IEL - Unit 1 PDF

Unit 1 Power Supplies, Transistors, Amplifiers Syllabus Power Supplies –Block diagram, Half-wave rectifier, Full-wave rectifiers and filters, Voltage regulators, Output resistance and voltage regulation, Voltage multipliers. Transistor: BJT structure and operation (npn), circuit symbol, configurat...

Unit 1 Power Supplies, Transistors, Amplifiers Syllabus Power Supplies –Block diagram, Half-wave rectifier, Full-wave rectifiers and filters, Voltage regulators, Output resistance and voltage regulation, Voltage multipliers. Transistor: BJT structure and operation (npn), circuit symbol, configurations, relation between transistor currents. Amplifiers – Definition, Types of amplifier, gain, Input-Output Resistance, Multi-stage amplifier; BJT as a switch: Cut-off and saturation modes. Text Book Mike Tooley, ‘Electronic Circuits, Fundamentals & Applications’, 4th Edition, Elsevier, 2015. DOI https://doi.org/10.4324/9781315737980. eBook ISBN9781315737980. Basic Electronics- Devices, circuits and IT fundamentals- By Santiram Kal- PHI, 2012 Pre requisites Semiconductor basics What is semiconductor? Why do we need semiconductor? Types of semiconductor- Intrinsic and Extrinsic Extrinsic- Doping, p type, n type- trivalent and pentavalent Majority and minority charge carriers Semiconductor diode Two-terminal unilateral device which allows the flow of current in only one direction Anode and cathode are the two terminals Diode offers low resistance hence permits current flow from Anode to Cathode It offers high resistance or restricts the flow of current from Cathode to Anode It can be biased (applying voltage across terminals of diode) in two ways: Forward bias and Reverse bias Diode is a pn junction which permits current flow when forward biased and blocks current when reverse biased Unbiased Diode No voltage applied across the junction Majority holes on p side start diffusing into n side Majority free electrons on n side start diffusing into p side Positive immobile ions are formed on n side and negative ions on p side near the junction, this is called depletion region In equilibrium condition, depletion region widens up to a point where no further electrons or holes can cross the junction. This acts as a barrier. Potential difference across depletion region is called barrier potential/junction potential/built-in voltage/cut-in potential Net current in unbiased diode is Zero. Forward biased diode P region is connected to positive and n region is connected to negative of dc supply Negative of the battery pushes free electrons across the depletion region, provided the applied voltage exceeds barrier voltage. Similarly, negative of the battery pushes holes against barrier from p to n region Barrier voltage for Si diode is 0.7V and Ge diode is 0.3V. Due to this width of the depletion region reduces and barrier potential also reduces Majority carriers cross the junction Hence current starts flowing from p to n side (Anode to cathode terminal)-Forward current Reverse biased diode P region is connected to negative and n region to positive of the dc voltage Negative of the battery attracts holes in p region and positive of the battery attracts electrons in n region Majority charge carriers move away from the junction Depletion region widens and barrier potential increases Resistance of diode is high Due to increased barrier potential, free electrons on p side are attracted towards positive while holes towards negative of the battery There is a very small reverse current due to the flow of minority carriers Reverse current is constant though reverse voltage is increased upto a limit. It is called reverse saturation current. Minority charge carriers are thermally generated hence this current is temperature dependant Reverse saturation current is in the order of micro amperes for Ge and few nano amperes for Si diodes Current voltage or I-V characteristics of diode First quadrant indicates the behaviour of diode when forward biased Current is nearly zero when forward voltage is less than knee or barrier voltage As forward voltage exceeds barrier voltage, current increases exponentially Third quadrant indicates the characteristics of reverse biased diode As the reverse voltage is increased, reverse current increases initially but after a small voltage becomes constant equal to reverse saturation current. Though reverse voltage is increased the reverse current remains constant. At reverse breakdown voltage, breakdown of diode occurs and current increases sharply damaging the diode. Diode approximations Ideal diode model Constant voltage drop model  DC power supply Power supply is a device that