12 power amplifiers
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12.0 Power Amplifiers 1 of 37
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12.0 Power Amplifiers 2 of 37
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12.0 Power Amplifiers 3 of 37
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12.0 Power Amplifiers 4 of 37
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12.0 Power Amplifiers 5 of 37 Class A Power Calculations Pi(dc) = Vcc*Ico Po(ac) = Icrms
2*Rc = (Icp/2 ½ )2*Rc = (Vcp/2 ½ )2/Rc = (Vcpp/2*2 ½ )2/Rc = (Vcpp)2/8Rc = (Icpp)2*Rc /8 and since, Vcpp = Icpp *Rc then, Po(ac) = Vcpp* Icpp /8
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12.0 Power Amplifiers 6 of 37 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 7 of 37 The 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 Po = Vrms2/R R = 20/8mA = 2500 = (5/20.5)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
Vce 20v 15v
8ma
Qpt
0
0 0
2ma
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12.0 Power Amplifiers 8 of 37 For the circuit below determine,
C1
1
Q1
Q2N2222
R1
1k
10.70mA
V220Vdc
R2
350k
55.13uA
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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12.0 Power Amplifiers 9 of 37 The collector-emitter voltage swing is,
Time
0s 2ms 4ms 6ms 8ms 10msV(R1:1)
7V
8V
9V
10V
11V
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12.0 Power Amplifiers 10 of 37
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12.0 Power Amplifiers 11 of 37
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12.0 Power Amplifiers 12 of 37
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12.0 Power Amplifiers 13 of 37
The signal output swing is nearly 2x the supply voltage. This is the efficiency improvement.
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12.0 Power Amplifiers 14 of 37 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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12.0 Power Amplifiers 15 of 37
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12.0 Power Amplifiers 16 of 37
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12.0 Power Amplifiers 17 of 37 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%.
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12.0 Power Amplifiers 18 of 37 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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12.0 Power Amplifiers 19 of 37 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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12.0 Power Amplifiers 20 of 37
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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12.0 Power Amplifiers 21 of 37 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 5pt Loadline 100 5pt 12V
5:1
4 ohms
0
0
200ma
0
400ma
300ma
100ma
5 10 15 20 Vce
2ma
4ma
6ma
8ma
10ma
12ma
0
Ic
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12.0 Power Amplifiers 22 of 37 The Class A transformer-coupled power amplifier circuit schematic and simulation results are given below. The schematic with DC collector current is,
TX1
V220Vdc
R1
4
Q1
Q2N2222
55.58mAC1
1
0
0
R2
75k
VAMPL = 0.004
Q1 collector voltage waveform is,
Time
18.0us 18.5us 19.0us 19.5us 20.0usV(TX1:2)
18V
19V
20V
21V
22V
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12.0 Power Amplifiers 23 of 37 The collector current waveform is,
Time
18.0us 18.5us 19.0us 19.5us 20.0usI(TX1:1)
48mA
52mA
56mA
60mA
64mA
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12.0 Power Amplifiers 24 of 37 4 ohm resistor voltage waveform is,
Time
18.0us 18.5us 19.0us 19.5us 20.0usV(TX1:3)
-200mV
-100mV
0V
100mV
200mV
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
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12.0 Power Amplifiers 25 of 37 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. Pi = Vcc * Vcc/2R = Vcc
2/2R
Po = (0.4Vcc/(21/2
))2/R
2. A 1:1 transformer-load power amplifier has a voltage output swing peak voltage equal to 0.8Vcc. Pi = Vcc*Vcc/R = Vcc
2/R
Po = (0.8Vcc/21/2
)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 Po = (0.7Vcc/2
1/2)
2/R
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12.0 Power Amplifiers 26 of 37 Class B amplifier - eliminate the bias current.
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12.0 Power Amplifiers 27 of 37
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12.0 Power Amplifiers 28 of 37
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12.0 Power Amplifiers 29 of 37 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 = 2* VLp/RL π Pi(dc) = 2*Vcc* VLp/RL π Output Power Po(ac) = VLrms
2/RL = VLp
2/2RL = VLpp
2/8RL
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12.0 Power Amplifiers 30 of 37 Efficiency % = 100%*Po(ac)/Pi(dc) = 100%*( VLp
2/2RL)/( 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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12.0 Power Amplifiers 31 of 37
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12.0 Power Amplifiers 32 of 37 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 33 of 37
Summary of Power Amplifiers 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 Ipp OutputPower
Avg Current
Voltage Supply
Input Power
%efficiency
Class A Vcc Vcc/R Vcc2/8R Vcc/2R Vcc Vcc2/2R 25Class A Transformer
2Vcc 2Vcc/R’ 4Vcc2/8R’ Vcc/R Vcc Vcc2/R’ 50
Class B 2Vcc 2Vcc/R 4Vcc2/8R 2Vcc/πR Vcc 2Vcc2/πR 100π/4
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12.0 Power Amplifiers 34 of 37 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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12.0 Power Amplifiers 35 of 37
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12.0 Power Amplifiers 36 of 37
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12.0 Power Amplifiers 37 of 37