Electronics-II Frequency Response PDF

Because learning changes everything. ® ELE-3413: Electronics-II LO1 – Frequency Response Electronic Principles Ninth Edition Albert Malvino, David J. Bates, Patrick E. Hoppe Updated by: Asad Hindash © 2021 McGraw Hill. All rights reserved. Authorized only for instructor...

Because learning changes everything. ® ELE-3413: Electronics-II LO1 – Frequency Response Electronic Principles Ninth Edition Albert Malvino, David J. Bates, Patrick E. Hoppe Updated by: Asad Hindash © 2021 McGraw Hill. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill.  Learning Outcome CLO1: Determine the frequency response and noise performance of single and multi-stage amplifiers. Sub outcome 1: Review of Bipolar Junction Transistors (Operation and DC biasing). Review of BJT Amplifiers. Sub outcome 2: Effects of external Bypass Capacitors & Internal Transistor Capacitance, Amplifier Low Frequency Response. Sub outcome 3: Amplifier high Frequency Response. Total Frequency Response of Amplifier. Frequency Response of multistage amplifiers. Lab1: Frequency Analysis of BJT Amplifiers. © McGraw-Hill 2  DC Analysis - Base Biased Transistor Solve for all VRB voltages and IB = V= VCC − VRC currents. RB CE 4.3V V = 12 V − 4.03V IB = CE 470 kΩ VCE = 7.97 V βdc = 200 I B = 9.15μA I C = β dc I B KVL I C 200 × 9.15μA = 0 VBB − VRB − VBE = I C = 1.83mA VRB = 4.3V V= RB VBB − VBE I B = 9.15μA V= RB 5V − 0.7 V VRC= I C × RC I C = 1.83mA VRB = 4.3V VRC 1.83mA × 2.2kΩ = VRC = 4.03V VRC = 4.03V 𝑉𝑉𝐶𝐶𝐶𝐶 = 7.97 𝑉𝑉 Note: Use 2nd approx., VBE = 0.7 V © McGraw-Hill 3  Introduction of Common Emitter Amplifier The circuit diagram shows the working of the common emitter amplifier circuit and it consists of voltage divider biasing, used to supply the base bias voltage as per the necessity. The voltage divider C2 biasing has a potential divider with two resistors are connected in a C1 way that the midpoint is used for supplying base bias voltage. 𝛽𝛽 = ∆𝐼𝐼𝐶𝐶 ⁄∆𝐼𝐼𝐵𝐵 𝐴𝐴𝑉𝑉 = 𝛽𝛽𝑅𝑅𝐶𝐶 ⁄∆𝑅𝑅𝐵𝐵 CE The first step in AC analysis of Common Emitter amplifier circuit is to draw the AC equivalent circuit by reducing all DC sources to zero and shorting all the capacitors. The below figure shows the AC equivalent circuit. © McGraw-Hill 4 CE Amplifier Frequency Response The voltage gain of a CE amplifier varies with signal frequency. It is because the reactance of the capacitors in the circuit changes with signal frequency and hence affects the output voltage. The curve drawn between voltage gain and the signal frequency of an amplifier is known as frequency response. The below figure shows the frequency response of a typical CE amplifier. At Low Frequencies (< FL) The reactance of coupling capacitor C2 is relatively high and hence very small part of the signal will pass from the amplifier stage to the load. Moreover, CE cannot shunt the RE effectively because of its large reactance at low frequencies. These two factors cause a drops off of voltage gain at low frequencies. At