ObjectiveSchematicWhy is it called one-sided?Finding the Thevenin output resistanceEven and odd excitation methodFinding the Thevenin voltageEven and odd excitation methodThevenin Equivalent Circuit for One-sided Differential Amplifier Objective It is shown how to find the Thevenin equivalent circuit of the unsymmetrical, one-sided differential amplifier using even and odd symmetry excitations and the principle of superposition. Schematic -711.9mVQ_nQn109.910mAQ_nQn209.910mA0POUT0+R2{R_C}0V5.045V+-V30{V_A}DOT-MODEL:B_f = 100I_s = 10fAV_af = 50V0+R1{R_C}0VSweep+-ACV501VPARAMETERS:R_C = 500I_M = 20mAV_CC = 10VV_A = 0V10.00VP+-V31{V_CC}.model Q_n NPN (Is={I_s} Bf={B_f} Vaf={V_af})0I1{I_M} FIGURE 1 The one-sided differential amplifier for which the small-signal Thevenin equivalent is sought NAME Q_Qn20 Q_Qn10MODEL Q_n Q_nIB 9.00E-05 9.00E-05IC 9.91E-03 9.91E-03VBE 7.12E-01 7.12E-01VBC -5.05E+00 -5.05E+00VCE 5.76E+00 5.76E+00BETADC 1.10E+02 1.10E+02GM 3.83E-01 3.83E-01RPI 2.87E+02 2.87E+02RX 0.00E+00 0.00E+00RO 5.55E+03 5.55E+03CBE 0.00E+00 0.00E+00CBC 0.00E+00 0.00E+00CJS 0.00E+00 0.00E+00BETAAC 1.10E+02 1.10E+02CBX/CBX20.00E+00 0.00E+00FT/FT2 6.10E+18 6.10E+18 FIGURE 2 PSPICE output file for circuit of Figure 1 Create Date 2/6/06 by J R Brews Page 1 2/6/2006Why is it called one-sided? The circuit is called one-sided because it is excited only on one side, so it is neither symmetrical nor asymmetrical because the excitation is neither even nor odd symmetry. Finding the Thevenin output resistance A straightforward algebraic analysis of the one-sided circuit is a mess. So a “trick” solution is used. The idea behind it is shown in the following figures. 020.46V0V0VF2GAIN = 110TH+R2{r_O}86.31mAI11A+R1{r_O}3.665mA0+R4{r_PI}405.3uA0PARAMETERS:r_O = 5.55kr_PI = 287R_C = 500116.3mV+R3{r_PI}405.3uA+R6{R_C}958.3mA479.1V+R5{R_C}40.92mA00V+-E 0V41.32mAF1GAIN = 110 FIGURE 3 The small-signal circuit with the test current applied at the output Figure 3 shows the standard approach to finding the Thevenin resistance by putting a test source at the output. Using a one amp current source has the advantage that the Thevenin resistance is just the voltage across the current source, that is RTH = 479.1 Ω in this case. Even and odd excitation method Because the circuit is linear, we can add any two solutions to get a third solution (the principle of superposition). Therefore, we make the problem easy to solve algebraically by using the sum of an even and an odd solution, as shown in Figure 4. The even and odd mode excitations can be added to produce the one-sided excitation. The corresponding addition of the output voltages using the SUMMER produces the same Thevenin resistance as Figure 3. Create Date 2/6/06 by J R Brews Page 2 2/6/2006E_TH+R3o{r_PI}0A+-Eo 0V41.32mAE_TH+R4o{r_PI}0A+R4e{r_PI}405.3uA000V249.8V229.3V+R2o{r_O}0PARAMETERS:r_O = 5.55kr_PI = 287R_C = 500+R6e{R_C}+R5o{R_C}F1eGAIN = 110-++-E1GAIN = 1O_TH+R6o{R_C}116.3mV+R3e{r_PI}405.3uAF2eGAIN = 110I2o0.5A+R2e{r_O}0O_THI1e0.5AF1oGAIN = 110I2e0.5A00I1o0.5A+R1o{r_O}0SUMMER0-229.3V0+R5e{R_C}479.1V+-Ee 0V6.939e-18ATHF2oGAIN = 110+R1e{r_O}00249.8V0 FIGURE 4 Even mode excitation (top) and odd mode excitation (bottom) can be added to produce Figure 3 with zero current injected into the left side and 1A injected into the right side Finding the Thevenin voltage The Thevenin voltage can be found the same way. The standard approach is shown in Figure 5. The Thevenin voltage is VTH = 87.9 V. Create Date 2/6/06 by J R Brews Page 3 2/6/20060VTH+R2{r_O}15.75mAF2GAIN = 11087.90V0V+-E 0V177.5mA+R5{R_C}175.8mA499.8mV+R3{r_PI}1.743mA+R4{r_PI}1.741mA-87.90V0+R6{R_C}175.8mA+-V_in1V1.743mA1.000V0+R1{r_O}15.93mAPARAMETERS:r_O = 5.55kr_PI = 287R_C = 5000F1GAIN = 110 FIGURE 5 Standard approach to finding the Thevenin voltage as the open-circuit voltage at the output for a 1V input Even and odd excitation method Again we use superposition of an even and an odd excitation to simplify the analysis. Again, because the circuit is linear, we can add any two solutions to get a third solution (the principle of superposition). Therefore, we make the problem easy to solve algebraically by using the sum of an even and an odd solution, shown in Figure 6. The even and odd mode excitations can be added to produce the one-sided excitation. The corresponding addition of the output voltages using the SUMMER produces the same Thevenin voltage as Figure 5 Create Date 2/6/06 by J R Brews Page 4 2/6/2006+R3e{r_PI}810.6nA+-V20.5V810.6nA+-V30.5V810.6nATH00+R5e{R_C}PARAMETERS:r_O = 5.55kr_PI = 287R_C = 500+R6o{R_C}O_TH+R3o{r_PI}1.742mAF1eGAIN = 1100F2eGAIN = 110+R2e{r_O}89.98uAE_THO_TH+R4o{r_PI}1.742mAF1oGAIN = 110+R4e{r_PI}810.6nA00+R5o{R_C}-77.34e-18V0+-Eo 0V177.5mAE_TH0+R6e{R_C}-++-E1GAIN = 1F2oGAIN = 110+R1o{r_O}+-V5-0.5V1.742mA87.90V+R1e{r_O}89.98uA+R2o{r_O}87.90V0405.3uV+-V40.5V1.742mA499.8mVSUMMER+-Ee 0V27.76e-18A FIGURE 6 Even mode excitation (top) and odd mode excitation (bottom) can be added to produce Figure 5 with zero volts on the right side and 1V applied to the left side Create Date 2/6/06 by J R Brews Page 5