Transcript
78
Analogue Building Blocks Part 3: Output Stages (Power Amplifiers)
for Analogue Circuit Fundamentals
by
Prof. Michael Tse
September 2014
Basic problem
Consider a common-emitter amplifier. It provides a good gain, but it fails to drive a low-resistance (high-power) load.
+15V
Rc = 700Ω
vo
11.6k
vi
Rload
3.4k
270
Quick calculation:
Biasing: VB = 3.4 V, VE ≈ 2.7V
IC ≈ IE = 2.7/270 = 10 mA
VC = 15 – 0.01Rc = 8 V
Small-signal:
gm = IC/26mV = 0.38 A/V
gain = gmRc = 266 or 48dB
Rout = Rc = 700 Ω
Rin = rπ || 3.4k || 11.6k
Equivalent circuit:
700
vi
Rin
±
266vi
vo
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Rload
What happens if the load Rload is only 10Ω?
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Solution
Never load the CE amplifier directly! Use an output stage or buffer.
+15V
Requirements of output stage:
•
Rc = 700Ω
vo
11.6k
•
vi
±
Rload
gain=1
3.4k
•
Rout,CE
270
Output resistance should be very small.
Input resistance should be very much greater than the output resistance of the CE-amplifier.
Gain = 1 (simply copy voltage).
additional output stage
CE amplifier
700
vi
Rin
±
266vi
0.1
vo
1M
±
1vo
vout
Rload = 10Ω
How to design this output stage?
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Objectives of output stage or power amplifier
Rout
vin
Rin
±
1vin
vout
Rload
An output stage is designed to
- deliver current to a load (Rout << Rload)
- reproduce the voltage as faithfully as possible (low distortion)
- seamlessly load the previous amplifier stage (Rin >> Rout,CE of previous stage)
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Emitter follower output stage (Class A power amplifier)
Emitter follower
+V
Implementation
+V
+
+
vin
vin
–
–
Q1
Iout
+
vout
RL
–
+
RL
vout
–
Q2
–V
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current source
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Large-signal analysis of Class A power amplifier
+V
+
vin
–
Q1
Iout
+
Also,
RL
vout
IQ
–
–V
Hence,
We see that the output voltage is very linearly related to the input voltage. This gives the Class A amplifier a very desirable property— very low distortion.
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Large-signal analysis of Class A power amplifier
vout
Q1 saturates
Vcc–VCE1(sat)
slope = 1
vin
Vbe1
Q2 saturates
–(Vcc–VCE2(sat))
We note that the good linearity is only maintained if both Q1 and Q2 are kept in active region.
If Q1 saturates, the output voltage clips at the top.
If Q2 saturates, the output voltage clips at the bottom.
Operating range without clippings caused by saturation:
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Clippings due to too much voltage swing
vin
vout
When vin goes too high, Q1 saturates around Vcc–VCE1(sat). The output waveform clips at this max voltage.
Vcc–VCE1(sat)
vin
vout
Vcc–VCE2(sat)
When vin goes too low, Q2 saturates around –(Vcc–VCE2(sat)). The output waveform clips at this min voltage.
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Severe clipping due to large load current
When the output current is too high (the load demands too much current), Q1 will be cut-off during negative swing because the current source cannot give enough negative output current and maintain Q1 in active region at the same time.
+V
+
vin
Q1
Iout
–
IQ
+
RL
vout
–
During +ve swing: Iout > 0, and Q1 has to give Iout+ IQ and hence surely active.
During –ve swing: Iout < 0, and Q1 is cutoff if | Iout| is just equal to IQ. At this point, we have
Iout = –IQ
Ic1 = 0
Q1 is cutoff
vout = –IQRL
The output clips at –IQRL which is a smaller negative swing than that limited by saturation!
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Severe clipping due to large load current
vout
Q1 saturates
Vcc–VCE1(sat)
slope = 1
vin
Vbe1
Q1 cutoff
–IQRL
vin
vout
–IQRL
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Clipping analysis
The basic clipping is due to saturation of Q1 and Q2. This clipping happens anyway and is roughly at ±(Vcc –VCE(sat)) ≈ ± Vcc (since VCE(sat) is usually just 0.2V).
Moreover, if the load current is too large, severe clipping may occur. Thus, we should check the load and determine whether severe clipping occurs or not.
Intuitively, severe clipping would occur only if such clipping gives a smaller swing limit. In other words, if IQRL < Vcc –VCE(sat), we have severe clipping.
CASE 1:
If
RL >
Vcc − VCE(sat) IQ
(small load current, RL large), no severe clipping.
Vcc − VCE(sat) IQ
(large load current, RL small), severe clipping occurs.
CASE 2:
€ If
RL <
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Graphical clipping analysis
Ic1
(no severe clipping)
IQ
(severe clipping)
VCE(sat)
Vcc
VCE1
2Vcc –VCE(sat)
Max load without severe clipping
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Power dissipation and efficiency
Recall the basic equation of power dissipation in any device:
+ v –
i
For the load resistor, assuming no clipping occurs, i.e., iout and vout are sinewaves, we have the average power output:
At max load without severe clipping,
the max power output is
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Power dissipation and efficiency
To find the total power drawn from the power supply:
+Vcc
IQ
RL
–Vcc
Average current and voltage drawn from supply by the class A output stage is
Iav = IQ
Vav = 2Vcc
Thus, power supplied to the output stage is
Psupply = 2VccIQ
Hence, the max efficiency is
ONLY 25%!!
