nashelsky_15_ca - Using Design Center, the circuit of a...

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Unformatted text preview: Using Design Center, the circuit of a series—fed class A amplifier is drawn as shown in Fig. 15.28. Figure 15.29 shows some of the analysis output. Edit the transistor model for values of only BF = 90 and IS = 213—15. This keeps the transistor model more ideal so that PSpice calculations better match those below. The dc bias of the collector voltage is shown to be i Class-A operation 7 ~ A a _..)./.0 Series—fed class A amplifier. Vc.(dc) = 12.47 V Series-fed Class-A Amplifier ““ CIRCUIT DESCRIPTION t’ltttitt.”t"tt¢‘$‘tttitttttitt“ttl“¢ttitt¢tt u" BITMODELPARAMETERS Q2N3904-Xl NPN IS 2000000515 BF 90 ”” SMALL SIGNAL BIAS SOLUTION tunnunuunnuntuuuuu"autumn-u” NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( VL) 12.4670 ($N_0001) 22.0000 (sN_0002) 0.0000 (SN_0003) .8146 VOLTAGE SOURCE CURRENTS NAME CURRENT V_Vi 0.000E+00 v_vcc -9.639E-02 TOTAL POWER DISSIPATION 2.12E+00 WATTS **” BIPOLAR JUNCTION TRANSISTORS NAME CLQl MODEL Q2N3 90+x1 1B 1.06E-03 IC 9535—02 VBE 8. 1512—01 VBC -i.17E+01 VCE 1.25E+01 BETADC 9.00E+01 GM 3.69E+00 RPI 2.44E+01 R0 10054-12 BETAAC 9.00E+Ol FT 5.87E+l9 A T Analysis ou:put for the circuit of Fig. 15128. With transistor beta set to 90, the ac gain is calculated as follows: [E = [C = 95 mA (from analysis output of PSpice) r. = 26 mV/95 mA = 0.27 D For a gain of A. = ~Rc/rg = —100/0.27 = —370 The output voltage is then V0 = A‘.V,- = (—370) - 10 mV = —3.7 V(peak) The output waveform obtained using probe is sh0wn in Fig. 15.30. For a peak—to-peak output of the peak output is V0(p-p) = 15.6V ~ 8.75 V = 6.85 v V0(p) = 6.85 V/2 = 3.4 v Chapter 15 Power Amplifiers 12V r__ _____________ __-___ sett—Js§;1§37__-_ I l I 8V+ -------------------- ---------- -E ---- -—+ 1 .935 1 . 5ms 2 . Oms 1.9.3va) Figure 15.30 Probe outp: 5:: Time the circuit of Fig. 1528. which compares well with that calculated above. From the circuit output analysis, the input power is Pi = VCCIC = (22 V) - (95 mA) 2 2.09 W From the probe ac data, the output power is P0(ac) = VO(p—p)2/[8 - RL] = (6.85)2/[8 - 100] = 58 mW The efficiency is then %n = Po/Pi - 100% = (58 mW/2.09 W) - 100% = 2.8% A larger input signal would increase the ac power delivered to the load and increase the efficiency (the maximum being 25%). Program 15.2—Quasi-Corriplememary Amplifier Figure 15.31 shows a quasi—complementary push-pull class B power amplifier. For the input of Vi = 20 V(p), the output waveform obtained using probe is shown in Fig. 15.32. Figure 15.31 Quasi— complementary class B power amplifier. 15.10 Computer Analysis 785 I I I I I I | I I I I l I I I | I I I I N o < + I I I I l I I l I I I I I I 1 I I I I ,__ l I I I I I I I J I I I I l I l I I I I I I I .1. Time Figure 13.32, Probe output of the circuit in Fig. 15.31, The resulting ac output voltage is seen to be V0(p-p) = 33.7 V so that P0 = V§(p-p)/(8 - RL) = (33.7 V)2/(8 - 8 Q) = 17.7 w The input power for that amplitude signal is P1- = Vccldc = Vcc[(2/w)(Va(p-p)/2)/RL] = (22 V) - [(2/w)(33.7 V/2)/8] = 29.5 w The circuit efficiency is then %n = Po/P,- 100% = (17.7 W/29.5 W) ~ 100% = 60% v‘. v e, Fr r' +2, —. , *9“, w... ‘ ..- <- qu- H / “ups -. »¢,;.. :' v—‘flfia .l"flt:}/s‘-.:‘ :3] "i ‘Vg- [111;KL’1_..\;:J~:.;. I M Figure 15.33 shows an op—amp push-pull amplifier providing ac output to an 8-0 load. As shown, the op-amp provides a gain of A. = —RF/R1 = —47 141/1st = —2.6 ;22V 2N3904 T 4 VII kHz/0 Deg —: fil— . 2N3906 Figure £5.33 Op-arnp class B amplifier. Chapter 15 Power Amplifiers 1 3'Arisir: Erwin: blisuiilélm x [ml-e] Fig. 15.33.ewb 1.022.053.054.115.146.177.20000E}m Tune (seconds) the circuit of Fig. 15.33. For the input, V,- = 1 V, the output is Vo(p) = AM = -2.6 - (1 V) = —2.6 v Figure 15.34 shows the oscilloscope display of the output voltage. The output power, input power, and circuit efficiency are then calculated to be P0 = V§(p—p)/(8 - RL) = (20.4 V)2/(8 - 8 Q) = 6.5 w The input power for that amplitude signal is Pi = VCCIdc = VCC[(2/7T)(V0(P'P)/2)/RLi = (12 V) - [(2/77) - (20.4 V/2)/8] = 9.7W The circuit efficiency is then Figure 13.34 Probe output for %n = PO/Pi- 100% = (6.5 W/9.7 W) . 100% = 67% MATH CAD The calculations for the class B power amplifier of Example 15.7 and for the class B power amplifier of Example 15.12 are shown below. Using Mathcad, one can enter any desired value of VCC, RL, and VL peak, with all the calculations immediately providing the new results. Class-B Power Amplifier (Example 15.7) vcc -.= 30 RL1=16 VLpeak := 20 ‘_ Vchak “‘pcak " RL ILpeak = 1.25 [do 2:2 mm" ldc = 0.796 3.14159 Pidc :=VCC-ldc Pidc = 23.873 2 Poac t= VLpeak Poac = l2.5 (2.12m / ‘1 n := 5 P03“ {-100 n = 52.36 1‘ Pidc ,1 Class-B Power Amplifier (Example 15.12) VCC :=25 RL :4 1/ , 2“ maxPidc IAN/i (3.l4159)‘RL ,1 7‘» .V “,1 maxPoac := ‘ CC ’ (2~RL) n ._ ;' maxPoac ‘1 1 max Pidc ,9 lOO PZQ :=maxPidc — maxPoac The harmonic distortion calculations of Examples 15.13 and 15.14 are shown for a select set of values for A1 through A4. 15.10 Computer Analysis maxPidc = 99.472 maxPoac = 78125 n = 78.54 P2Q = 21.347 1787 ...
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This note was uploaded on 05/25/2011 for the course ELECTRONIC 311 taught by Professor Umutsezen during the Fall '11 term at Hacettepe Üniversitesi.

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nashelsky_15_ca - Using Design Center, the circuit of a...

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