12.0 Power Amplifiers
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Class A Power Calculations
Pi(dc)
=
Vcc*Ico
Po(ac)
=
Icrms2*Rc
=
(Icp/2 ½ )2*Rc
=
(Vcp/2 ½ )2/Rc
=
(Vcpp/2*2 ½ )2/Rc
=
(Vcpp)2/8Rc
=
(Icpp)2*Rc /8
=
Icpp *Rc
=
Vcpp* Icpp /8
and since,
Vcpp
then,
Po(ac)
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12.0 Power Amplifiers
Maximum Power Output and Efficiency
If the max output voltage swing is,
Vcppmax = Vcc and then,
Pomax(ac) = (Vcc)2/8Rc
In order for this to be true the bias situation is that
VCQ
=
Vcc/2 and therefore,
ICQ
=
Vcc/2Rc
Power input to the circuit is then,
Pimax(dc) =
Vcc*Vcc/2Rc
The efficiency is then,
% = Pomax(ac)/Pimax(dc) = 25%
But, if the bias is not in the center, the power output is as calculated as
Po(ac) = Vcpp* Icpp /8 and
Pi(dc) = Vcc*Ico
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12.0 Power Amplifiers
The
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Class A amplifier has a load line and Q-point as shown. Determine the power situation.
a) What is the Pi?
b) What is the maximum Po when the amp is operated at this Q-point?
Pi = VCC*ICQ = 20V * 2mA = 40mW
R = 20/8mA = 2500
Po = Vrms2/R
0.5 2
= (5/2 ) /2500 =
= 5mW or,
Po = (Vcemax – Vcemin)(Icmax – Icmin) /8
= (20-10)(4 - )/8 = 5mW
c) What is the maximum Po when the amp is operated at a Q-point of 10v and 4ma?
Po = (20 – 0)(8 – 0)/8 = 20mW and Pi = 20V*4mA = 80mW
Ic
8ma
0
Qpt
2ma
0
0
15v
20v
Vce
12.0 Power Amplifiers
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For the circuit below determine,
10.70mA
R1
55.13uA
R2
1k
350k
C1
1
V2
Q1
20Vdc
Q2N2222
VAMPL = 4m
0
a) Pi(dc) , ignore the base current
Pi = 20V*10.7mA = 0.214W
b) Po(ac)
Po = (10.6 – 7.8)2 /1k = 0.952mW
( 2*21/2 )2
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The collector-emitter voltage swing is,
11V
10V
9V
8V
7V
0s
2ms
4ms
6ms
V(R1:1)
Time
8ms
10ms
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The signal output swing is nearly 2x the supply voltage. This is the efficiency improvement.
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Class A Transformer Power Calculations
VCEpp
=
VCEmax – VCEmin
ICpp
=
ICmax – Icmin
Po(ac)
=
(VCEmax – VCEmin)( ICmax – Icmin)/8
Efficiency of amplifier
Pi(dc)
=
Vcc*ICQ
% efficiency = 50*((VCEmax – VCEmin) /(VCEmax + VCEmin))2
If VCEmin = 0, then efficiency is max at 50%
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IB = 6mA & Ib p-p = 8mA, swing of +- 4mA
for N1//N2 = 3 & RL = 8
RL’ = 72 ohms
Then the slope of 1/72 gives an intersection of current axis at,
Ic = 140mA + 10V/72 = 140mA + 139mA = 279mA
Looking at the graph & setting the max Ib = 6mA + 4mA = 10mA........results in a max of 255mA by
eyeballing the plot.
The minimum current is when Ib = 6mA – 4mA = 2mA and is 25mA again from estimating the plot.
from the plot the min and max Vce’s are found to be 1.7V and 18.3V.
Notice that the signal swing is nearly 2X the supply voltage.....this where the efficiency gain occurs.
So, the power output is,
Po(ac) = (18.3 – 1.7)(255mA – 25mA)/8 = 0.477W
Pi(dc) = Vcc * Icq = 10V*140mA = 1.4W
& the efficiency is 0.477/1.4 = 34.1%
The max for a class A transformer coupled amp is 50%.
12.0 Power Amplifiers
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The resistance load of a transistor amplifier is plotted on the transistor characteristics. Determine,
a) Input power, P(dc)
b) Maximum output power
Q-point
P(dc) = (160mA)(25) = 4 W
R
= 25/0.4 = 62.5 ohms
P(ac) = Vp2/2R = (25-15)2/2*62.5 = 0.8 W
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The load reflected back to the transistor is 50 ohms. Determine the following,
a) Load line
b) Minimum and maximum output voltages
c) Maximum power efficiency for this bias condition
Q-point
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Q-point
eff = 50%(VCEmax – VCEmin)( VCEmax + VCEmin)/8
= 50%[(17.5 – 2.5)/(17.5 + 2.5)]2
= 28.125%
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For the transformer-coupled Class A amplifier below the dc base current is 6ma and the signal base current
swings from 8ma to 4ma. The turns ratio is 5:1 and the load resistance is 4 ohms.
a) Find and plot the Q-point on the collector characteristics below.
12V & 180mA
5pt
b) What is the input power including both the base and collector currents?
(180+6)*10 = 1860mW = 1.86W
5pt
c) What is the power delivered to the load?
