Electronic Circuits - Part 2 ppt

Teacher: Dr LUU THE VINH & The chapter outline is shown below The chapter outline is shown below Basic Stages Basic Stages Emitter Follower Emitter Follower Push−Pull Stage Push−Pull Stage and Improved and Improved Variants Variants Large-Signal Large-Signal Considerations Considerations Omissionof Omission of PNP Transistor PNP Transistor High-Fidelity High-Fidelity Design Design Heat Heat Dissipation Dissipation Power Ratings Power Ratings ThermalRunaway Thermal Runaway PA - Power Amplifier Efficiency and Efficiency and PA Classes PA Classes Efficiency of PAs Efficiency of PAs Classes of PAs Classes of PAs 13.1 General Considerations •• The reader may wonder why the amplifier stages studied in The reader may wonder why the amplifier stages studied in previous chapters are not suited to high-power applications previous chapters are not suited to high-power applications Suppose we wish to deliver W to an 8Ω- speaker Suppose we wish to deliver W to an 8Ω- speaker Approximating the signal with a sinusoid of peak amplitude Approximating the signal with a sinusoid of peak amplitude VPP,we express the power absorbed by the speaker as: V , we express the power absorbed by the speaker as: (13.1) Where VPP = p2 denotes the root mean square (rms) value Where V = p2 denotes the root mean square (rms) value of the sinusoid and RLL represents the speaker impedance of the sinusoid and R represents the speaker impedance For RLL= 8Ω and Pout = W, For R = 8Ω and Pout = W, VP = 4V Also, the peak current flowing through the speaker is given by Also, the peak current flowing through the speaker is given by Important observations 1) The resistance that must be driven by the amplifier is 1) The resistance that must be driven by the amplifier is much lower than the typical values (hundreds to much lower than the typical values (hundreds to thousands of ohms) seen in previous chapters thousands of ohms) seen in previous chapters 2) The current levels involved in this example are much 2) The current levels involved in this example are much greater than the typical currents (milliamperes) greater than the typical currents (milliamperes) encountered in previous circuits encountered in previous circuits 3) The voltage swings delivered by the amplifier can hardly 3) The voltage swings delivered by the amplifier can hardly be viewed as “small” signals, requiring a good be viewed as “small” signals, requiring a good understanding of the large-signal behavior of the circuit understanding of the large-signal behavior of the circuit 4) The power drawn from the supply voltage, at least 1W, is 4) The power drawn from the supply voltage, at least 1W, is much higher than our typical values much higher than our typical values 5) A transistor carrying such high currents and sustaining 5) A transistor carrying such high currents and sustaining several volts (e.g., between collector and emitter) several volts (e.g., between collector and emitter) dissipates a high power and, as a result, heats up Highdissipates a high power and, as a result, heats up Highpower transistors must therefore handle high currents power transistors must therefore handle high currents and high temperature and high temperature 12.1 General Considerations •• Based on the above observations, we can predict the parameters of Based on the above observations, we can predict the parameters of interest in the design of power stages: interest in the design of power stages: (1) “Distortion,” i.e., the nonlinearity resulting from large-signal operation A (1) “Distortion,” i.e., the nonlinearity resulting from large-signal operation A high-quality audio amplifier must achieve a very low distortion so as to high-quality audio amplifier must achieve a very low distortion so as to reproduce music with high fidelity In previous chapters, we rarely dealt reproduce music with high fidelity In previous chapters, we rarely dealt with distortion with distortion (2) “Power efficiency” or simply “efficiency,” denoted by η and defined as: (2) “Power efficiency” or simply “efficiency,” denoted by η and defined as: For example, a cellphone power amplifier that consumes W from the For example, a cellphone power amplifier that consumes W from the battery to deliver W to the antenna provides η≈33,3% In previous battery to deliver W to the antenna provides η≈33,3% In previous chapters, the efficiency of circuits was of little concern because the chapters, the efficiency of circuits was of little concern because