High Frequencies (> FH) The reactance of coupling capacitor C2 is very small and it behaves as a short circuit. This increases the loading effect of the amplifier stage and serves to reduce the voltage gain. Moreover, at high frequencies, the capacitive reactance of base-emitters junction is low which increases the base current. This frequency reduces the current amplification factor β. Due to these two reasons, the voltage gain drops off at a high frequency. At Mid Frequencies (FL to FH) The voltage gain of the amplifier is constant. The effect of the coupling capacitor C2 in this frequency range is such as to maintain a constant voltage gain. Thus, as the frequency increases in this range, the reactance of CC decreases, which tends to increase the gain. However, at the same time, lower reactance means higher almost cancel each other, resulting in a uniform fair at mid- frequency. We can observe the frequency response of any amplifier circuit is the difference in its performance through changes within the input signal’s frequency because it shows the frequency bands where the output remains fairly stable. The circuit bandwidth can be defined as the frequency range either small or big among ƒH & ƒL. © McGraw-Hill 5  Voltage Divider Bias - VDB DC and AC Equivalents DC Circuit AC Circuit h Model AC Circuit T Model © McGraw-Hill 6  Summary of Transistor Bias Circuits npn transistors are shown. Supply voltage polarities are reversed for pnp transistors. VOLTAGE-DIVIDER BIAS EMITTER- FEEDBACK BIAS BASE BIAS COLLECTOR- EMITTER FEEDBACK BIAS BIAS © McGraw-Hill 7  Summary of the Common-Emitter Amplifier CIRCUIT WITH VOLTAGE-  Input is at the base. Output is at the collector. DIVIDER BIAS  There is a phase inversion from input to output.  C1 and C3 are coupling capacitors for the input and output signals.  C2 is the emitter-bypass capacitor.  All capacitors must have a negligible reactance at the frequency of operation, so they appear as shorts.  Emitter is at ac ground due to the bypass capacitor. EQUIVALENT CIRCUITS AND FORMULAS © McGraw-Hill 8  Summary of the Common-Emitter Amplifier SWAMPED AMPLIFIER WITH RESISTIVE LOAD © McGraw-Hill 9  Frequency Analysis of CE Amplifiers In this learning outcome, we will study the effect of frequency on the gain of CE amplifiers. The analysis can be split into four parts: © McGraw-Hill 10  DC Analysis of CE Amplifiers Follow the steps taken in Electronics I i. All the capacitors 𝐶𝐶1, 𝐶𝐶2, and 𝐶𝐶3 are open under DC operation since the frequency is zero. Note that the impedance response of a capacitor 𝐶𝐶 is given as 1/𝑗𝑗𝜔𝜔. Figure (b) sketch the circuit with all capacitors open. ii. Notice that 𝑅𝑅𝐸𝐸1 and 𝑅𝑅𝐸𝐸2 are in series and they act as a single resistance 𝑅𝑅𝐸𝐸 = 𝑅𝑅𝐸𝐸1 + 𝑅𝑅𝐸𝐸2. Taking this action yield to figure (c). © McGraw-Hill 11  DC Analysis of CE Amplifiers Follow the steps taken in Electronics I i. Assume stiff-analysis, which means the base current is zero, 𝐼𝐼𝐵𝐵 = 0 𝐴𝐴 𝑅𝑅2 ii. The base-voltage can be computed using voltage divider. 