supplies electric power to a load. A step-down transformer of appropriate turns ratio is used to convert high voltage from the mains to a low voltage (9V, 12V, 15V, 20V, 30V). This is achieved by varying the turns ratio on the transformer. The a.c. output from the transformer secondary is then rectified using conventional silicon rectifier diodes to produce an unsmoothed (sometimes referred to as pulsating d.c.) output. The output is smoothed and filtered before being applied to a circuit which will regulate (or stabilize) the output voltage so that it remains relatively constant in spite of variations in both load current and incoming mains voltage. DC power supply The iron-cored step-down transformer feeds a rectifier arrangement (often based on a bridge circuit). The output of the rectifier is then applied to a high-value reservoir capacitor. The capacitor helps to smooth out the voltage pulses produced by the rectifier. A stabilizing circuit (often based on a series transistor regulator and a Zener diode voltage reference) provides a constant output voltage. A SIMPLE DC SUPPLY 13  RECTIFIERS A rectifier is a device that converts alternating current (ac) to direct current (dc). Rectifiers Semiconductor diodes are commonly used to convert alternating current (a.c.) to direct current (d.c), in which case they are referred to as rectifiers. Types- Half-wave rectifier, Full-wave rectifier, Bridge rectifier Half-wave rectifier uses single diode and operates on only either positive or negative half-cycles of the supply Full-wave rectifier uses two diodes with centre tap transformer and operates in both positive and negative half cycles Bridge rectifier uses four diodes and operates in both positive and negative half cycles Vrms – Root mean square voltage Vm (= Vpeak) – peak voltage Vavg (= Vdc) – Average voltage Half-wave rectifier Mains voltage (220 to 240 V) is applied to the primary of a step-down transformer (T1). The secondary of T1 steps down the 240 V r.m.s. to 12 V r.m.s. (the turns ratio of T1 will thus be 240/12 or 20:1). D1 will be forward biased during each positive half-cycle (relative to common) and will effectively behave like a closed switch. D1 will be reverse biased during each negative half-cycle and will effectively behave like a open switch. Half-wave rectifier- Working The switching action of D1 results in a pulsating output voltage which is developed across the load resistor (RL). Mains supply and output developed across RL both have same frequency 50 Hz. During the positive half-cycle, the diode will drop the 0.6 V to 0.7 V forward threshold voltage normally associated with silicon diodes. However, during the negative half-cycle the peak a.c. voltage will be dropped across D1 when it is reverse biased. This is an important consideration when selecting a diode for a particular application. Half-wave rectifier- Working Assuming that the secondary of T1 provides 12 V r.m.s., the peak voltage output from the transformer’s secondary winding will be given by: The peak voltage applied to D1 will thus be approximately 17 V. The negative half-cycles are blocked by D1 and thus only the positive half-cycle appear across RL. Actual peak voltage across RL will be the 17 V positive peak being supplied from the secondary on T1, minus the 0.7 V forward threshold voltage dropped by D1. Positive half-cycle pulses having a peak amplitude of 16.3 V will appear across RL. Half-wave rectifier- Waveforms Problem 1 A mains transformer having a turns ratio of 44:1 is connected to a 220 V r.m.s. mains supply. If the secondary output is applied to a half-wave rectifier, determine the peak voltage that will appear across a load. Ans: Reservoir and smoothing circuits Improvement in Half-wave rectifier circuit is possible by adding the capacitor, C1, to ensure that the output voltage remains at, or near, the peak voltage even when the diode is not conducting. When the primary voltage is first applied to T1, the first positive half-cycle output from the secondary will charge C1 to the peak value seen across RL. Hence C1 charges to 16.3 V at the peak of the positive half-cycle. Because C1 and RL are in parallel, the voltage across RL will be the same as that across C1. The time required for C1 to charge to the maximum (peak) level is determined by the charging circuit time constant (the series resistance multiplied by the capacitance value). Half-wave rectifier with capacitor filter The series resistance comprises the secondary winding resistance together with the forward resistance of the diode and the (minimal) resistance of the wiring and connections. Hence C1 charges very rapidly as soon as D1 starts to conduct. The time required for C1 to discharge is, in contrast, very much greater. The discharge time constant is determined by the capacitance value and the load resistance, RL. In practice, RL is very much larger than the resistance of the secondary circuit and hence C1 takes an appreciable time to discharge. During this time, D1 will be reverse biased and will thus be held in its non-conducting state. As a consequence, the only discharge path for C1 is through RL.  