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TUTORIAL 4 – Questions & Answers Q1. The power amplifier in Figure 1 is required to deliver 10 W to the load resistance RL at a single frequency sinusoidal input signal, vI. Both QN and QP are matched transistors and have the following parameters; ISQ = 1.45 x 10-12 A and = 50 D1 and D2 are identical diodes with ISD = 1.45 x 10-12 A . The diode current is assumed to follow the exponential law; I D I SDeVD / VT and the transistor collector current is assumed to follow the exponential law; I C I SQeVBE /VT where; VT 26 mV TUTORIAL 4 – Questions & Answers Q1. (cont’d) (a) Determine the value of Ibias to ensure the diode current iD 1 mA at any time. (b) Calculate the collector currents of QN and QP when vO = 0 . You may assume that iBN = iBP = 0 under this condition. (c) Determine the required amplitude of the input voltage, vI. TUTORIAL 4 – Questions & Answers Q1. (cont’d) Figure 1 TUTORIAL 4 – Questions & Answers Q1 – Solution (a) The value of Ibias to ensure the diode current iD 1 mA at any time. The r.m.s value of output voltage is given by; vO rms R 2 P 10 W vO rms 10 8 W 8.944 V TUTORIAL 4 – Questions & Answers Q1 – Solution (a) (cont’d) The peak value is; vO p 2 8.944 12.65 V The peak load current is; iL p VO p RL 12.65 8 1.581 A TUTORIAL 4 – Questions & Answers Q1 – Solution (a) (cont’d) This load current is supplied by QN i.e. iN. Hence; iN p iCN p 1.581 A and iBN p iCN p 1.581 31.6 mA 50 TUTORIAL 4 – Questions & Answers Q1 – Solution (a) (cont’d) To guarantee a minimum of 1 mA diode current, select I bias 33 mA TUTORIAL 4 – Questions & Answers Q1 – Solution (b) The collector currents of QN and QP when vO = 0. When vO = 0, the whole of Ibias flows through the diodes (assuming iBN is negligible). ID ISe VD / VT or; ID VD VT ln IS TUTORIAL 4 – Questions & Answers Q1 – Solution (b) (cont’d) Substituting values; 33 10 3 VD 26 ln 12 1.45 10 620 mV Since the diodes are identical, VBB 2VD 1.24 V TUTORIAL 4 – Questions & Answers Q1 – Solution (b) (cont’d) and; VBEN VEBP VBB 620 mV 2 iCN 0 iCP 0 I SQeVBE / VT 1.45 10 12 e 620/ 26 32.9 mA TUTORIAL 4 – Questions & Answers Q1 – Solution (c) The required amplitude of the input voltage, vI. From (a), the amplitude of vO is 12.65 V and the corresponding base current iBn is 31.6 mA. The diode current is therefore; iD I bias iBN 33 31.6 1.4 mA TUTORIAL 4 – Questions & Answers Q1 – Solution (c) (cont’d) And at this value of diode current; ID VD VT ln I SD 1.4 103 26 ln 12 1.45 10 538 mV VBB 2VD 1.075 V TUTORIAL 4 – Questions & Answers Q1 – Solution (c) (cont’d) The peak load current occurs at the peak value of the output voltage which corresponds to the peak value of the input voltage. This current approximately equals the emitter current iN. Hence; iL iN 1.581 A and iCN 1.581 1 50 1.581 1.55 A 50 1 TUTORIAL 4 – Questions & Answers Q1 – Solution (c) (cont’d) Using the equation vBEN I C I S eVBE / VT iCN VT ln I SQ 1.55 26 ln 12 1.45 10 720 mV TUTORIAL 4 – Questions & Answers Q1 – Solution (c) (cont’d) vI VBB vBEN vO Substituting values; vI 1.075 0.72 12.65 vI 12.3 V TUTORIAL 4 – Questions & Answers Q2. Class-AB power amplifier in Figure 2 employs input buffer transistors Q1 and Q2 to establish the required quiescent bias current. Q1, Q2, Q3 and Q4 are matched transistors. Derive an expression for the approximate current gain, Ai where; iO Ai iI State any assumption made in your derivation. TUTORIAL 4 – Questions & Answers Q2. (cont’d) Figure 2 TUTORIAL 4 – Questions & Answers Q2 – Solution Applying KCL at the input node; iI iB 2 iB1 Assuming iB3 is negligibly small compared to iE1; V VEB1 vI iB1 1 1 R TUTORIAL 4 – Questions & Answers Q2 – Solution Assuming iB4 is negligibly small compared to iE2; iB 2 vI VBE 2 V 1 2 R TUTORIAL 4 – Questions & Answers Q2 – Solution Hence; iI iB 2 iB1 vI VBE 2 V V VEB1 vI 1 2 R 1 1 R Since all the transistors are matched; 1 2 ; vI VBE V V VEB vI iI 1 R 2vI