𝑉𝑉𝐵𝐵 = 𝑅𝑅 𝑉𝑉𝐶𝐶𝐶𝐶 1 +𝑅𝑅2 iii. Compute the emitter voltage 𝑉𝑉𝐸𝐸 = 𝑉𝑉𝐵𝐵 − 𝑉𝑉𝐵𝐵E = 𝑉𝑉𝐵𝐵 − 0.7 𝑉𝑉 iv. Compute the emitter current 𝐼𝐼𝐸𝐸 = 𝑅𝑅𝐸𝐸 𝐸𝐸 25 𝑚𝑚𝑚𝑚 v. Compute the transistor’s internal resistance 𝑟𝑟𝑒𝑒′ = 𝐼𝐼𝐸𝐸 Temperature dependent and is based on an ambient temperature of 20°C. What is the gain in DC? © McGraw-Hill 12  Mid-band AC Gain Analysis In Mid-band gain figure (a), we assume the frequency is high enough that makes all the external capacitor impedances 1/𝑗𝑗𝜔𝜔𝐶𝐶1, 1/𝑗𝑗𝜔𝜔𝐶𝐶2, and 1/𝑗𝑗𝜔𝜔𝐶𝐶3 to act as a short circuit (very small), figure (b). Remember, in AC analysis, all the DC sources are grounded. Taking the two assumptions above yields to the circuit in figure (c). (a) 𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 𝑉𝑉𝑏𝑏 𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 To compute the gain, 𝐴𝐴𝑣𝑣(𝑚𝑚𝑚𝑚𝑚𝑚) = = × 𝑉𝑉𝑠𝑠 𝑉𝑉𝑠𝑠 𝑉𝑉𝑏𝑏 𝑉𝑉𝑏𝑏 𝑅𝑅1 𝑅𝑅2 𝛽𝛽 𝑟𝑟𝑒𝑒′ +𝑅𝑅𝐸𝐸𝐸 = 𝑉𝑉𝑠𝑠 𝑅𝑅𝑠𝑠 +𝑅𝑅1 𝑅𝑅2 𝛽𝛽 𝑟𝑟𝑒𝑒′ +𝑅𝑅𝐸𝐸𝐸 𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑅𝑅𝐶𝐶 𝑅𝑅𝐿𝐿 𝑅𝑅𝑐𝑐 =− =− =− 𝑉𝑉𝑏𝑏 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑟𝑟𝑒𝑒′ +𝑅𝑅𝐸𝐸𝐸 𝑟𝑟𝑒𝑒′ +𝑅𝑅𝐸𝐸𝐸 where 𝑅𝑅𝑐𝑐 = 𝑅𝑅𝐶𝐶 𝑅𝑅𝐿𝐿 , Thus 𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 𝑉𝑉𝑏𝑏 𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 𝑅𝑅1 𝑅𝑅2 𝛽𝛽 𝑟𝑟𝑒𝑒′ +𝑅𝑅𝐸𝐸𝐸 𝑅𝑅𝐶𝐶 𝑅𝑅𝐿𝐿 𝐴𝐴𝑣𝑣(𝑚𝑚𝑚𝑚𝑚𝑚) = = × =− × 𝑉𝑉𝑠𝑠 𝑉𝑉𝑠𝑠 𝑉𝑉𝑏𝑏 𝑅𝑅𝑠𝑠 +𝑅𝑅1 𝑅𝑅2 𝛽𝛽 𝑟𝑟𝑒𝑒′ +𝑅𝑅𝐸𝐸𝐸 𝑟𝑟𝑒𝑒′ +𝑅𝑅𝐸𝐸𝐸 𝑉𝑉𝑏𝑏 Notice if 𝑅𝑅𝑠𝑠 = 0, too small then =1 𝑉𝑉𝑠𝑠 𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 𝑅𝑅𝐶𝐶 𝑅𝑅𝐿𝐿 𝐴𝐴𝑣𝑣(𝑚𝑚𝑚𝑚𝑚𝑚) = = =− 𝑉𝑉𝑠𝑠 𝑉𝑉𝑏𝑏 𝑟𝑟𝑒𝑒′ +𝑅𝑅𝐸𝐸𝐸 In decibel 𝐴𝐴𝑣𝑣(𝑚𝑚𝑚𝑚𝑚𝑚) 𝑑𝑑𝑑𝑑 = 20 log 𝐴𝐴𝑣𝑣(𝑚𝑚𝑚𝑚𝑚𝑚) (b) (c) © McGraw-Hill 13  A Trick to Simplify AC Analysis © McGraw-Hill 14  Effect of Coupling Capacitors Coupling capacitors are in series with the signal and are part of a high-pass filter network. They affect the low-frequency response of the amplifier. © McGraw-Hill 15  Low Frequency Analysis, C1 What if the frequency is not high enough to make 1/𝑗𝑗𝜔𝜔𝐶𝐶1, 1/𝑗𝑗𝜔𝜔𝐶𝐶2, and 1/𝑗𝑗𝜔𝜔𝐶𝐶3 negligible? In this case, their effect has to be considered. To take the effect of 𝐶𝐶1 into account, short all the other capacitors along with the AC source, and compute the equivalent resistor looking from the terminals of 𝐶𝐶1. 𝑅𝑅𝑒𝑒𝑒𝑒(𝐶𝐶𝐶) = 𝑅𝑅𝑠𝑠 + 𝑅𝑅1 𝑅𝑅2 ‖𝛽𝛽 𝑟𝑟𝑒𝑒′ + 𝑅𝑅𝐸𝐸𝐸 1 The cutoff frequency of the high pass filter formed by 𝐶𝐶1 is 𝑓𝑓𝐶𝐶𝐶 = 2𝜋𝜋𝐶𝐶1 𝑅𝑅𝑒𝑒𝑒𝑒(𝐶𝐶𝐶) 𝑓𝑓𝐶𝐶𝐶 sometimes referred to as 𝑓𝑓𝐶𝐶𝑙𝑙(𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖) to indicate input lower-cutoff frequency. © McGraw-Hill 16  Low Frequency Analysis, C2 To take the effect of 𝐶𝐶2 into account, short all the other capacitors along with the AC source, and compute the equivalent resistor looking from the terminals of 𝐶𝐶2. 