Half-wave rectifier with capacitor filter C1 is referred to as a reservoir capacitor. It stores charge during the positive half-cycles of secondary voltage and releases it during the negative half-cycles. C1 will discharge by a small amount during the negative half-cycle periods from the transformer secondary. Small variation in dc output voltage is ripple Since ripple is undesirable we must take additional precautions to reduce it. One obvious method of reducing the amplitude of the ripple is that of simply increasing the discharge time constant. Discharge time constant can be increased by increasing the value of C1 or by increasing the resistance value of RL. Usually RL cant be changed. Increasing the value of C1 is a more practical alternative and very large capacitor values (often in excess of 4,700 μF) are typical. Refinement to the circuit to reduce ripple (use of R-C smoothing filter) This circuit employs two additional components, R1 and C1, which act as a filter to remove the ripple. The value of C2 is chosen so that the component exhibits a negligible reactance at the ripple frequency (50 Hz for a half-wave rectifier or 100 Hz for a full-wave rectifier) The amount of ripple is reduced by an approximate factor equal to: Problem 2 The R–C smoothing filter in a 50 Hz mains operated half-wave rectifier circuit consists of R1 = 100 Ω and C 2 = 1,000 μF. If 1 V of ripple appears at the input of the circuit, determine the amount of ripple appearing at the output. Ans: C2 Full-wave rectifiers A better rectifier arrangement would make use of both positive and negative half-cycles. Improvement over half-wave rectifiers They are not only more efficient but are significantly less demanding in terms of the reservoir and smoothing components. Two types: phase type and the bridge rectifier type. Bi-phase rectifier circuits Mains voltage (240 V) is applied to the primary of the step-down transformer (T1) which has two identical secondary windings, each providing 12 V r.m.s. (the turns ratio of T1 will thus be 240/12 or 20:1 for each secondary winding). On positive half-cycles, point A will be positive with respect to point B. Similarly, point B will be positive with respect to point C. In this condition D1 will allow conduction while D2 will not allow conduction. On negative half-cycles, point C will be positive with respect to point B. Similarly, point B will be positive with respect to point A. In this condition D2 will allow conduction while D1 will not allow conduction. Equivalent circuits during positive and negative half-cycle Bi-phase rectifier circuits with capacitor filter The current is routed through the load in the same direction on successive half-cycles. Pulsating output voltage being developed across the load resistor (RL). Frequency of the output is 100 Hz. This doubling of the ripple frequency allows us to use smaller values of reservoir and smoothing capacitor to obtain the same degree of ripple reduction. Peak voltage produced by each of the secondary windings will be approximately 17 V and the peak voltage across RL will be 16.3 V If C1 is added at the output, it charges to approximately 16.3 V at the peak of the positive half-cycle and holds the voltage at this level when the diodes are in their non-conducting states. OPERATION OF BI-PHASE RECTIFIER D1 FWD Biased Vin T1 A + + VL R - L B - t + - C D2 REV Biased  OPERATION OF BI-PHASE RECTIFIER D1 REV Biased vin T1 A - + vout R L B + - t - C + D2 FWD Biased Bi-phase rectifier circuits- waveforms The time required for C1 to charge to the maximum (peak) level is determined by series resistance which comprises of secondary winding resistance together with the forward resistance of the diode and the (minimal) resistance of the wiring and connections. Hence C1 charges very rapidly as soon as either D1 or D2 starts to conduct. The time required for C1 to discharge is, in contrast, very much greater. The discharge time contrast is determined by the capacitance value and the load resistance, RL which is large. C1 takes an appreciable time to discharge. During this time, D1 and D2 will be reverse biased and held in a non-conducting state, thus only discharge path for C1 is through RL. Bridge rectifier circuits This arrangement avoids the need to have two separate secondary windings. It uses 4 diodes. Mains voltage (240 V) is applied to the primary of a step-down transformer (T1). The secondary winding provides 12 V r.m.s. (approximately 17 V peak) and has a turns ratio of 20:1 On positive half-cycles, point A will be positive with respect to point B. In this condition D1 and D2 will allow conduction while D3 and D4 will not allow conduction. On negative half-cycles, point B will be positive with respect to point A. In this condition D3 and D4 will allow conduction while D1 and D2 will not allow conduction.  