Transcript Power

POWER AMPLIFIER
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AGENDA
Power amplifier
Signals / definitions
DC power supply
AC signal power
Efficiency
Classes
Class A type CE
Class A type CC
Push Pull
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Cross over distortion
Vbe Multiplier
Thermal runaway
Emitter resistors
Temperature dependent
Bias voltage
Short circuit protection
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Power Amplifier
Power amplifier = current amplifier
Rloudspeaker = 8 Ohm,
P
for 50 Watt >> I = 2,5 Ampere
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I R
Power IN = DC power
Power out = AC power
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Signals / definitions 1/3 DC supply power
DC-power =
DC value for supply voltages,
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And Average for current waves
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Signals / definitions 2/3 AC Signal power

U
U signal 
2

I signal
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I

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Signals / definitions 3/3 efficiency
 = efficiency / for definition see below
AC signalpower delivered to load

DC  power from DC source
So,
POWER SUPPLY delivers DC or average power
And the
LOAD gets SIGNAL power (use root mean square value)
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ɳ
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Classes
Class-A Output device(s) conduct through 360 degrees of
input cycle (never switch off) - A single output device is
possible. The device conducts for the entire waveform
Class-B Output devices conduct for 180 degrees (1/2 of
input cycle) - for audio, two output devices in "push-pull"
must be used
Class-AB Halfway (or partway) between the above two
examples (181 to 200 degrees typical) - also requires
push-pull operation for audio.
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Classes
Class-C Output device(s) conduct for less than 180
degrees (100 to 150 degrees typical) - Radio Frequencies
only - cannot be used for audio! This is the sound heard
when one of the output devices goes open circuit in an
audio amp!
Class-D Quasi-digital amplification. Uses pulse-widthmodulation of a high frequency (square wave) carrier to
reproduce the audio signal.
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Class A type CE
The configuration is a common emitter
The load / loudspeaker is in the collector
If there is no signal there is a quecient
current to adjust Vc = ½ Vcc for
symmetrical use of voltage range
If there is no signal, the power supply
should deliver power to the circuit
The efficiency is very low
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The lost power is just dissipation:
heating of components: so you need to
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cool for high power
Class A
Vcc = 40 Volt
Rload = 8 Ohm
Quesientpoint
Uc = 20 volt
I bias = 2,5 A
I signal max = 2,5 A
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Vcc I
q
AC signalpower  2

2
2
DC power  Vcc  I q
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Vcc  I q
 4
 0,25
Vcc  I q
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Class A type CC
The configuration is a common collector
The load / loudspeaker is connected to
the emitter
If there is no signal there is a quecient
current to adjust Ve = ½ Vcc for
symmetrical use of voltage range
If there is no signal, the power supply
should deliver power to the circuit
The efficiency is very low
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The lost power is just dissipation:
heating of components: so you need to
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cool for high power
Vcc = 40 Volt
Class A
Rload = 8 Ohm
Quesientpoint Uc = 20 volt
Isignal max = 2,5 Ampere
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Vcc I
q
AC signalpower  2

2
2
DC power  Vcc  I q
1
Vcc  I q
4

 0,25
Vcc  I q
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Push Pull idea
If you need water;
open Q1
If you deliver water;
Open Q2
Q1 and Q2 never
opens at the same
time
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Push Pull with transistors
Here the BJT are
complementary
(NPN and PNP)
Each device amplify
the opposite halves
of the input signal
At the output you
get the total signal.
excellent efficiency
But small mismatch
between the two
halves of the signal
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Push Pull in realisation
Vcc
Vcc = +20 Volt
Vee = -20 Volt
Rload = 8 Ohm
Uin = 0 Volt
Uout = 0 Volt
Vee
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So no bias current
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
Push Pull in realisation
Vcc

Vcc
I
Psignal 

2
2
20 2,5


 25 Watt
2
2

Pdc  2  Vcc 
 24 
Vee
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2,5

I

 32 Watt
25
 
 0, 78
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Cross-over distortion
Dead zone = 1,4 Volt
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Eliminating Cross-Over
Distortion
Shift NPN 0,7 Volt to left
Shift PNP 0,7 Volt to right
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Eliminating Cross-Over
Distortion
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Eliminating Cross-Over
Distortion
Shift
1,4 Volt
Or
Shift
2,8 Volt
Or
Something
else …
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Vbe multiplier
Base current is negligible, so:
VBE
I R1  I R 2 
R2
VBE
VBE R1
VR1  I R 2 R1 
R2
Vbias
Vbias
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 R1  R2 
RV
1 BE
 VR1  VR 2 
 VBE  
 VBE
R2
 R2 
 R1  R2 

 0.7
 R2 
Inverse
voltage divider !!!
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Vbe multiplier
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Thermal runaway
Fit a bigger heatsink.
Use series
emitter-resistors.
Use a temperature
dependent bias voltage.
The latter two are preferred
methods. Both introduce
negative feedback.
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Emitter resisters
2Vbias  VBE1  VEB 2  2VRE
By symmetry:
VBE1  VEB 2  Vbias  VRE
 Vbias  I C RE
So, if IC rises, VBE falls and
IC is reduced.
Note RE should be small
compared with RL to minimise
power wasted.
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Temperature
dependent
bias voltage
If junction temperature rises
but IC stays the same, VBE
must fall causing Vbias to fall
also.
Negative thermal feedback
achieved if the transistor is in
close contact with the output
devices.
Especially suitable for
integrated circuits where
close thermal contact is
guaranteed.
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Short circuit protection
Current limitation
Rx is a current
measuring resistor
T1
Rx
If URx> 0,7 the T1
switches ON
Then output current
in limited !!
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