𝑅𝑅𝑠𝑠 𝑅𝑅1 𝑅𝑅2 𝑅𝑅𝑒𝑒𝑒𝑒(𝐶𝐶𝐶) = + 𝑟𝑟𝑒𝑒′ + 𝑅𝑅𝐸𝐸𝐸 𝑅𝑅𝐸𝐸𝐸 𝛽𝛽 1 The cutoff frequency of the high pass filter formed by 𝐶𝐶2 is 𝑓𝑓𝐶𝐶𝐶 = 2𝜋𝜋𝐶𝐶2 𝑅𝑅𝑒𝑒𝑒𝑒(𝐶𝐶𝐶) 𝑓𝑓𝐶𝐶2 sometimes referred to as 𝑓𝑓𝐶𝐶𝑙𝑙(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) to indicate bypass lower-cutoff frequency. © McGraw-Hill 17  Low Frequency Analysis, C3 Remember in amplification/active mode, the BC junction is reverse biased and its equivalent resistant is very high. Consequently we can consider it open circuit while we compute the equivalent resistance of C3. To take the effect of 𝐶𝐶3 into account, short all the other capacitors along with the AC source, and compute the equivalent resistor looking from the terminals of 𝐶𝐶3. 𝑅𝑅𝑒𝑒𝑒𝑒(𝐶𝐶𝐶) = 𝑅𝑅𝐶𝐶 + 𝑅𝑅𝐿𝐿 1 The cutoff frequency of the high pass filter formed by 𝐶𝐶3 is 𝑓𝑓𝐶𝐶𝐶 = 2𝜋𝜋𝐶𝐶3 𝑅𝑅𝑒𝑒𝑒𝑒(𝐶𝐶𝐶) 𝑓𝑓𝐶𝐶3 sometimes referred to as 𝑓𝑓𝐶𝐶𝑙𝑙(𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜) to indicate output lower-cutoff frequency. © McGraw-Hill 18  Low Frequency Response The bypass RC circuit response can be found by observing the charge/discharge paths. For this circuit, there is one path through 𝑅𝑅𝐸𝐸𝐸 A second path goes through 𝑅𝑅𝐸𝐸1 , 𝑟𝑟𝑒𝑒′ and the parallel combination of bias and source resistances (source resistance not shown in figure). The total resistance of the paths can be found by: 𝑅𝑅𝑠𝑠 𝑅𝑅1 𝑅𝑅2 𝑅𝑅𝑒𝑒𝑒𝑒𝑢𝑢𝑢𝑢𝑢𝑢 = + 𝑟𝑟𝑒𝑒′ + 𝑅𝑅𝐸𝐸𝐸 𝑅𝑅𝐸𝐸𝐸 𝛽𝛽 © McGraw-Hill 19  Low-Frequency Response © McGraw-Hill 20  Low-Frequency Response © McGraw-Hill 21  Low-Frequency Response Add Rs to Rin © McGraw-Hill 22  Low-Frequency Response © McGraw-Hill 23  Phase-Shift of the Input RC Circuit The phase angle of the input RC circuit is expressed as: 𝑋𝑋𝐶𝐶𝐶 𝜃𝜃 = tan−1 𝑅𝑅𝑖𝑖𝑖𝑖 Phase angle vs frequency for the output RC circuit The input RC circuit causes the base voltage to lead the input voltage below midrange by an amount equal to the phase angle, θ. © McGraw-Hill 24  Phase-Shift of the Input RC Circuit The phase angle of the input RC circuit is expressed as: 𝑅𝑅𝑠𝑠 ‖𝑅𝑅1 ‖𝑅𝑅2 ‖𝛽𝛽𝑠𝑠𝑠𝑠 𝑟𝑟𝑒𝑒′ + 𝑅𝑅𝐸𝐸𝐸 𝜃𝜃 = tan 𝑋𝑋𝐶𝐶(𝑡𝑡𝑡𝑡𝑡𝑡) 1 And 𝑋𝑋𝐶𝐶(𝑡𝑡𝑡𝑡𝑡𝑡) = 2𝜋𝜋𝜋𝜋𝐶𝐶𝑖𝑖𝑖𝑖,𝑡𝑡𝑡𝑡𝑡𝑡 The output of the circuit lags the input. At the critical frequency, the phase angle is 45° with the signal voltage at the base of the transistor lagging the input signal. As the frequency increases above fc, the phase angle increases above 45° and approaches 90° when the frequency is sufficiently high.. © McGraw-Hill 25  Phase-Shift of the Output RC Circuit The phase angle of the output RC circuit is expressed as: 𝑋𝑋𝐶𝐶𝐶 𝜃𝜃 = tan−1 𝑅𝑅𝐶𝐶 + 𝑅𝑅𝐿𝐿 Phase angle vs frequency for the output RC circuit θ ≅ 0° for the midrange frequencies and approaches 90° as the frequency approaches zero (XC3 approaches infinity). At the critical frequency fc, the phase shift is 45°. © McGraw-Hill 26  The Bode Plot The Bode plot is a plot of decibel voltage gain versus frequency. The frequency axis is logarithmic; the decibel gain is plotted on a linear scale. The −3dB point is the critical frequency. A ten-times change in frequency is called a decade (-20dB/decade). A two-times change in frequency is called an octave. (-6dB/octave). © McGraw-Hill 27  Example: Voltage Gain Calculations © McGraw-Hill 28  Composite Bode of a BJT Amplifier Response Composite Bode plot of a BJT amplifier response for three low-frequency RC circuits with different