Operation of Bridge Rectifier vin T A+ D4 + 1 RE D1 FWD V t D2 + vout B FWD - D3 - REV RL t -  Operation of Bridge Rectifier A- vin T1 FW D4 D D1 REV t + vout D2 REV D3 B+ FW RL D - t Equivalent circuits during positive and negative half cycles Once again, the result is that current is routed through the load in the same direction on successive half-cycles. Once again, the peak output voltage is approximately 16.3 V (i.e. 17 V less the 0.7 V forward threshold voltage). Bridge rectifier circuits with reservoir capacitor Reservoir capacitor (C1) can be added to maintain the output voltage when the diodes are not conducting. C1 charges to approximately 16.3 V at the peak of the positive half-cycle and holds the voltage at this level when the diodes are in their non-conducting states. R–C and L–C ripple filters can be added to bi-phase and bridge rectifier circuits in exactly the same way as those shown for the half-wave rectifier arrangement  What is a Voltage Regulator? A voltage regulator provides a constant DC output voltage that is independent of AC line voltage variations, load current and temperature. The input to a voltage regulator comes from the filtered output of a rectifier derived from an AC voltage. Voltage regulators Regulator circuit using Zener diode Rs is included to limit the zener current to a safe value when the load is disconnected When a load (RL) is connected, the zener current (IZ) will fall as current is diverted into the load resistance (it is usual to allow a minimum current of 2 mA to 5 mA in order to ensure that the diode regulates). The output voltage (VZ) will remain at the zener voltage until regulation fails at the point at which the potential divider formed by RS and RL, produces a lower output voltage that is less than VZ. Equations where VIN is the unregulated input voltage. Problem 3 A 5 V zener diode has a maximum rated power dissipation of 500 mW. If the diode is to be used in a simple regulator circuit to supply a regulated 5 V to a load having a resistance of 400 Ω, determine a suitable value of series resistor for operation in conjunction with a supply of 9 V. Output resistance and voltage regulation In a perfect power supply, the output voltage would remain constant regardless of the current taken by the load, but in practice output voltage falls as the load current increases. Power supply has internal resistance (ideally this should be zero). This internal resistance appears at the output of the supply. The regulation of a power supply is given by the relationship: Ideally, the value of regulation should be very small. Simple shunt zener diode regulators are capable of producing values of regulation of 5% to 10%. More sophisticated circuits based on discrete components produce values of between 1% and 5% and integrated circuit regulators often provide values of 1% or less. Problem 4 The following data were obtained during a test carried out on a d.c. power supply: (i) Load test: Output voltage (no-load) = 12 V, Output voltage (2 A load current) = 11.5 V (ii) Regulation test: Output voltage (mains input, 220 V) = 12 V, Output voltage (mains input, 200 V) = 11.9 V Determine (a) the equivalent output resistance of the power supply and (b) the regulation of the power supply. Ans: Voltage multipliers: Voltage doubler Increasing the output of simple half-wave rectifier C1 will charge to the positive peak secondary voltage while C2 will charge to the negative peak secondary voltage. Since the output is taken from C1 and C2 connected in series the resulting output voltage is twice that produced by one diode alone.  