Electronic Circuits - Part 2 ppt

2 The current levels involved in this example are much greater than the typical currents milliamperes encountered in previous circuits.. 2 The current levels involved in this example are

Teacher: Dr LUU THE VINH

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The chapter outline is shown below.

Omission of PNP Transistor

 High-Fidelity Design

Large-Signal Considerations

Omission of PNP Transistor

 High-Fidelity Design

Heat Dissipation

Power Ratings

Thermal Runaway

Heat Dissipation

13.1 General Considerations

• 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

speaker as:

• 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:

Where VP = p2 denotes the root mean square (rms) value

of the sinusoid and RL represents the speaker impedance For RL = 8 and Pout = 1 W,

Also, the peak current flowing through the speaker is given by

V P = 4V

(13.1)

Important observations

1) The resistance that must be driven by the amplifier is

much lower than the typical values (hundreds to thousands of ohms) seen in previous chapters

2) The current levels involved in this example are much

greater than the typical currents (milliamperes) encountered in previous circuits

3) The voltage swings delivered by the amplifier can hardly

be viewed as “small” signals, requiring a good understanding of the large-signal behavior of the circuit 4) The power drawn from the supply voltage, at least 1W,

is much higher than our typical values

5) A transistor carrying such high currents and sustaining

several volts (e.g., between collector and emitter) dissipates a high power and, as a result, heats up High-power transistors must therefore handle high currents and high temperature

1) The resistance that must be driven by the amplifier is

much lower than the typical values (hundreds to thousands of ohms) seen in previous chapters

2) The current levels involved in this example are much

greater than the typical currents (milliamperes) encountered in previous circuits

3) The voltage swings delivered by the amplifier can hardly

be viewed as “small” signals, requiring a good understanding of the large-signal behavior of the circuit

4) The power drawn from the supply voltage, at least 1W,

is much higher than our typical values

5) A transistor carrying such high currents and sustaining

several volts (e.g., between collector and emitter) dissipates a high power and, as a result, heats up High-power transistors must therefore handle high currents and high temperature

12.1 General Considerations

• Based on the above observations, we can predict the parameters of

interest in the design of power stages:

(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 reproduce music with high fidelity In previous chapters, we rarely dealt with distortion.

(2) “Power efficiency” or simply “efficiency,” denoted by  and defined as:

• Based on the above observations, we can predict the parameters of

interest in the design of power stages:

(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 reproduce music with high fidelity In previous chapters, we rarely dealt with distortion.

(2) “Power efficiency” or simply “efficiency,” denoted by  and defined as:

For example, a cellphone power amplifier that consumes 3 W from the battery to deliver 1 W to the antenna provides 33,3% In previous chapters, the efficiency of circuits was of little concern because the absolute value of the power consumption was quite small (a few milliwatts).

For example, a cellphone power amplifier that consumes 3 W from the battery to deliver 1 W to the antenna provides 33,3% In previous chapters, the efficiency of circuits was of little concern because the absolute value of the power consumption was quite small (a few milliwatts).

(3) “Voltage rating.” As suggested by Eq (13.1), higher power levels or load resistance values translate to large voltage swings and (possibly) high supply voltages Also, the transistors in the output stage must exhibit breakdown voltages well above the output voltage swings.

(3) “Voltage rating.” As suggested by Eq (13.1), higher power levels or load resistance values translate to large voltage swings and (possibly) high supply voltages Also, the transistors in the output stage must 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 considered a good candidate for driving “heavy” loads, i.e., low impedances As shown in Chapter 5, the small- signal gain of the follower is given by

With its relatively low output impedance, the emitter follower may be considered a good candidate for driving “heavy” loads, i.e., low impedances As shown in Chapter 5, the small- signal gain of the follower is given by

We may therefore surmise that for, say, RL=8, a gain near unity can be obtained if 1/gm<< RL,

e.g., 1/gm =0,8 , requiring a collector bias

current of 32.5 mA We assume >>1.

We may therefore surmise that for, say, RL=8, a gain near unity can be obtained if 1/gm<< RL,

e.g., 1/gm =0,8 , requiring a collector bias

current of 32.5 mA We assume >>1.

