Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
TUTORIAL 3 – Questions & Answers Q1 : The circuit shown in Figure 1 has transistor parameters = 120 and VA = . (a) Determine the small signal parameters, gm and r for both transistors. (b) Determine the overall small-signal voltage gain, Av = vo/vs. (c) Determine the input resistance Ris and the output resistance Ro. TUTORIAL 3 – Questions & Answers Q1 (cont’d) Figure 1 TUTORIAL 3 – Questions & Answers Q1 – Solution (a) The small signal parameters, gm and r for both transistors. DC analysis : Stage 1 RB1 R1 R2 67.3 12.7 10.68 k R2 VBB1 VCC R1 R2 12.7 12 1.905 V 67.3 12.7 TUTORIAL 3 – Questions & Answers Q1 – Solution (a) (Cont’d) VBB1 VBE1 I B1 RB1 1 1 RE1 1.905 0.7 10.68k 1 1202k 4.77 μA I C1 1 I B1 120 4.77μ 0.572 mA TUTORIAL 3 – Questions & Answers Q1 – Solution (a) (Cont’d) I C1 0.572m g m1 22 mA/V VT 26m r 1 1 g m1 1VT I C1 120 26m 0.572m r 1 5.45 k VA ro1 I C1 0.572m TUTORIAL 3 – Questions & Answers Q1 – Solution (a) (Cont’d) DC analysis : Stage 2 RB 2 R3 R4 15 45 11.25 k VBB 2 R4 VCC R3 R4 45 12 9V 15 45 TUTORIAL 3 – Questions & Answers Q1 – Solution (a) (Cont’d) I B2 VBB 2 VBE 2 RB 2 1 2 RE 2 9 0.7 11.25k 1 1201.6k 40.5 μA IC 2 2 I B2 120 40.5μ 4.86 mA TUTORIAL 3 – Questions & Answers Q1 – Solution (a) (Cont’d) gm2 I C 2 4.86m 0.187 A/V VT 26m r 2 2 gm2 2VT IC 2 120 26m 4.86m r 2 642 VA ro 2 I C 2 4.86m TUTORIAL 3 – Questions & Answers Q1 – Solution (b) The overall small-signal voltage gain, Av = vo/vs. First, consider the 2nd stage; V 2 Vo g m 2V 2 RE 2 // RL r 2 1 g m 2 RE 2 // RL V 2 r 2 V 2 1 2 RE 2 // RL r 2 TUTORIAL 3 – Questions & Answers Q1 – Solution (b) (Cont’d) and; Vo ' V 2 Vo 1 V 2 g m 2 RE RL V 2 r 2 V 2 r 2 1 2 RE RL r 2 TUTORIAL 3 – Questions & Answers Q1 – Solution (b) (Cont’d) Vo Av 2 Vo ' V 2 1 2 RE 2 RL r 2 r 2 1 2 RE 2 RL Vr 2 2 TUTORIAL 3 – Questions & Answers Q1 – Solution (b) (Cont’d) 1 2 RE 2 RL Av 2 r 2 1 2 RE 2 RL 1 1201.6 0.25 0.642 1 1201.6 0.25 0.976 TUTORIAL 3 – Questions & Answers Q1 – Solution (b) (Cont’d) To find Av1, first find Rib2 and Ri2; TUTORIAL 3 – Questions & Answers Q1 – Solution (b) (Cont’d) Rib 2 r 2 1 2 RE 2 RL 0.642 1 120 1.6 0.25 26.8 k TUTORIAL 3 – Questions & Answers Q1 – Solution (b) (Cont’d) Ri 2 R3 R4 Rib2 15 45 26.8 7.9 k Next , replace the 2nd stage with its equivalent input impedance, Ri2; TUTORIAL 3 – Questions & Answers Q1 – Solution (b) (Cont’d) Vo ' gm1V 1 RC1 Ri 2 Vs V 1 Vo ' Av1 g m1 RC1 Ri 2 22m10 7.9 k 97.1 Vs Av Av1 Av 2 97.1 0.976 94.8 TUTORIAL 3 – Questions & Answers Q1 – Solution (c) The input and the output resistances TUTORIAL 3 – Questions & Answers Q1 – Solution (c) (Cont’d) Ris