OPERATION OF VOLTAGE DOUBLER D FWD 1 Biased vi T n + 1 + V +C p -1 t - C 2 - D REV 2 Biased  Operation of Voltage Doubler D REV 1 Biased vi T n + 1 + V +C p -1 t - V +C p -2 - D FWD 2 Biased  OPERATION OF VOLTAGE DOUBLER D REV 1 Biased vi T n + 1 + vou 2 t V +C V p -1 p t - t V +C p -2 - D FWD 2 Biased Voltage tripler Voltage tripler ∙ During the first positive half cycle of AC, Diode D1 get forward biased and capacitor C1 get charged through the D1. Capacitor C1 get charged up to the peak voltage of AC i.e. Vpeak. ∙ During the negative half cycle of the AC, Diode D2 conducts and D1 reverse biased. D1 blocks the discharging of capacitor C1. Now the capacitor C2 charge with the combined voltage of capacitor C1 (Vpeak) and the negative peak of the AC voltage (Vpeak). So the capacitor C2 charge up to 2Vpeak. ∙ During the second positive half cycle, D2 gets reverse biased and D3 conducts. So capacitor C2 charges the capacitor C3 up to the same voltage as itself, which is 2Vpeak. Meanwhile, Diode D1 conducts to charge Capacitor C1 up to Vpeak. ∙ Now the capacitor C1 and C3 are in series and voltage across C1 is Vpeak and voltage across C3 is 2Vpeak, so the voltage across the series connection of C1 and C3 is Vpeak+2Vpeak = 3Vpeak, that’s how we get the triple voltage of the peak value of AC. Transistors BJT – BIPOLAR JUNCTION TRANSISTORS  flow Since the sandwiched p-type material is very thin and has a low conductivity, a very small number of these carriers will take this path of high resistance to the base terminal. From eq.(2), neglecting ICBO , Ic = αIE From eq (1) , Ic = α(Ic+Ib) Ic = (α/1-α)IB Ic =βIB Where, β= (α/1-α) (where β is called as common emitter current gain and typically it ranges b/w 25 to 300) In a common emitter transistor circuit, if β = 100 and IB = 50μA, compute the values of α, IE and IC. Calculate α and β if IC is measured as 1mA and base current is 25µA. Also determine the new base current to give IC of 5mA. A emitter current of transistor in Common base configuration is IE=25mA and IC=23mA. Calculate base current, common base dc current gain and common emitter dc current gain. The following current measurements are made on transistor: IC= 12.42mA, IB= 200µA. Determine a new IC level when IB is 150 µA.  Common-base configuration The input signal is applied between the transistors base and the emitter terminals, while the corresponding output signal is taken from between the base and the collector terminals as shown.  Common-emitter configuration The input signal is applied between the base and the emitter, while the output is taken from between the collector and the emitter as shown.  Common-collector configuration The input signal is applied between the base and the collector, while the output is taken from between the emitter and the collector as shown. BJT – Regions of operation Transistor as a switch When the base emitter junction is open or reverse biased (i.e., IB=0) no collector current(IC) flows, the transistor is said to be OFF. The input and Base are grounded ( 0V ) Base-Emitter voltage VBE < 0.7V Base-Emitter junction is reverse biased Base-Collector junction is reverse biased Transistor is “fully-OFF” ( Cut-off region ) No Collector current flows ( IC = 0 ) VOUT = VCE = VCC = ”1″ Transistor operates as an “open switch” Transistor as a switch When the base emitter junction is forward biased, IB flows which results in IC. At some point of IB , IC becomes saturated, i.e., it does not increase further and becomes independent of IB. At this point, the transistor may be treated as fully ON as it is conducting in saturation region. The input and Base are connected to VCC Base-Emitter voltage VBE > 0.7v Base-Emitter junction is forward biased Base-Collector junction is forward biased Transistor is “fully-ON” ( saturation region ) Max Collector current flows ( IC = Vcc/RL ) VCE = 0 ( ideal saturation ) VOUT = VCE = ”0″ Transistor operates as a “closed switch” Amplifiers Amplifiers are electronic circuits that increase the strength of a signal (voltage/current/power) It uses electric power from a power supply to increase the amplitude of a signal applied to its input terminals, producing a proportionally greater amplitude signal at its output. The amount of amplification provided by an amplifier is measured by its gain: the ratio of output voltage, current, or power to input. Amplification is fundamental to modern electronics, and amplifiers are widely used in almost all electronic equipment. Types of amplifier AC coupled amplifiers: stages are coupled together in such a way that DC levels are isolated and only the AC components of a signal are transferred from stage to stage. DC coupled amplifiers: stages are coupled together in such a way that stages are not isolated to DC potentials. Both AC and DC signal components are transferred from stage to stage. Large-signal amplifiers: Large-signal amplifiers are designed to cater for appreciable voltage and/or current levels (typically from 1 V to 100 V or more). Example: Power amplifiers (audio) Small-signal