But, let us analyze the circuit’s behavior in delivering large voltage swings (e.g 4 VP ) to heavy loads To this end, consider the follower 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 and positive power supplies, allowing Vout to be centered around zero For Vin  0,8 V, we have

Vout  0 and IC 32,5 mA If Vin rises from 0.8 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 13.1(b)].

But, let us analyze the circuit’s behavior in delivering large voltage swings (e.g 4 VP ) to heavy loads To this end, consider the follower 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 and positive power supplies, allowing Vout to be centered around zero For Vin  0,8 V, we have

Vout  0 and IC 32,5 mA If Vin rises from 0.8 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 13.1(b)].

• Now suppose Vin begins from +0,8 V and gradually goes down [Fig 13.1(c)] We expect

Vout to go below zero and hence part of I1 to flow

from RL For example, if Vin  0,7 V, t h e n Vout 

0,1 V, and RL carries a current of 12.5 mA That is,

IC1  IE1 =20 mA Similarly, if Vin  0,6 V, t h e n

Vout  0,2 V, IRL  25 mA, and hence IC1  7,5 mA

In other words, the collector current of continues to fall.

• Now suppose Vin begins from +0,8 V and gradually goes down [Fig 13.1(c)] We expect

Vout to go below zero and hence part of I1 to flow

from RL For example, if Vin  0,7 V, t h e n Vout 

0,1 V, and RL carries a current of 12.5 mA That is,

IC1  IE1 =20 mA Similarly, if Vin  0,6 V, t h e n

Vout  0,2 V, IRL  25 mA, and hence IC1  7,5 mA

In other words, the collector current of continues to 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.

Since, Eqs (13.5) and (13.6) can be combined to yield

(13.5)

(13.6)

Beginning with a guess Vout = -0,2V and after a few iterations, we

obtain

Note from (13.6) that IC1 6,13 mA To determine the value of Vin that yields IC1  0,01I1 =0,325 mA,

we eliminate Vout from Eqs (13.5) and (13.6):

Note from (13.6) that IC1 6,13 mA To determine the value of Vin that yields IC1  0,01I1 =0,325 mA,

we 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 Vin rises

because Q1 can carry both I1 and the current drawn by RL

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 Fig 13.1(a) cannot deliver voltage swings

as large as U4 V to an 8- speaker How can we remedy the situation? Noting that Vout; min =

I1RL, we can increase I1 to greater than 50 mA so that for Vout = 4 V,Q1 still remains on.

This solution, however, yields a higher power

dissipation and a lower efficiency.

• Our foregoing study reveals that the follower of

Fig 13.1(a) cannot deliver voltage swings

as large as U4 V to an 8- speaker How can we remedy the situation? Noting that Vout; min =

I1RL, we can increase I1 to greater than 50 mA so that for Vout = 4 V,Q1 still remains on.

This solution, however, yields a higher power

dissipation and a lower efficiency.

Considering the operation of the emitter

follower in the previous section, we postulate that the performance can be improved if I1 increases

only when needed In other words, we envision an arrangement where in I1 increases as Vin becomes more negative and vice versa Shown in Fig.13.3(a)

is a possible realization of this idea Here, the

constant current source is replaced with a emitter follower so that, as begins to turn off, “kicks in” and allows to track

Considering the operation of the emitter

follower in the previous section, we postulate that

the performance can be improved if I1 increases

only when needed In other words, we envision an

more negative and vice versa Shown in Fig.13.3(a)

is a possible realization of this idea Here, the

constant current source is replaced with a emitter

follower so that, as begins to turn off, “kicks in” and allows to track

13.3 Push-Pull Stage

Called the “push-pull” stage, this circuit merits a

detailed study We note that if Vin is sufficiently

positive, Q1 operates as an emitter follower, Vout =

Vin - VBE1, and Q2 remains off [Fig.13.3(b)] because its base-emitter junction is reverse-biased By

symmetry, if Vin is sufficiently

negative, the reverse occurs [Fig 13.3(c)] and Vout

= Vin+ [VBE2] Wesay Q1 “pushes” current

Into RL in the former case and Q2 “pulls” current

from RL in the latter.

Called the “push-pull” stage, this circuit merits a

detailed study We note that if Vin is sufficiently

positive, Q1 operates as an emitter follower, Vout =

Vin - VBE1, and Q2 remains off [Fig.13.3(b)] because its base-emitter junction is reverse-biased By

symmetry, if Vin is sufficiently

negative, the reverse occurs [Fig 13.3(c)] and Vout

= Vin+ [VBE2] Wesay Q1 “pushes” current

Into RL in the former case and Q2 “pulls” current

from RL in the latter.