critical frequencies. Total response is shown by the blue curve. Comment: the arrangement of the cutoff frequencies could be different based on what has been computed for each example. Composite Bode plot of an amplifier response where all RC circuits have the same fcl (Blue is ideal; red is actual.) © McGraw-Hill 29  Bode Plot Plotter Multisim has a fictitious instrument called the Bode plotter. This is the previous BJT amplifier. The Bode plotter allows you to see the Bode plot directly. By selecting the proper scales, you can magnify the response. Move the cursor to the point where the total response is – 3dB from midband and read fc. © McGraw-Hill 30  Low-Frequency Response © McGraw-Hill 31  Low-Frequency Response 𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 𝑅𝑅𝐶𝐶 𝑅𝑅𝐿𝐿 𝐴𝐴𝑣𝑣(𝑚𝑚𝑚𝑚𝑚𝑚) = = =− ′ 𝑉𝑉𝑠𝑠 𝑉𝑉𝑏𝑏 𝑟𝑟𝑒𝑒 + 𝑅𝑅𝐸𝐸𝐸 © McGraw-Hill 32  High-Frequency Analysis, Internal Capacitors The internal capacitors of formed by base-collector and base-emitter junctions, 𝐶𝐶𝑏𝑏𝑐𝑐 and 𝐶𝐶𝑏𝑏𝑒𝑒 are very small, in terms of pico-farads. Their impedance effects, 1/𝑗𝑗𝜔𝜔𝐶𝐶𝑏𝑏𝑐𝑐 and 1/𝑗𝑗𝜔𝜔𝐶𝐶𝑏𝑏𝑒𝑒 appear at very high frequencies in terms of Giga Hertz. To make the analysis easy, we will use the Miller Theorem The internal capacitors act as a low pass filters At lower frequencies, the internal capacitances have a very high reactance because of their low capacitance value (usually only a few picofarads) and the low frequency value. Therefore, they look like opens and have no effect on the transistor’s performance. As the frequency goes up, the internal capacitive reactances go down, and at some point they begin to have a significant effect on the transistor’s gain. © McGraw-Hill 33  Miller’s Theorem Miller’s theorem states that, for inverting amplifiers, the capacitance between the input and output is equivalent to separate input and output capacitances to ground. Av is the absolute value of the voltage gain of the amplifier at midrange frequencies, and C represents Cbc. For the input capacitance, the gain has a large effect on the equivalent capacitance, which is an important consideration when using inverting amplifiers. Miller’s theorem is used to simplify the analysis of inverting amplifiers at high frequencies where the internal Transistor capacitances are important. © McGraw-Hill 34  High-Frequency Response In the high frequency response of amplifiers, the coupling and bypass capacitors are treated as effective shorts and do not appear in the equivalent circuit. © McGraw-Hill 35  High-Frequency Response The high frequency response of inverting amplifiers is primarily determined by the transistor’s internal capacitance and the Miller effect. The equivalent high-frequency AC circuit is shown for a voltage- divider biased CE amplifier with a fully bypassed emitter resistor. © McGraw-Hill 36  High-Frequency Response, Input Cutoff For the fully bypassed case, such as the one shown in the text in Example 10-11, the ac emitter resistance (𝑟𝑟𝑒𝑒′ ) is multiplied by 