mode signal, the right side of the circuit is equivalent to a heavily swamped CE amplifier. Since 𝑅𝐸 is always much greater than 𝑟′𝑒 , the swamped voltage gain is approximately: RC Av(CM) = 2R F With typical values of R C and 𝑅𝐸 , the common-mode voltage gain is usually less than 1. Common-Mode Rejection Ratio The common-mode rejection ratio (CMRR) is defined as the voltage gain divided by common- mode voltage gain. In symbols: Av CMMR = Av(CM) For instance, if Av = 200 and Av(CM) = 0.5, CMRR = 400. The higher the CMRR, the better. A high CMRR means that the diff amp is amplifying the wanted signal and discriminating against the common-mode signal. Data sheets usually specify CMRR in decibels, using the following formula for the decibel conversion: CMRRdB = 20 log CMRR Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 36 Activities/Assessments 1. Objective type quiz and problem solving a. Use the answer sheet below for the objective type test. b. Answers must be handwritten on the space provided. c. Submission is on the next lecture period. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 37 LESSON 3: Linear Op-Amp Applications The output of a linear op-amp circuit has the same shape as the input signal. If the input is sinusoidal, the output is sinusoidal. At no time during the cycle does the op amp go into saturation. There are a variety of linear op-amp circuit applications including inverting amplifiers, noninverting amplifiers, differential amplifiers, instrumentation amplifiers, current boosters, controlled current sources, and automatic gain control circuits. Inverting-Amplifier Circuits The inverting amplifier is the most basic op-amp circuit. It uses negative feedback to stabilize the overall voltage gain. The reason we need to stabilize the overall voltage gain is because 𝐴𝑉𝑂𝐿 is too high and unstable to be of any use without some form of feedback. For instance, the 741C has a minimum 𝐴𝑉𝑂𝐿 of 20,000 and a maximum 𝐴𝑉𝑂𝐿 of more than 200,000. An unpredictable voltage gain of this magnitude and variation is useless without feedback. Hence, one advantage of this amplifier is that its voltage gain equals the ratio of the feedback resistance to the input resistance. High-Impedance Probe The figure below shows a high-impedance probe that can be used with a digital multimeter. Because of the virtual ground in the first stage, the probe has an input impedance of 100 MV at low frequencies. The first stage is an inverting amplifier with a voltage gain of 0.1. The second stage is an inverting amplifier with a voltage gain of either 1 or 10. The circuit gives you the basic idea of the 10:1 probe. It has a very high input impedance and an overall voltage gain of either 0.1 or 1. In the X10 position of the switch, the output signal is attenuated by a factor of 10. In the X1 position, there is no attenuation of the output signal. The basic circuit shown here can be improved by adding more components to increase the bandwidth. Noninverting-Amplifier Circuits The noninverting amplifier is another basic op-amp circuit. Advantages include stable voltage gain, high input impedance, and low output impedance. Here are some applications. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 38 Audio Distribution Amplifier The following figure shows an ac-coupled noninverting amplifier driving three voltage followers. This is one way to distribute an audio signal to several different outputs. The closed-loop voltage gain and bandwidth of the first stage are given by the familiar equations shown in the figure below. For the values shown, the closed-loop voltage gain is 40. If 𝑓𝑢𝑛𝑖𝑡𝑦 is 1 MHz, the closed-loop bandwidth is 25 kHz. Incidentally, an op amp like an LM348 is convenient to use in a circuit like the one above because the LM348 is a quad 741—four 741s in a 14-pin package. One of the op amps can be the first stage, and the others can be the voltage followers. Voltage Reference The MC1403 is a special-function IC called a voltage reference, a circuit that produces an extremely accurate and stable output voltage. For any positive supply voltage between 4.5 and 40 V, it produces an output voltage of 2.5 V with a tolerance of ±1 percent. The temperature coefficient is only 10 ppm/°C. The abbreviation ppm stands for “parts per million” (1 ppm is equivalent to 0.0001 percent). Therefore, 10 ppm/°C produces a change of only 2.5 mV for a 100°C change in temperature (10 × 0.0001 percent × 100 × 2.5 V). The point is that the output voltage is ultra-stable and equal to 2.5 V over a large temperature range. The only problem is that 2.5 V may be too low a voltage reference for many applications. For instance, suppose we want a voltage reference of 10 V. Then, one solution is to use an MC1403 and a noninverting amplifier as shown in the following figure. With the circuit values shown, the voltage gain is: 30 𝑘Ω Av = +1=4 10 𝑘Ω and the output voltage is: Vout = 4(2.5 𝑉) = 10 𝑉 Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 39 Because the closed-loop voltage gain of the noninverting amplifier is only 4, the output voltage will be a stable voltage reference of 10 V Inverter/Non-inverter Circuits The following are circuit applications where the input signal drives both inputs of the op am simultaneously. When an input signal drives both inputs, we get both inverting and non-inverting amplification at the same time. This produces some interesting results because the output is the superposition of two amplified signals. The total voltage gain with an input signal driving both sides of the op amp equals the voltage gain of the inverting channel plus the voltage gain of the noninverting channel: Av = Av(inv) + Av(non) Switchable Inverter/Non-inverter The figure below shows an op amp that can function as either an inverter or a non-inverter. With the switch in the lower position, the noninverting input is grounded and the circuit is an inverting amplifier. Since the feedback and input resistances are equal, the inverting amplifier has a closed- loop voltage gain of: −𝑅 Av == −1 𝑅 When the switch is moved to the upper position, the input signal drives both the inverting and the noninverting inputs simultaneously. The voltage gain of the inverting channel is still: Av(inv) = −1 The voltage gain of the noninverting channel is: 𝑅 Av(non) =+1 = 2 𝑅 The total voltage gain is the superposition or algebraic sum of the two gains: Av = Av(inv) + Av(non) = −1 + 2 = 1 The circuit is a switchable inverter/non-inverter. It has a voltage gain of either 1 or -1, depending on the position of the switch. In other words, the circuit produces an output voltage with the same Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 40 magnitude as the input voltage, but the phase can be switched between 0° and -180°. As an example, if vin is +5 V, vout will be either +5 V or -5 V depending on the switch position. Inverter with Adjustable Gain When the variable resistor of the figure below is zero, the noninverting input is grounded and the −R circuit becomes an inverting amplifier with a voltage gain of 2. R1 When the variable resistor is increased to R 2 , equal voltages drive the noninverting and inverting inputs of the op-amp (common-mode input). Due to common-mode rejection, output voltage is approximately zero. Example: 1. Using the figure above, what are the values of the maximum and minimum voltage gain if R1 = 1.2 kΩ and R 2 = 91 kΩ? Solution: The maximum voltage gain is: −91 𝑘𝛺 𝐴𝑣 = = −75.8 1.2 𝑘𝛺 The minimum voltage gain is zero. Sign Changer The circuit below is called a sign changer, a rather unusual circuit because its voltage gain can be varied from 21 to 1. Here is the theory of operation: When the wiper is all the way to the right, the noninverting input is grounded and the circuit has a voltage gain of: Av = −1 Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 41 When the wiper is all the way to the left, the input signal drives the noninverting input as well as the inverting input. In this case, the total voltage gain is the superposition of the inverting and noninverting voltage gains: Av(non) = 2 Av(inv) = −1 Av = Av(inv) + Av(non) = 1 In summary, when the wiper is moved from right to left, the voltage gain changes continuously from -1 to 1. At the crossover point (wiper at center), a common-mode signal drives the op amp and the output is ideally zero. Differential Amplifiers One of the most important characteristics of a differential amplifier is its CMRR because the typical input signal is a small differential voltage and a large common-mode voltage. Basic Differential Amplifier The following figure shows an op amp connected as a differential amplifier. The resistor R1′ has the same nominal value as R1 but differs slightly in value because of tolerances. For instance, if the resistors are 1 kΩ ±1 percent, R1 may be as high as 1010 Ω and R1′ may be as low as 990 Ω, and vice versa. Similarly, R 2 and R 2′ are nominally equal but may differ slightly because of tolerances. In the figure, the desired input voltage vin is called the differential input voltage to distinguish it from the common-mode input voltage vin(CM). This circuit amplifies the differential input voltage vin to get an output voltage of vout. Using the superposition theorem, it can be shown that: vout = Av vin where −𝑅2 𝐴𝑣 = 𝑅1 This voltage gain is called the differential voltage gain to distinguish it from the common-mode voltage gain 𝐴𝑣(𝐶𝑀). By using precision resistors, we can build a differential amplifier with a precise voltage gain. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 42 A differential amplifier is often used in applications in which the differential input signal vin is a small dc voltage (millivolts) and the common-mode input signal is a large dc voltage (volts). As a result, the CMRR of the circuit becomes a critical parameter. For instance, if the differential input signal is 7.5 mV and the common-mode signal is 7.5 V, the differential input signal is 60 dB less than the common-mode input signal. Unless the circuit has a very high CMRR, the common-mode output signal will be objectionably large. CMRR of the Op Amp In the following figure, two factors determine the overall CMRR of the circuit. 