the absolute value of the power consumption was quite small (a few absolute value of the power consumption was quite small (a few milliwatts) milliwatts) (3) “Voltage rating.” As suggested by Eq (13.1), higher power levels or (3) “Voltage rating.” As suggested by Eq (13.1), higher power levels or load resistance values translate to large voltage swings and (possibly) load resistance values translate to large voltage swings and (possibly) high supply voltages Also, the transistors in the output stage must high supply voltages Also, the transistors in the output stage must exhibit breakdown voltages well above the output voltage swings exhibit breakdown voltages well above the output voltage swings 13.2 Emitter Follower as Power Amplifier With its relatively low output impedance, the emitter follower may be With its relatively low output impedance, the emitter follower may be considered a good candidate for driving “heavy” loads, i.e., low considered a good candidate for driving “heavy” loads, i.e., low impedances As shown in Chapter 5, the small-signal gain of the impedances As shown in Chapter 5, the small-signal gain of the follower is given by follower is given by We may therefore surmise that for, say, RLL=8Ω, a We may therefore surmise that for, say, R =8Ω, a gain near unity can be obtained if 1/gm1 32.5 mA We assume β>>1 But, let us analyze the circuit’s behavior in delivering large voltage But, let us analyze the circuit’s behavior in delivering large voltage swings (e.g VPP )) to heavy loads To this end, consider the follower swings (e.g V to heavy loads To this end, consider the follower shown in Fig 13.1(a), where I1 serves as the bias current source To shown in Fig 13.1(a), where I1 serves as the bias current source To simplify the analysis, we assume the circuit operates from negative simplify the analysis, we assume the circuit operates from negative and positive power supplies, allowing Vout to be centered around zero and positive power supplies, allowing Vout to be centered around zero For Vinin≈ 0,8 V, we have Vout ≈ and ICC≈32,5 mA If Vininrises from 0.8 For V ≈ 0,8 V, we have Vout ≈ and I ≈32,5 mA If V rises from 0.8 V to 4.8 V, the emitter voltage follows the base voltage with a relatively V to 4.8 V, the emitter voltage follows the base voltage with a relatively constant difference of 0.8 V, producing a 4-V swing at the output [Fig constant difference of 0.8 V, producing a 4-V swing at the output [Fig 13.1(b)] 13.1(b)] •• Now suppose Vininbegins from +0,8 V and gradually goes down [Fig Now suppose V begins from +0,8 V and gradually goes down [Fig 13.1(c)] We expect 13.1(c)] We expect Vout to go below zero and hence part of I1 to flow from RLL.For Vout to go below zero and hence part of I1 to flow from R For example, if Vinin≈ 0,7 V, tth e n Vout ≈ 0,1 V, and RLLcarries a current of example, if V ≈ 0,7 V, h e n Vout ≈ 0,1 V, and R carries a current of 12.5 mA That is, IC1 ≈ IE1 =20 mA Similarly, ififVin ≈ 0,6 V, tth e n Vout 12.5 mA That is, IC1 ≈ IE1 =20 mA Similarly, Vin ≈ 0,6 V, h e n Vout ≈ 0,2 V, IRL ≈ 25 mA, and hence IC1 ≈ 7,5 mA In other words, the ≈ 0,2 V, IRL ≈ 25 mA, and hence IC1 ≈ 7,5 mA In other words, the collector current of continues to fall collector current of continues to fall 13.7 Heat Dissipation Since the output transistors in a power amplifier carry a finite current Since the output transistors in a power amplifier carry a finite current and sustain a finite voltage and sustain a finite voltage for part of the period, they consume power and hence heat up If the for part of the period, they consume power and hence heat up If the junction temperature rises junction temperature rises excessively, the transistor may be irreversibly damaged Thus, the excessively, the transistor may be irreversibly damaged Thus, the “power rating” (the maximum “power rating” (the maximum allowable power dissipation) of each transistor must be chosen allowable power dissipation) of each transistor must be chosen properly in the design process properly in the design process 13.7.1 Emitter Follower Power Rating •• Let us first compute the power dissipated by Q1 in the simple emitter Let us first compute the power dissipated by Q1 in the simple emitter follower of Fig 13.25, assuming that the circuit delivers a sinusoid of follower of Fig 13.25, assuming that the circuit delivers a sinusoid of VPPsinωtto a load resistance RLL V sinωt to a load resistance R 13.7.1 Emitter Follower Power Rating •• Recall from Section 13.2 that I11 ≥ VPP/RL