Electronic Circuits - Part 2

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The reader may wonder why the ampliﬁer stages studied in previous chapters are not suited to high-power applications. Suppose we wish to deliver 1 W to an 8- speaker. Approximating the signal with a sinusoid of peak amplitude VP, we express the power absorbed by the speaker as:

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Nội dung Text: Electronic Circuits - Part 2

Teacher: Dr. LUU THE VINH
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The chapter outline is shown below. The chapter outline is shown below. Basic Stages Large-Signal Heat Efficiency and Efficiency and Basic Stages Large-Signal Heat PA Classes Considerations Dissipation Dissipation PA Classes Considerations Emitter Follower Emitter Follower Omission of  Power Ratings Efficiency of PAs Omission of Power Ratings Efficiency of PAs Push−Pull Stage Push−Pull Stage PNP Transistor Thermal Runaway  Classes of PAs PNP Transistor Thermal Runaway Classes of PAs High-Fidelity High-Fidelity and Improved Design Design and Improved Variants Variants PA - Power Amplifier
13.1 General Considerations •• The reader may wonder why the ampliﬁer stages studied The reader may wonder why the ampliﬁer stages studied in previous chapters are not suited to high-power in previous chapters are not suited to high-power applications. Suppose we wish to deliver 1 W to an 8Ω- applications. Suppose we wish to deliver 1 W to an 8Ω- speaker. Approximating the signal with a sinusoid of peak speaker. Approximating the signal with a sinusoid of peak amplitude VPP, we express the power absorbed by the amplitude V , we express the power absorbed by the speaker as: 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 = 1 W, For R = 8Ω and Pout = 1 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, 4) The power drawn from the supply voltage, at least 1W, is much higher than our typical values. is 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. High- dissipates a high power and, as a result, heats up. High- power 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 3 W from the For example, a cellphone power ampliﬁer that consumes 3 W from the battery to deliver 1 W to the antenna provides η≈ 33,3%. In previous battery to deliver 1 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), (3) “Voltage rating.” As suggested by Eq. (13.1), higher power levels or load resistance values higher power levels or load resistance values translate to large voltage swings and (possibly) high translate to large voltage swings and (possibly) high supply voltages. Also, the transistors in the output supply voltages. Also, the transistors in the output stage must exhibit breakdown voltages well above stage must exhibit breakdown voltages well above the output voltage swings. the output voltage swings.
13.2 Emitter Follower as Power Ampliﬁer With its relatively low output impedance, the With its relatively low output impedance, the emitter follower may be considered a good emitter follower may be considered a good candidate for driving “heavy” loads, i.e., low candidate for driving “heavy” loads, i.e., low impedances. As shown in Chapter 5, the small- impedances. As shown in Chapter 5, the small- signal gain of the follower is given by signal gain of the 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 But, let us analyze the circuit’s behavior in delivering large voltage swings (e.g. 4 VPP )) to heavy delivering large voltage swings (e.g. 4 V to heavy loads. To this end, consider the follower shown in loads. To this end, consider the follower shown in Fig. 13.1(a), where I1 serves as the bias current Fig. 13.1(a), where I1 serves as the bias current source. To simplify the analysis, we assume the source. To simplify the analysis, we assume the circuit operates from negative and positive power circuit operates from negative and positive power supplies, allowing Vout to be centered around zero. supplies, allowing Vout to be centered around zero. For Vin ≈ 0,8 V, we have Vout ≈ 0 and IIC ≈ 32,5 mA. For Vin ≈ 0,8 V, we have Vout ≈ 0 and C ≈ 32,5 mA. If Vin rises from 0.8 V to 4.8 V, the emitter voltage If Vin rises from 0.8 V to 4.8 V, the emitter voltage follows the base voltage with a relatively constant follows the base voltage with a relatively constant difference of 0.8 V, producing a 4-V swing at the difference of 0.8 V, producing a 4-V swing at the output [Fig. 13.1(b)]. output [Fig. 13.1(b)].