PL =~ 1/8 *(18 – 7)*(240mA – 125mA) = 0.18W
Loadline 100
5pt
5pt
12V
4 ohms
0
Ic
5:1
400ma
12ma
300ma
0
10ma
8ma
200ma
6ma
4ma
100ma
2ma
0
0
5
10
15
20
Vce
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The Class A transformer-coupled power amplifier circuit schematic and simulation results are given below.
The schematic with DC collector current is,
TX1
R1
R2
4
75k
C1
0
Q1
1
V2
20Vdc
55.58mA
Q2N2222
VAMPL = 0.004
0
Q1 collector voltage waveform is,
22V
21V
20V
19V
18V
18.0us
V(TX1:2)
18.5us
19.0us
Time
19.5us
20.0us
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The collector current waveform is,
64mA
60mA
56mA
52mA
48mA
18.0us
I(TX1:1)
18.5us
19.0us
Time
19.5us
20.0us
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4 ohm resistor voltage waveform is,
200mV
100mV
0V
-100mV
-200mV
18.0us
V(TX1:3)
18.5us
19.0us
19.5us
Time
Determine the following,
a) Pi(dc) ignore the base current
Pi = 20V*55.58mA = W
b) Po(ac) using the collector voltage & current waveform values.
Po = (1/8)(21.5V – 18.4V)*(61.8mA – 49.8mA) = W
c) Po(ac) using the load voltage waveform values (show calculations).
Po = [(200mV + 185mV)/2*2 ½ ]2/4
20.0us
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Assume the power amplifiers in the questions below are operating at a Q-point such that maximum power
output could be obtained if desired.
Determine for each power amplifier circuit the following,
a) Power Input
b) Power Output
Show your work; i.e., the formulas/algebra you used for each problem.
1. A resistor-load power amplifier has a voltage output swing peak voltage equal to 0.4Vcc.
2
Pi = Vcc * Vcc/2R = Vcc /2R
1/2 2
Po = (0.4Vcc/(2 )) /R
2. A 1:1 transformer-load power amplifier has a voltage output swing peak voltage equal to 0.8Vcc.
2
Pi = Vcc*Vcc/R = Vcc /R
1/2 2
Po = (0.8Vcc/2 ) /R
3. A Class B (push-pull) power amplifier has a voltage output swing peak voltage equal to 0.7Vcc.
Pi = Vcc*2(0.7)Vcc/πR
1/2 2
Po = (0.7Vcc/2 ) /R
12.0 Power Amplifiers
Class B amplifier - eliminate the bias current.
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Class B Power Calculations
Input Power
=
Pi(dc)
=
Vcc*Idc
For each half cycle current goes through the power supply, Vcc.
For a half wave rectifier the average current is,
Idc
=
Ip/π
and since this current flows on each half cycle the average current is
Idc
=
2Ip/π
Pi(dc)
=
2* VLp/RL π
=
2*Vcc* VLp/RL π
Output Power
Po(ac)
=
VLrms2/RL
=
VLp2/2RL
=
VLpp2/8RL
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Efficiency
%
=
100%*Po(ac)/Pi(dc)
=
100%*( VLp2/2RL)/( 2* VLp/RL π)
100%*π*(VLp/4Vcc)
and if VLp = Vcc/2 then
%
=
π/4 =
78.5%
Transistor Power
P2Q =
Pi(dc) - Po(ac)
PQ =
½ *P2Q
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Power Amplifiers
Resistor Load
Pi = Vcc * Vcc/2R = Vcc2/2R
Po = (Vcc/(2*21/2 ))2/R = Vcc2/8R
%efficiency max = 25%
Transformer Load 1:1 turns ratio
Pi = Vcc*Vcc/R = Vcc2/R
Po = (Vcc/21/2 )2/R = Vcc2/2R
%efficiency max = 50%
Push-Pull
Pi = 2Vcc*Vcc/πR
Po = (Vcc/21/2)2/R = Vcc2/2R
%efficiency max = (1/2)/(2/ π) = 25π %
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12.0 Power Amplifiers
Summary of Power Amplifiers
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Draw 3 circuits
Output power - voltage and current
Signal swing
(peak-to-peak / 2)/sqrt of 2 for rms
Ipp*Vpp/8
Input power
Average Current*Average Voltage Supply
Vpp
Class A
Vcc
Class A
2Vcc
Transformer
Class B
2Vcc
Ipp
Vcc/R
2Vcc/R’
Output
Power
Vcc2/8R
4Vcc2/8R’
Avg
Current
Vcc/2R
Vcc/R
Voltage
Supply
Vcc
Vcc
2Vcc/R
4Vcc2/8R
2Vcc/πR Vcc
Input
Power
Vcc2/2R
Vcc2/R’
%efficiency
2Vcc2/πR
100π/4
25
50
12.0 Power Amplifiers
Amplifier Distortion
Harmonic Distortion
% nth harmonic distortion
= % Dn where n = 2, 3, 4,……..
= 100%* |An|/|A1|
% THD
= square root of sum of squares of Dn
Power of Signal Having Distortiion
P1
=
I12*Rc/2
P
=
(I12+ I22+ I22 + …..)Rc/2
=
(1 + THD2)P1
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