𝛽𝛽𝑎𝑎𝑎𝑎 obtain the equivalent input resistance at the transistor’s base. © McGraw-Hill 37  High-Frequency Response, Input Cutoff Combining the capacitors in parallel and Thevenizing forms an equivalent basic RC low-pass filter: 1 𝑓𝑓𝑐𝑐 = 2𝜋𝜋 𝑅𝑅𝑠𝑠 𝑅𝑅1 𝑅𝑅2 𝛽𝛽𝑟𝑟𝑒𝑒′ 𝐶𝐶𝑡𝑡𝑡𝑡𝑡𝑡 © McGraw-Hill 38  High-Frequency Response, Input Cutoff If there is an unbypassed emitter resistor, such as RE1 in the earlier example, it is shown in the emitter circuit and acts to increase 𝑟𝑟𝑒𝑒′ and thus reduce fc. © McGraw-Hill 39  High-Frequency Response, Input Cutoff If there is an unbypassed emitter resistor (RE1 in this case), the Thevenin resistance is modified to © McGraw-Hill 40  High-Frequency Response, Output Cutoff Frequency For the output cutoff frequency at high frequencies. Take into consideration that the BC junction is reverse biased. The cutoff frequency is: 1 𝑓𝑓ℎ𝑐𝑐(𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜) = 2𝜋𝜋 𝑅𝑅𝐿𝐿 𝑅𝑅𝐶𝐶 𝐶𝐶𝑜𝑜𝑜𝑜𝑜𝑜(𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀) © McGraw-Hill 41  Miller’s Theorem © McGraw-Hill 42  High-Frequency Response, Input Cutoff © McGraw-Hill 43  High Frequency Analysis of CE Amplifiers © McGraw-Hill 44  High Frequency Analysis of CE Amplifiers © McGraw-Hill 45  High Frequency Analysis of CE Amplifiers © McGraw-Hill 46  Phase-Shift of the Output RC Circuit (with Miller effect) The phase angle of the output RC circuit is expressed as: 𝑅𝑅𝑐𝑐 𝜃𝜃 = tan−1 𝑋𝑋𝐶𝐶𝑜𝑜𝑜𝑜𝑜𝑜(𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀) 1 And 𝑋𝑋𝐶𝐶𝑜𝑜𝑜𝑜𝑜𝑜(𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀) = 2𝜋𝜋𝜋𝜋𝐶𝐶𝑜𝑜𝑜𝑜𝑜𝑜,𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 © McGraw-Hill 47  Total Amplifier Frequency Response In general, the overall frequency response is the combination of three lower critical frequencies due to coupling and bypass capacitors and two upper critical frequencies due to internal capacitances. The ideal Bode plot for a typical amplifier is: Comment: the arrangement of the cutoff frequencies could be different based on what has been computed for each example. The lower dominant cutoff frequency 𝑓𝑓𝑙𝑙(𝑑𝑑om) is the highest among all the lower cutoff frequencies 𝑓𝑓𝑐𝑐l(𝑜𝑜utput), 𝑓𝑓𝑐𝑐l(𝑏𝑏ypass) , and 𝑓𝑓𝑐𝑐l(input). The higher dominant cutoff frequency 𝑓𝑓h(𝑑𝑑om) is the lowest among all the higher cutoff frequencies 𝑓𝑓𝑐𝑐l(𝑜𝑜utput) , and 𝑓𝑓𝑐𝑐l(input) © McGraw-Hill 48  Total Amplifier Frequency Response © McGraw-Hill 49  Gain-Bandwidth Product The product of the voltage gain and the bandwidth is always constant when the roll-off is -20 dB/decade. This characteristic is called the gain-bandwidth product. Let’s assume that the dominant lower critical frequency of a particular amplifier is much less than the dominant upper critical frequency. 