RB1 r 1 R1 R2 r 1 67.3 12.7 5.45 3.61 k TUTORIAL 3 – Questions & Answers Q1 – Solution (c) (Cont’d) The small-signal equivalent circuit of the complete amplifier; TUTORIAL 3 – Questions & Answers Q1 – Solution (c) (Cont’d) The equivalent circuit may be simplified as follows; TUTORIAL 3 – Questions & Answers Q1 – Solution (c) (Cont’d) To find Ro, set Vs to zero and apply Vx at the output port; TUTORIAL 3 – Questions & Answers Q1 – Solution (c) (Cont’d) Since Vs is set to zero, RB1 and r1 are shorted, V1 and gmV1 are both equal to zero. The equivalent circuit becomes; I e 2 I x I b 2 g mV 2 TUTORIAL 3 – Questions & Answers Q1 – Solution (c) (Cont’d) 1 Vx V 2 I x g m V 2 I x 1 2 RE 2 r 2 r 2 r 2 V 2 Vx r 2 Ro1 Vx Vx I x 1 2 RE 2 r 2 Ro1 TUTORIAL 3 – Questions & Answers Q1 – Solution (c) (Cont’d) 1 r 2 Ro1 1 2 RE 2 1 2 Ix Vx Vx r 2 Ro1 RE 2 RE 2 r 2 Ro1 TUTORIAL 3 – Questions & Answers Q1 – Solution (c) (Cont’d) I x r 2 Ro1 1 2 RE 2 r 2 Ro1 RE 2 Vx Vx r 2 Ro1 RE 2 I x r 2 Ro1 1 2 RE 2 TUTORIAL 3 – Questions & Answers Q1 – Solution (c) (Cont’d) r 2 Ro1 RE 2 Vx r 2 Ro1 1 2 RE 2 || Rx 1 2 I x r 2 Ro1 R E2 1 2 TUTORIAL 3 – Questions & Answers Q1 – Solution (c) (Cont’d) Substituting values; Ro1 RC1 R3 R4 10 15 45 5.3 k 0.642 5.3 Rx 1.6 || 1.6 || 0.049 48 1 120 TUTORIAL 3 – Questions & Answers Q2 : For the circuit in Figure 2, assume 1 = 120, 2 = 150, VA = and VBE (ON) = 0.7 V. Find: (a) The small-signal parameters, gm and r for both transistors. (b) The overall small-signal voltage gain Av = vo/vs. TUTORIAL 3 – Questions & Answers Q2 (cont’d) Figure 2 TUTORIAL 3 – Questions & Answers Q2 – Solution (a) The small-signal parameters, gm and r for both transistors. VBB R2 (VCC VEE ) VEE R1 R2 45 5 (5) 5 80 45 1.4 V RBB R1 R2 80(45) 28.8 K R1 R2 80 45 TUTORIAL 3 – Questions & Answers Q2 – Solution (a) (Cont’d) To find IB1, perform KVL around the baseemitter loop of Q1; VBB RBB I B1 VBE1 RE1 I E1 V Rearrange the equation, RBB I B1 1 1I B1 RE1 VBB V VBE1 TUTORIAL 3 – Questions & Answers Q2 – Solution (a) (Cont’d) Substituting values; 28.8 103 I B1 120 12 103 I B1 1.4 5 0.7 Hence; I B1 10.7A and; I C1 1I B1 1.285mA TUTORIAL 3 – Questions & Answers Q2 – Solution (a) (Cont’d) To find IB2; The voltage drop across the resistor RC1 equals the voltage drop across RE2 plus the voltage VEB2. RE 2 I E 2 VEB2 RC1I1 where; I1 I C1 I B 2 I E 2 2 1I B 2 TUTORIAL 3 – Questions & Answers Q2 – Solution (a) (Cont’d) 2 1RE 2 I B 2 VEB2 RC1 I C1 I B 2 Susbtituting values; 150 13 103 I B 2 0.7 4 103 1.285 103 I B 2 TUTORIAL 3 – Questions & Answers Q2 – Solution (a) (Cont’d) I B 2 9.716 μA I C 2 2 I B 2 1.457 mA Thus, finding the transistor parameters; g