Transcript Class A Output Stage - Recap

Class A Output Stage - Recap
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Class A output stage is a simple linear current
amplifier.
It is also very inefficient, typical maximum
efficiency between 10 and 20 %.
Only suitable for low power applications.
High power requires much better efficiency.
Why is class A so inefficient ?
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Single transistor can only conduct in one
direction.
D.C. bias current is needed to cope with
negative going signals.
75 % (or more) of the supplied power is
dissipated by d.c.
Solution : eliminate the bias current.
Class B Output Stage
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Q1 and Q2 form two unbiased
emitter followers
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Q1 only conducts when the input is
positive
Q2 only conducts when the input is
negative
Conduction angle is, therefore,
180°
When the input is zero, neither
conducts
i.e. the quiescent power
dissipation is zero
Class B Current Waveforms
Iout
time
IC1
time
IC2
time
Class B Efficiency
Average power drawn from the
positive supply:
P  ve   VS I C1
IC1
A/RL
A sin()
0
1
I C1 
2
2
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2 Phase, 

1 A
A
0 I C1  d  2 0 RL sin  d  RL
 P  ve  
VS A
RL
By symmetry, power drawn from +ve and –ve
supplies will be the same. Total power, therefore:
PS  P  ve   P ve   2 P ve  
2VS A
RL
Load power:
2
vout
A2 sin 2 t  A2
PL 


RL
RL
2 RL
Efficiency:
PL
A2 RL
A
 

PS 2 RL 2VS A 4VS
NB. A  VS     / 4  78.5%
Power Dissipation
To select appropriate output transistors, the maximum
power dissipation must be calculated.
2VS A A2
PD  PS  PL 

RL 2 RL
Just need to find the maximum value of PD to select
transistors/heatsinks
E.g. VS = 15 V, RL = 100 W
1.5
PL
PS
PD
Power [W]
1
2VS A
PS 
RL
A2
PL 
2 RL
0.5
0
0
5
10
Peak Output Amplitude, A [V]
15
Maximum Power Dissipation
PD is a quadratic function of A,
2VS A A2
PD 

RL 2 RL
dPD
0
maximum when:
dA

2VS A

0
RL RL
 PD (max)
 A
2VS

4VS2
2VS2
2VS2
 2  2  2
 RL  RL  RL
Efficiency / Power Dissipation
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Peak efficiency of the class B output stage
is 78.5 %, much higher than class A.
Unlike class A, power dissipation varies
with output amplitude.
Remember, there are two output devices
so the power dissipation is shared
between them.
Design Example
Design a class B amplifier which will deliver up to 25 W
into a 4 W load.
A2
PL  25 
 25  A  14.1 V
2 RL
Supply voltages must be larger than Amax so choose
Vs = 15V.
2VS2
PD  max   2  11.4 W  2  5.7 W
 RL
Each of the two output transistors must be able to safely
dissipate up to 5.7 Watts. Using a TIP120 & TIP 125:
TJ  TA   JA PD  TJ   JA PD  TA
 JA PD max  TA  TJ max
 5.7 JA  25  150
  JA  22 C/W
But, with JC = 1.92 °C/W
 JA   JC   CA  1.92   CA
 CA  20 C/W
i.e. Either two heatsinks rated at less than 20°C/W are
required or a single heatsink rated at less than 10°C/W.
Suggested heatsink
Dimensions, 50mm x 50mm x 9.5mm
Accommodates two devices
Rating 6.5°C/W
Cost 60p inc VAT
Cross-Over Distortion
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A small base-emitter voltage is
needed to turn on a transistor
Q1 actually only conducts when
vin > 0.7 V
Q2 actually only conducts when
vin < -0.7 V
When 0.7 > vin > -0.7, nothing
conducts and the output is zero.
i.e. the input-output relationship
is not at all linear.
Actual Input-Output Curve
vout
vout  vin  VBE
-VBE
+VBE
vout  vin  VBE
vin
Effect of Cross-Over Distortion
Audio Demo
Undistorted original
Class B amplifier output
Class B Summary
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A class B output stage can be far more
efficient than a class A stage (78.5 %
maximum efficiency compared with 25 %).
It also requires twice as many output
transistors…
…and it isn’t very linear; cross-over
distortion can be significant.