amplifiers: Small-signal amplifiers are designed to cater for low-level signals (normally less than 1 V and often much smaller). Small-signal amplifiers have to be specially designed to combat the effects of noise. Example: instrumentation amplifiers Audio frequency amplifiers: operate in the band of frequencies that is normally associated with audio signals (e.g. 20 Hz to 20 kHz). Types of amplifier Wideband amplifiers: capable of amplifying a very wide range of frequencies, typically from a few tens of hertz to several megahertz. Wideband amplifiers are usually untuned; that is, their ac load is resistive. Radio frequency amplifiers: operate in the band of frequencies that is normally associated with radio signals (e.g. from 100 kHz to over 1 GHz). They are frequency selective. They are restricted to narrow band of frequencies. Narrowband amplifiers are usually tuned RF amplifiers, which means that their ac load is a high-Q resonant tank tuned to a radio station or television channel. Low-noise amplifiers: Low-noise amplifiers are designed so that they contribute negligible noise (signal disturbance) to the signal being amplified. These amplifiers are usually designed for use with very small signal levels (usually less than 10 mV or so). Amplifier characteristics - Gain It indicates amount of amplification Gain is simply the ratio of output voltage to input voltage, output current to input current, or output power to input power Amplifier characteristics - Input and output resistance Input resistance is the ratio of input voltage to input current and it is expressed in ohms. It is resistive in the mid band frequency band. In other cases it is complex quantity, then it is referred as input impedance considering the effect of capacitance in parallel with it. Output resistance is the ratio of open-circuit output voltage to short-circuit output current and is measured in ohms. In the presence of reactive component it is referred to as output impedance. Input and output resistance are internal to the amplifier Problem 5 An amplifier produces an output voltage of 2 V for an input of 50 mV. If the input and output currents in this condition are, respectively, 4 mA and 200 mA, determine: (a) the voltage gain; (b) the current gain; (c) the power gain.  Assignment 1. An amplifier produces an output voltage of 5 V for an input of 20 mV. If the input and output currents of the amplifier are 5mA and 200mA respectively. Determine voltage gain and power gain in dB. 2. The following measurements were made during a test on an amplifier: Vin = 250 mV, Iin = 2.5 mA, Vout = 10 V, Iout = 400 mA Determine: (a) the voltage gain; (b) the current gain; (c) the power gain; (d) the input resistance. 3. An amplifier has a power gain of 13.79dB and identical input and output resistances of 600 Ω. Determine the input voltage required to produce an output of 10 V. Multi-stage amplifiers In order to provide sufficiently large values of gain, it is frequently necessary to use a number of interconnected stages within an amplifier. The overall gain of an amplifier with several stages (i.e. a multi-stage amplifier) is simply the product of the individual voltage gains. Bandwidth of a multistage amplifier will be less than the bandwidth of each individual stage. An increase in gain can only be achieved at the expense of a reduction in bandwidth. R–C coupling The stages are coupled together using capacitors having a low reactance at the signal frequency and resistors. Can be used in audio applications. As it uses cheaper coupling devices such as resistors, capacitors, it is low-cost and economical. The main disadvantage is it has very poor impedance matching characteristics. It provides low voltage and power gain. L–C coupling Inductors have a high reactance at the signal frequency. This type of coupling is generally only used in RF and high-frequency amplifiers. (helps in tuning and impedance matching) Transformer coupling Transformers are used to couple the different stages of amplifier. Transformer coupling is used in high power audio amplifiers. It provides a very good impedance matching property. These amplifiers have high efficiency and low losses. These amplifiers are costly because of using the transformer as a coupling device. These amplifiers have poor frequency response, the gain decreases with an increase in frequency. Direct coupling Direct coupling: DC levels are preserved This circuit can amplify both the AC and DC signals. It does not use any coupling elements and hence the circuit is very simple and easy to make. The cost is very low. It has a very low bandwidth. The operating point is not stable.