m1 I C1 1.285 49.42 mA/V VT 26 r 1 1VT I C1 120(26) 2428 1.285 TUTORIAL 3 – Questions & Answers Q2 – Solution (a) (Cont’d) g m2 I C 2 1.457 56.0 mA/V VT 26 r 2 2VT IC2 150(26) 2676.7 1.457 TUTORIAL 3 – Questions & Answers Q2 – Solution (b) The overall small-signal voltage gain Av = vo/vs. A A1 A2 g m1 RC1 // r 2 g m 2 RC 2 // RL A g m1 g m 2 RC1 // r 2 RC 2 // RL 49.4(56)( 4 2.676)(1.5 4.5) 4990 V/V TUTORIAL 3 – Questions & Answers Q3 : Figure 3 shows an amplifier in cascode configuration where a commonbase stage (Q2) is being driven by a common-emitter stage (Q1). (a) Draw the circuit under dc condition and determine the quiescent currents (ICQ1 and ICQ2) of both transistors. (b) Draw the small-signal equivalent circuit of the complete amplifier and perform an ac analysis to show that the approximate overall small-signal voltage gain is given by the expression; Av vo g m1 RC // RL vi (c) If the peak-to-peak value of the input voltage is 30 mV, determine the approximate peak-to-peak value of the output voltage . Assume; 1 2 150; VBE1active VBE 2active VBE active 0.7 V; VA1 VA2 VA ; VT 26 mV and VCC 15 V TUTORIAL 3 – Questions & Answers Q3 (cont’d) Figure 3 TUTORIAL 3 – Questions & Answers Q3 – Solution TUTORIAL 3 – Questions & Answers Q4 : Determine the input impedance Ri and output impedance Ro of the twostage amplifier shown in Figure 4. Assume VCC = +5 V, VEE = -5 V, 1 = 127, 2 = 200, VBE(active) for both transistors is 0.7 V and VT = 26 mV. TUTORIAL 3 – Questions & Answers Q4 (Cont’d) Figure 4 TUTORIAL 3 – Questions & Answers Q4 – Solution VBB VCC R2 VEE VEE R1 R2 47 5 5 5 1.8 V 100 47 R1 R2 RB R1 R2 100 47 32 k 100 47 TUTORIAL 3 – Questions & Answers Q4 – Solution Taking the base-emitter loop of Q1; VBB RB I B1 VBE1 RE1I E1 VEE 0 Or; RB I B1 1 1RE1I B1 VBB VEE VBE1 Substituting values; 32 103 I B1 127 12 103 I B1 1.8 5 0.7 I B1 2.6 9 μA 3 288 10 TUTORIAL 3 – Questions & Answers Q4 – Solution I C1 1 I B1 127 9 10 6 1.15 mA For the RC1 - collector of Q1 - base of Q2 - loop; RE 2 I E 2 VEB2 RC1I1 I 1 I C1 I B 2 I E 2 2 1I B 2 2 1RE 2 I B 2 VEB2 RC1 I C1 I B 2 TUTORIAL 3 – Questions & Answers Q4 – Solution Substituting values; 200 12 103 I B 2 0.7 5 103 1.15 103 I B 2 I B2 5.05 12.4 μA 3 407 10 IC 2 2 I B2 200 12.4 10 6 2.48 mA TUTORIAL 3 – Questions & Answers Q4 – Solution g m1 I C1 1.15 44.2 mA/V VT 26 r 1 g m2 1 g m1 127 2.87 k 0.0442 I C 2 2.48 95.4 mA/V VT 26 r 2 2 g m2 200 2.1 k 0.0954 TUTORIAL 3 – Questions & Answers Q4 – Solution Rib r 1 1 1 RE1 2.87 1 1272 258.87 k TUTORIAL 3 – Questions & Answers Q4 – Solution Ri RB Rib 32 258.87 28.5 k TUTORIAL 3 – Questions & Answers Q4 – Solution Ro RC 2 1.5 k