1. CMRR of the op amp itself. (741C’s minimum CMRR is 70 dB) 2. Tolerance of the resistors Calculating CMRR In a circuit like the one above, resistors with a tolerance of ±0.1 percent are commonly used. When 𝑅1 = 𝑅2 , the differential voltage gain is: −𝑅2 𝐴𝑣 = = −1 𝑅1 And the common-mode voltage gain is: 𝐴𝑣(𝐶𝑀) = ±2(0.1%) = ±2(0.001) = ±0.002 The CMRR has a magnitude of: |𝐴𝑣 | 1 𝐶𝑀𝑅𝑅 = = = 500 |𝐴𝑣(𝐶𝑀) | 0.002 which is equivalent to 54 dB. Instrumentation Amplifiers An instrumentation amplifier has a large voltage gain, a high CMRR, low input offsets, low temperature drift, and high input impedance. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 43 Basic Instrumentation Amplifier The figure below shows the classic design used for most instrumentation amplifiers. The output op amp is a differential amplifier with the voltage gain of unity. The resistors used in this output stage are usually matched to within ±0.1 percent or better. This means that the CMRR of the output stage is at least 54 dB. Precision resistors are commercially available from less than 1 Ω to more than 10 MΩ, with tolerances of ±0.01 to ±1 percent. If we use matched resistors that are within ±0.01 percent of each other, the CMRR of the output stage can be as high as 74 dB. Also, temperature drift of precision resistors can be as low as 1 ppm/°C. Example: 1. In the previous figure, 𝑅1 = 1 kΩ, 𝑅2 = 100 kΩ, and 𝑅 = 10 kΩ. What is the differential voltage gain of the instrumentation amplifier? What is the common- mode voltage gain if the resistor tolerances in the second stage are ±0.01 percent? If 𝑣𝑖𝑛 = 10 mV and 𝑣𝑖𝑛(𝐶𝑀) = 10 V, what are the values of the differential and common-mode output signals? Solution: The voltage gain of the preamp is: 100 𝑘Ω 𝐴𝑣 = + 1 = 101 1 𝑘Ω Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 44 Since the voltage gain of the second stage is -1, the voltage gain of the instrumentation amplifier is -101. The common-mode voltage gain of the second stage is: 𝐴𝑣(𝐶𝑀) = ±2(0.01%) = ±2(0.0001) = ±0.0002 A differential input signal of 10 mV will produce an output signal of: 𝑣𝑜𝑢𝑡 = ±0.0002(10 𝑉) = ±2 𝑚𝑉 Current Boosters The short-circuit output current of an op amp is typically 25 mA or less. One way to get more output current is to use a power op amp like the LM675 or LM12. These op amps have short- circuit output currents of 3 and 10 A. Another way to get more short-circuit output current is to use a current booster, a power transistor or other device that has a current gain and a higher current rating than the op amp. Unidirectional Booster The figure below shows one way to increase the maximum load current. The output of an op amp drives an emitter follower. The closed-loop voltage gain is: 𝑅2 𝐴𝑣 = +1 𝑅1 In this circuit, the op amp no longer has to supply the load current. Instead, it only has to supply base current to the emitter follower. Because of the current gain of the transistor, the maximum load current is increased to: 𝐼𝑚𝑎𝑥 = 𝛽𝑑𝑐 𝐼𝑆𝐶 where ISC is the short-circuit output current of the op amp. This means that an op amp like a 741C can have a maximum output current of 25 mA increased by a factor of 𝛽𝑑𝑐. The circuit can drive low-impedance loads because the negative feedback reduces the output impedance of the emitter follower by a factor of 1 + 𝐴𝑉𝑂𝐿 𝐵. Since the emitter follower already has a low output impedance, the closed-loop output impedance will be very small. Instructional Material for Fundamentals of Mixed Signals and Sensors (CMPE 30153) 45 Bidirectional Current The disadvantage of the current booster shown in the previous figure is its unidirectional load current. The figure below shows one way to get a bidirectional load current. An inverting amplifier drives a Class-B push-pull emitter follower. In this circuit, the closed-loop voltage gain is: −𝑅2 𝐴𝑣 = 𝑅1 When the input voltage is positive, the lower transistor is conducting and the load voltage is negative. When the input voltage is negative, the upper transistor is conducting and the output voltage is positive. In either case, the maximum output current is increased by the current gain of the conducting transistor. Since the Class-B push-pull emitter follower is inside the feedback loop, the closed-loop output impedance is very small. Example: 1. In the previously given circuit, 𝑅1 = 1 kΩ and 𝑅2 = 51 kΩ. If a 741C is used for the op amp, what is the voltage gain of the circuit? What is the closed-loop output impedance? What is the shorted-load of the circuit if each transistor has a current gain of 125? Solution: The voltag

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