•• Now suppose Vin begins from +0,8 V and gradually Now suppose Vin begins from +0,8 V and gradually goes down [Fig. 13.1(c)]. We expect goes down [Fig. 13.1(c)]. We expect Vout to go below zero and hence part of I1 to ﬂow Vout to go below zero and hence part of I1 to ﬂow from RLL. For example, if Vin ≈ 0,7 V, tt h e n Vout ≈ from R . For example, if Vin ≈ 0,7 V, h e n Vout ≈ 0,1 V, and RLL carries a current of 12.5 mA. That is, 0,1 V, and R carries a current of 12.5 mA. That is, IIC1 ≈ IIE1 =20 mA. Similarly, if Vin ≈ 0,6 V, tt h e n C1 ≈ E1 =20 mA. Similarly, if Vin ≈ 0,6 V, h e n Vout ≈ 0,2 V, IIRL ≈ 25 mA, and hence IIC1 ≈ 7,5 mA. Vout ≈ 0,2 V, RL ≈ 25 mA, and hence C1 ≈ 7,5 mA. In other words, the collector current of continues to In other words, the collector current of continues to fall. fall.
13.2 Emitter Follower as Power Ampliﬁer
What happens as Vin becomes more negative? Does Vout still track Vin? We observe that for a sufﬁciently low Vin, the collector current of Q1 drops to zero and RL carries the entire I1 [Fig. 13.1(d)]. For lower values of Vin,Q1 remains off and Vout = - I1RL = -260mV.
(13.5) (13.6) Since, Eqs. (13.5) and (13.6) can be combined to yield Beginning with a guess Vout = -0,2V and after a few iterations, we obtain
Note from (13.6) that IIC1≈ 6,13 mA. To determine the ≈ Note from (13.6) that C1 6,13 mA. To determine the value of Vin that yields IIC1 ≈ 0,01I1 =0,325 mA, we value of Vin that yields C1 ≈ 0,01I1 =0,325 mA, we eliminate Vout from Eqs. (13.5) and (13.6): eliminate Vout from Eqs. (13.5) and (13.6): Setting IC1+= 0,325mA, we obtain Note from (13.5) that Vout ≈ 257mV under this condition. Note from (13.5) that Vout ≈ 257mV under this condition.
Let us summarize our thoughts thus far. In the arrangement of Fig. 13.1(a), the output tracks he input 2 as V in rises because Q1 can carry both I1 and the current drawn by R L. On the other hand, as Vin falls, so does IC1, eventually turning Q1 off and leading to a constant output voltage even though the input changes. As illustrated in the waveforms of Fig. 13.2(a), the output is severely distorted. From another perspective, the input/output characteristic of the circuit, depicted in Fig. 13.2(b), begins to substantially depart from a straight line as Vin falls below approximately 0.4 V (from Example 13.1)
•• Our foregoing study reveals that the follower of Our foregoing study reveals that the follower of Fig. 13.1(a) cannot deliver voltage swings Fig. 13.1(a) cannot deliver voltage swings as large as U4 V to an 8- speaker. How can we as large as U4 V to an 8- speaker. How can we remedy the situation? Noting that Vout;; min = remedy the situation? Noting that Vout min = II1RL, we can increase I1 to greater than 50 mA so 1RL, we can increase I1 to greater than 50 mA so that for Vout = 4 V,Q1 still remains on. that for Vout = 4 V,Q1 still remains on. This solution, however, yields a higher power This solution, however, yields a higher power dissipation and a lower efﬁciency. dissipation and a lower efﬁciency.
13.3 Push-Pull Stage 13.3 Push-Pull Stage Considering the operation of the emitter follower Considering the operation of the emitter follower in the previous section, we postulate that the in the previous section, we postulate that the performance can be improved if I1 increases only performance can be improved if I1 increases only when needed. In other words, we envision an when needed. In other words, we envision an arrangement where in I1 increases as Vin becomes arrangement where in I1 increases as Vin becomes more negative and vice versa. Shown in Fig.13.3(a) more negative and vice versa. Shown in Fig.13.3(a) is a possible realization of this idea. Here, the is a possible realization of this idea. Here, the constant current source is replaced with a emitter constant current source is replaced with a emitter follower so that, as begins to turn off, “kicks in” and follower so that, as begins to turn off, “kicks in” and allows to track .. allows to track
Called the “push-pull” stage, this circuit merits a Called the “push-pull” stage, this circuit merits a detailed study. We note that if Vin is sufﬁciently detailed study. We note that if Vin is sufﬁciently positive, Q1 operates as an emitter follower, Vout = positive, Q1 operates as an emitter follower, Vout = Vin -- VBE1,, and Q2 remains off [Fig.13.3(b)] because Vin VBE1 and Q2 remains off [Fig.13.3(b)] because its base-emitter junction is reverse-biased. By its base-emitter junction is reverse-biased. By symmetry, if Vin is sufﬁciently symmetry, if Vin is sufﬁciently negative, the reverse occurs [Fig. 13.3(c)] and Vout = negative, the reverse occurs [Fig. 13.3(c)] and Vout = Vin+ [VBE2]. Wesay Q1 “pushes” current Vin+ [VBE2]. Wesay Q1 “pushes” current Into RLL in the former case and Q2 “pulls” current Into R in the former case and Q2 “pulls” current from RL in the latter. from RL in the latter.