to ensure can reach The Recall from Section 13.2 that I ≥ V /RL to ensure can reach The instantaneous power dissipated by Q11is given by ICC.VCEand its instantaneous power dissipated by Q is given by I VCE and its average value (over one period) equals: average value (over one period) equals: Where T = 2π/ω Since IC ≈IE = I1 + Vout/RL and VCE = VCC – Vout = VCC -Vpsinωt, We have: • At the other extreme,If VP ≈VCC,, then Example 13 Example Calculate the power dissipated by the current source I1 in Fig 13.25 Solution Solution The current source sustains a voltage equal to Vout- VEE = VPsinωt -VEE Thus, The value is, of course, positive because VEE < to accommodate negative swings at the output 13.7.2 Push-Pull Stage Power Rating We now determine the power dissipated by the output transistors in the push-pull stage (Fig.13.26) To simplify our calculations, we assume that each transistor carries a negligible current around and turns off for half of the period If Vout = VPPsinωt, then IIRL=(VP/RL)sinωtbut only for half of the If Vout = V sinωt, then RL=(VP/RL)sinωt but only for half of the cycle Also, the collector-emitter voltage of Q1 is given by cycle Also, the collector-emitter voltage of Q1 is given by VCC Vout = VCC VPPsinωt.The average power dissipated in Q1 VCC Vout = VCC V sinωt The average power dissipated in Q1 is therefore equal to is therefore equal to (13.42) (13.43 ) Where T =1/ω, and β is assumed large enough to allow the approximation IC ≈ IE Expanding the terms inside the integral and noting that (13.43 ) we have (13.47) •• For example, if VPP=4V, RLL=8Ω, andVCC =6V, then Q1 For example, if V =4V, R =8Ω, andVCC =6V, then Q1 dissipates 455 mW Transistor also consumes this amount dissipates 455 mW Transistor also consumes this amount of power if /VBE/ = VCC of power if /VBE/ = VCC •• Equation (13.47) indicates that for VPP≈ or VPP=4VCC/π,the power Equation (13.47) indicates that for V ≈ or V =4VCC/π, the power dissipated in Q1 approaches zero, suggesting that Pav must reach a dissipated in Q1 approaches zero, suggesting that Pav must reach a maximum between these extremes Differentiating Pav with respect to maximum between these extremes Differentiating Pav with respect to VPPand equating the result to zero, we have VPP= 2VCC/π and hence V and equating the result to zero, we have V = 2VCC/π and hence 13.7 Heat Dissipation •• Called a “heat sink” and shown in Fig 13.27, one such Called a “heat sink” and shown in Fig 13.27, one such means is formed as a metal structure (typically aluminum) means is formed as a metal structure (typically aluminum) with a large surface area and attached to the transistor or with a large surface area and attached to the transistor or chip package The idea is to “sink” the heat from the chip package The idea is to “sink” the heat from the package and subsequently dissipate it through a much package and subsequently dissipate it through a much larger surface larger surface Figure 13.27 Example of heat sink Figure 13.27 Example of heat sink 13.7.3 Thermal Runaway 13.7.3 Thermal Runaway •Interestingly, the use of diode biasing [Fig 13.28(b)] can prohibit thermal runaway If the diodes experience the same temperature change as the output transistors, thenV D1 + VD2 decreases as the temperature rises (because their bias current is relatively constant), thereby stabilizing the collector currents of Q1 and Q2 From another perspective, since D1 and Q1 form a current mirror,IC1 is a constant multiple of I1 if D1 and Q1 remain at the same temperature More accurately, we have for D1 and D2: (13.49) (13.50) • Similarly, for Q1 and Q2 : (13.51) (13.52) Equating (13.50) and (13.52) and assuming the same value of V T (i.e., the same temperature) for both expressions, we write Since ID1 ≈ ID2 ≈ I1, IC1 ≈ IC2, we observe that IC1 and IC2 “track” I1 so long as the IS values (which are temperature-dependent) also track 13.8 Efficiency ... //r? ?2 and a gm equal to gm1 + gm2 [Fig 13.17(c)] For this circuit, we have vπ1= v? ?2 = vN- vout and: [Fig 13.17(c)] For this circuit, we have vπ1= v? ?2 = vN- vout and: (13.18) Now,Q1 and Q2 operate... and 2rDD simplified circuit shown in Fig.13.17(a), where V = ∞ and 2r represents the total small-signal resistance of D1 and D2 Let represents the total small-signal resistance of D1 and D2 Let... carries a current of 12. 5 mA That is, IC1 ≈ IE1 =20 mA Similarly, ififVin ≈ 0,6 V, tth e n Vout 12. 5 mA That is, IC1 ≈ IE1 =20 mA Similarly, Vin ≈ 0,6 V, h e n Vout ≈ 0 ,2 V, IRL ≈ 25 mA, and hence