Figure 13.4 Push-pull stage characteristic.

Figure 13.5 Push-pull stage characteristic with dead zone.

approaches zero? The rough

characteristic in Fig 13.4

suggests that the two segments

cannot meet if they must remain

linear In other words, the overall

characteristic inevitably incurs

nonlinearity and resembles that

shown in Fig 13.5, exhibiting a

“dead zone” around

What happens as Vin

approaches zero? The rough

characteristic in Fig 13.4

suggests that the two segments

cannot meet if they must remain

linear In other words, the overall

characteristic inevitably incurs

nonlinearity and resembles that

shown in Fig 13.5, exhibiting a

“dead zone” around

Why does the circuit suffer from a dead zone? We make two observations:

• First, Q1 and Q2 cannot be on simultaneously: for Q1 to be on,Vin >Vout, but for Q2, Vin <Vout

• Second, if Vin=0, Vout must also be zero This can be proved by

• For example, if V out > 0 (Fig

13.6), then the current must be

provided by (from V CC ), requiring

VBE1>0 and hence Vout = Vin-VBE1< 0

That is, for Vin =0, both transistors

are off.

• For example, if V out > 0 (Fig

13.6), then the current must be

provided by (from V CC ), requiring

VBE1>0 and hence Vout = Vin-VBE1< 0

That is, for Vin =0, both transistors

are off.

Now suppose Vin begins to increase from

zero Since Vout is initially at zero, Vin

must reach at east VBE  600-700mV

before Q1 turns on The output therefore

remains at zero for Vin < 600mV,

exhibiting the dead zone depicted in Fig

13.5 Similar observations apply to the

dead zone for Vin<0.

Now suppose Vin begins to increase from

zero Since Vout is initially at zero, Vin

must reach at east VBE  600-700mV

before Q1 turns on The output therefore

remains at zero for Vin < 600mV,

exhibiting the dead zone depicted in Fig

13.5 Similar observations apply to the

dead zone for Vin<0. Figure 13.6 Push-pull stage with zero input voltage.

Figure 13.7 Gain of push-pull stage as a function of input.

• For Vin well above 600 mV, either Q1 or Q2 serves as

an emitter follower, thus producing a reasonable

sinusoid at the output Under this condition, the plot in Fig 13.5 indicates that Vout = Vin + /VBE2/ or Vin - VBE1

Within the dead zone, however, Vout  0 Illustrated in Fig.13.8, Vout exhibits distorted “zero crossings.” We also say the circuit suffers from “crossover distortion”`

Solution

an emitter follower, thus producing a reasonable

sinusoid at the output Under this condition, the plot in Fig 13.5 indicates that Vout = Vin + /VBE2/ or Vin - VBE1

Within the dead zone, however, Vout  0 Illustrated in Fig.13.8, Vout exhibits distorted “zero crossings.” We also say the circuit suffers from “crossover distortion”`

Figure 13.8 Input and output waveforms in the presence of dead zone.

Figure 13.8 Input and output waveforms in the presence of dead zone.

Méo xuyên tâm (crossover distortion)

Crossover distortion

13.4 Improved Push-Pull Stage

13.4.1 Reduction of Crossover Distortion

How the battery VB must be implemented?

How the battery VB must be implemented?

13.4.2 Addition of CE Stage

• The two current sources in Fig 13.14 can be

realized with pnp and npn transistors as depicted

in Fig 13.15(a) We may therefore decide to

apply the input signal to the base of one of the

current sources so as to obtain a greater gain

Illustrated in Fig 13.15(b), the idea is to employ Q4 as a common-emitter stage, thus providing

voltage gain from Vin to the base of Q1 and Q2 The CE stage is called the “predriver.”

• The two current sources in Fig 13.14 can be

realized with pnp and npn transistors as depicted

in Fig 13.15(a) We may therefore decide to

apply the input signal to the base of one of the

current sources so as to obtain a greater gain

Illustrated in Fig 13.15(b), the idea is to employ Q4 as a common-emitter stage, thus providing

voltage gain from Vin to the base of Q1 and Q2.

The CE stage is called the “predriver.”