Class A has a poor efficiency, but very low distortion — high-fidelity!
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Example of efficiency calculation
Suppose we wish to find the efficiency of the following Class A power amplifier.
+Vcc=+15V
Ic
IQ=500mA
–Vcc=–15V
Ic
IQ
Here,
IQ = 500 mA, Vcc = 15 V and VCE(sat) = 0.2 V
Suppose
RL = 80 Ω and vin(peak) = 12 V
Then,
vout(peak) = 12 V
iout(peak) = 150 mA which is less than IQ;
RL = 80 Ω
Hence, no clipping occurs.
Pout = 0.5 x 12 x 0.15 = 0.9 W
Pin = 2 x 15 x 0.5 = 15 W
Efficiency η = 6 %
15
30
VCE
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Class B push-pull power amplifier
The main disadvantage of Class A power amplifier is the poor efficiency because of the large dissipation even at zero output.
To improve efficiency, we may try to eliminate the dissipation at zero output. That means, we aim to design an amplifier whose dissipation is zero when vin = 0.
Q1
Solution:
Class B push-pull output stage
Use two complementary active devices.
+
+
Each conducts for half cycle.
vin
RL
vout
Q2
–
–
Basically, it is emitter follower circuit.
Q1 and Q2 are cutoff at vin = 0.
–Vcc
Actually, Q1 and Q2 are cutoff for |vin | < 0.7V
Q1 conducts when vin > 0.7V
Complementary push-pull amplifier
Q2 conducts when vin < –0.7V
Clipping occurs similar to class A if | vout | is
too large, exceeding the supply |Vcc–VCE(sat) |
+Vcc
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Class B push-pull power amplifier
vout
Note what happens when –0.7 < vin < 0.7!!
Vcc–VCE(sat)
+Vcc
Q1
+
–vbe2
vbe1
+
vin
Q2
–
vin
RL
vout
notch or deadband
–
–(Vcc–VCE(sat))
–Vcc
Complementary push-pull amplifier
When input is in this range, the output is zero. So, a sinewave input will be distorted near vin = 0. This is called crossover distortion.
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Crossover distortion in Class B amplifier
vout
clipping
Vcc–VCE(sat)
vout
–vbe2
vbe1
vin
crossover distortion
–(Vcc–VCE(sat))
clipping
vin
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Power dissipation and efficiency
Each device only conducts for half cycle, i.e., 180o. So, the current waveform for each device is a half-sinusoid.
Ic1
Ic2
Basically, in order to find the efficiency, we need to find
- the current supplied by ±Vcc and hence the input power.
- the power delivered to the load, I.e., the output power
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Power dissipation and efficiency
For simplicity of analysis, we ignore the crossover distortion.
The supply current is
Hence, the input power is
The power delivered to the load Pload
Hence, the efficiency is
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Maximum efficiency
The best efficiency is when the output voltage is the greatest possible without clipping.
Thus, the maximum efficiency is
MUCH BETTER THAN CLASS A AMPLIFIER!
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Class AB power amplifier
To eliminate the crossover distortion, we may bias the active devices so that they conduct a small quiescent current near the crossover vin = 0. Don’t let them cutoff!
Here, for vin < 0, Q2 turns on and and vout follows . But since the diodes maintain a constant total bias voltage across the BE junctions of both Q1 turns Q2, VBE of Q2 decreases by the same amount that VBE of Q1 increases. Thus, Q1 stays on, but only allows a very small current to flow.
Similarly, for vin > 0, we have the same situation, with Q1 and Q2 interchanged.
Efficiency is the same as Class B.
+Vcc
IQ
Q1
+
Q2
+
vin
RL
vout
–
–
–Vcc
vout
–VBE2
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vin
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Crossover distortion improvements in Class AB
Ic1=βIb1 (+ve)
Ic2=βIb2 (-ve)
Ic1=βIb1 (+ve)
Ic2=βIb2 (-ve)
Q1
–0.7
0.7
Q2
vbe1 (+ve)
vbe2 (–ve)
Q1
–0.7
0.7
Class B: Either Q1 or Q2 is cutoff;
Only one transistor is active.
vbe1 (+ve)
vbe2 (–ve)
Class AB:
Q2
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Both Q1 and Q2 are active at all times, with different vbe.
For large +vin, Q1 is largely biased (big current in Ic1) while Q2 is slightly biased (little current in Ic2); vice versa for –vin.
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Quasi-complementary push-pull output stage
Very popular!
Since pnp transistors have limited current capability, the Class B or AB shown previously is not suitable for power greater than several watts.
Solution: use a composite pnp made from a pnp and a high power npn.
Q3
Ic3
Q4
IQ
+
RL
vout
–
+
=
Ic
Q4: high current npn
+Vcc
vin
Ic
–
–Vcc
Q3: pnp
Effectively, this is like a high current pnp.
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Summary
Class A:
one device in active linear region
max efficiency = 25%
distortion due to clipping and large power output, otherwise very linear!
Class B:
two devices in push-pull configuration, each conducts for half cycle
max efficiency = 78%
distortion due to crossover deadband and clipping
Class AB:
two devices in push-pull, each conducts for half cycle and stays on all the time
carrying small current even when input is 0.
max efficiency = 78%
distortion due to clipping and a little due to crossover
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