𝑓𝑓𝑐𝑐𝑐𝑐(𝑑𝑑𝑑𝑑𝑑𝑑) ≪ 𝑓𝑓𝑐𝑐𝑐𝑐(𝑑𝑑𝑑𝑑𝑑𝑑) The bandwidth can then be approximated as 𝐵𝐵𝐵𝐵 = 𝑓𝑓𝑐𝑐𝑐𝑐(𝑑𝑑𝑑𝑑𝑑𝑑) − 𝑓𝑓𝑐𝑐𝑐𝑐(𝑑𝑑𝑑𝑑𝑑𝑑) ≅ 𝑓𝑓𝑐𝑐𝑐𝑐 The Unity-Gain Frequency 𝑓𝑓𝑇𝑇 = 𝐴𝐴𝑣𝑣(𝑚𝑚𝑚𝑚𝑚𝑚) × 𝐵𝐵𝐵𝐵 Simplified response curve where fcl(dom) The significance of the unity gain frequency, is negligible (assumed to be zero) fT , is that it always equals the midrange compared to fcu(dom). voltage gain times the bandwidth and is constant for a given transistor. For example, if a transistor datasheet specifies fT = 100 MHz, this means that the transistor is capable of producing a voltage gain of 1 up to 100 MHz, or a gain of 100 up to 1 MHz, or any combination of gain and bandwidth that produces a product of 100 MHz. © McGraw-Hill 50  Total Amplifier-Frequency Response The overall response can be viewed on the Bode plotter by choosing the appropriate scales. The overall response for the BJT example given previously is shown © McGraw-Hill 51  Multistage Amplifiers Frequency Response For multistage amplifiers, the individual stages have an effect on the overall response. 1. Each stage has a different dominant lower critical frequency and a different dominant upper critical frequency. 2. Each stage has the same dominant lower critical frequency and the same dominant upper critical frequency. In general, with different cutoff frequencies, the dominant lower cutoff frequency is equal to the highest fcl; the dominant upper critical frequency is equal to lowest fcu. When the critical frequencies for multistage amplifiers are equal, the lower critical frequency is higher than any one and 𝒇𝒇𝒄𝒄𝒄𝒄 is given by 𝒇𝒇′𝒄𝒄𝒄𝒄 = 𝟏𝟏 𝟐𝟐 𝒏𝒏 −𝟏𝟏 𝟏𝟏 and the upper critical frequency is given by 𝒇𝒇′𝒄𝒄𝒖𝒖 = 𝒇𝒇𝒄𝒄𝒄𝒄 𝟐𝟐 𝒏𝒏 − 𝟏𝟏 © McGraw-Hill 52  Key Terms Term Description Decibel A logarithmic measure of the ratio of one power level to another or one voltage to another. Midrange gain The gain that occurs for the range of frequencies between the lower and upper critical frequencies. Critical The frequency at which the response of an amplifier or filter is 3 dB less frequency than at midrange. Roll-off The rate of decrease in the gain of an amplifier above or below the critical frequencies. Decade A tem times increase or decrease in the value of a quantity such as frequency. Bode Plot An idealized graph of the gain in dB verses frequency used to graphically illustrate the response of an amplifier or filter. Bandwidth The characteristic of certain types of electronic circuits that specifies the usable range of frequencies that pass from input to output. © McGraw-Hill 53  Interactive Materials Coupling Capacitors https://www.youtube.com/watch?v=tZmXTADSmMs Coupling Capacitors https://www.youtube.com/watch?v=1OA5wk7N1hY Effect of External Capacitors https://www.youtube.com/watch?v=2UkuLxLu7AI BJT Amplifiers Low Frequency Response https://www.youtube.com/watch?v=lI6iBrZEGI8&list=PLZvLSclgk4yI0WqeVk2i1f9diaRivg8mh&index=4 CE Amplifier High Frequency Response https://www.youtube.com/watch?v=RM6BAeZsc5s&list=PLZvLSclgk4yI0WqeVk2i1f9diaRivg8mh&index=10 BJT Amplifier High Frequency Response (Miller Effect) https://www.youtube.com/watch?v=KzwhAjEIe1Y&list=PLZvLSclgk4yI0WqeVk2i1f9diaRivg8mh&index=5 BJT Frequency response in Multisim https://www.youtube.com/watch?v=qoCS1WIE7Ok © McGraw-Hill 54  Quiz (CBL) 1. For a CE amplifier, the emitter bypass capacitor affects the a. low-frequency response b. high-frequency response c. both of the above d. none of the above 2. For a CS amplifier, the gate-drain capacitance affects the a. low-frequency response b. high-frequency response c. both of the above d. none of the above 3. For an inverting amplifier, the Miller effect causes the equivalent capacitance to ground to appear a. smaller for both Cin and Cout b. smaller for Cin and larger for Cout c. larger for Cin and smaller for Cout d. larger for both Cin and Cout © McGraw-Hill 55  Quiz (CBL) 4. For the CE amplifier shown, the output low- frequency response is determined by a. (RC||RL) C3 b. (RC||RL) + C3 c. (RC+RL) C3 d. (RC+RL) + C3 5. For the CE amplifier shown, the resistor that is not part of the RC charge and discharge path for (C2 ) is a. R1 b. R2 c. RC d. RE © McGraw-Hill 56  Quiz (CBL) 6. The decibel is a ratio of two powers; for this reason the measurement unit is a. the volt b. the watt c. the volt-amp d. dimensionless 7. At the cutoff frequency for an amplifier, the power output compared to the midband power output is a. −2 dB b. −3 dB c. +2 dB d. +3 dB 8. The effect of an unbypassed emitter resistor on the upper cutoff frequency in a CE amplifier is a. to increase f cu b. to decrease f cu c. no effect © McGraw-Hill 57  Quiz (CBL) 9. The y-axis of a Bode Plot is used for the a. frequency scale b. power scale c. voltage scale d. decibel scale 10. The term bandwidth refers to those frequencies a. between the lower and upper critical frequencies b. above the upper critical frequency c. below the lower critical frequency d. none of the above Answers: 1. a 2. b 3. d 4. c 5. c 6. d 7. b 8. b 9. d 10. a © McGraw-Hill 58  Homework (CBL) Solve the following problems from the textbook. Chapter 10, Pages 559-562, Problems: 5, 6, 7, 9, 10, 11, 14, 19, 23, 24, 26, 29, 31 © McGraw-Hill 59  Key Formulas © McGraw-Hill 60  Because learning changes everything. ® www.mheducation.com © 2021 McGraw Hill. All rights reserved. Authorized only for instructor use in the classroom. © McGraw-Hill No reproduction or further distribution permitted without the prior written consent of McGraw-Hill.  Backup Slides Next, extra slides for students reference © McGraw-Hill 62  Low-Frequency Response © McGraw-Hill 63  Total Low-Frequency Response © McGraw-Hill 64  Effect of Bypass Capacitors © McGraw-Hill 65  Low-Frequency Response © McGraw-Hill 66  Low-Frequency Response © McGraw-Hill 67  Low-Frequency Response © McGraw-Hill 68  Low-Frequency Response © McGraw-Hill 69  References 1. https://www.electronics-tutorials.ws/filter/decibels.html 2. https://www.elprocus.com/common-emitter-amplifier-circuit-working/ 3. https://www.theengineeringprojects.com/2021/08/bjt-definition-symbol- working-characteristics-types-applications.html 4. https://components101.com/articles/understanding-bjt-transistor-and-how-to- use-it-in-your-circuit-designs © McGraw-Hill 70