V V 1 R VBE VEB V V TUTORIAL 4 – Questions & Answers Q2 – Solution Also, since V V 2vI V V 2vI iI 1 R 1 R V V TUTORIAL 4 – Questions & Answers Q2 – Solution Since the voltage gain of the amplifier is assumed unity, vO vI and; vO vI iO RL RL V V TUTORIAL 4 – Questions & Answers Q2 – Solution The current gain is; iO vI 2v I 1 R Ai iI RL 1 R 2 RL In deriving the above expression, the following assumptions were made: (a) All transistors are perfectly matched; (b) The voltage gain of the system is unity; and (c) The base currents of Q3 and Q4 are negligibly small as compared to the emitter currents of Q1 and Q2 respectively. TUTORIAL 4 – Questions & Answers Q3. All the transistors in class-AB power amplifier shown in Figure 3 are matched. If = 40, VBE(npn) = VEB(pnp) = 0.7 V and V+ = V = 12 V; (a) determine iE1, iE2, iB1 and iB2 when vI = 0; (b) determine iO, iE1, iE2, iB1, iB2 and iI when vI = 5 V; (c) use the results of part (b) to determine the current gain, Ai where; iO Ai iI TUTORIAL 4 – Questions & Answers Q3. (cont’d) Figure 3 TUTORIAL 4 – Questions & Answers Q3 – Solution (a) iE1, iE2, iB1 and iB2 when vI = 0; Since all transistors are perfectly matched, all their respective quiescent currents (when vI = 0) are equal i.e. iB1 iB 2 iB 3 iB 4 iC1 iC 2 iC 3 iC 4 i E1 i E 2 i E 3 i E 4 TUTORIAL 4 – Questions & Answers Q3 – Solution (a) (cont’d) V vEB1 iR R 12 0.7 250 45.2 mA TUTORIAL 4 – Questions & Answers Q3 – Solution (a) (cont’d) iR iE1 iB 3 iE 3 iE 1 1 iE1 iE 1 1 1 iE1 1 1 TUTORIAL 4 – Questions & Answers Q3 – Solution (a) (cont’d) iE 2 iE 1 1 iR 2 1 40 45.2 2 40 44.1 mA TUTORIAL 4 – Questions & Answers Q3 – Solution (a) (cont’d) iB 2 iB1 iE1 1 44.1 1 40 1.08 mA TUTORIAL 4 – Questions & Answers Q3 – Solution (b) iO, iE1, iE2, iB1, iB2 and iI when vI = 5 V; vO vI 5 V vO iO RL 5 0.625 A 8 iE 3 iO 0.625 A TUTORIAL 4 – Questions & Answers Q3 – Solution (b) (cont’d) Consider the Q1 – Q3 pair; iE 3 iB 3 1 0.625 15.2 mA 1 40 V vEB1 vI iR R 12 0.7 5 25.2 mA 250 TUTORIAL 4 – Questions & Answers Q3 – Solution (b) (cont’d) iE1 iR iB3 25.2 15.2 10 mA iE1 iB1 1 10 0.244 mA 1 40 TUTORIAL 4 – Questions & Answers Q3 – Solution (b) (cont’d) Consider the Q2 – Q4 pair; vI vBE 2 V iR R 5 0.7 12 250 65.2 mA Neglecting iB4; iE 2 iR 65.2 mA TUTORIAL 4 – Questions & Answers Q3 – Solution (b) (cont’d) iE 2 1 65.2 1.59 mA 1 40 iB 2 iI iB 2 iB1 1.59 0.244 1.35 mA TUTORIAL 4 – Questions & Answers Q3 – Solution (c) The current gain, Ai; Ai iO iI 0.625 463 53.3 dB 0.00135 Comparing with the expression of approximate gain; Ai 1 R 2 RL 1 40250 641 56.1 dB 28 TUTORIAL 4 – Questions & Answers Q4. All the transistors in class-AB power amplifier shown in Figure 4 are matched. The parameters of the transistors are: = 60 and IS = 5 x 10-13 A . Let V+ = 10 V and V = 10 V. (a) Determine the quiescent collector currents in the four transistors when vI = vO = 0. (b) If RL = 200 and vO(peak) = 6 V, determine the current gain, Ai and the voltage gain, Av of the circuit where; iO Ai iI and vO Av vI TUTORIAL 4 – Questions & Answers Q4. (cont’d) Figure 4 TUTORIAL 4 – Questions & Answers Q4 – Solution (a) The quiescent collector currents in the four transistors when vI = vO = 0. iE1 iB 3 3 mA iE 3 iE1 3 mA 1 Because the transistors are matched; iE1 iE 3 iE TUTORIAL 4 – Questions & Answers Q4 – Solution (a) (cont’d) Hence; 1 iE 1 3 mA 1 1 iE 3 2 1 60 3 2 60 2.952 mA TUTORIAL 4 – Questions & Answers Q4 – Solution (a) (cont’d) iE 1 2.952 48.4 μA 1 60 iB iC iB 60 48.4 μA 2.903 mA TUTORIAL 4 – Questions & Answers Q4 – Solution (b) The current gain and voltage gains when RL = 200 and vO(peak) = 6 V. vO 6 iO 30 mA RL 200 iE 3 iO peak 30 mA iE 3 iB 3 1 30 0.492 mA 1 60 TUTORIAL 4 – Questions & Answers Q4 – Solution (b) (cont’d) iE 1 3 i B 3 3 0.492 2.51 mA iE 1 1 2.51 41.1 μA 1 60 iB1 TUTORIAL 4 – Questions & Answers Q4 – Solution (b) (cont’d) Neglecting iB4; iE 2 3 mA iE 2 iB 2 1 3 49.2 μA 1 60 iI iB 2 iB1 49.2 41.1 8.1 μA TUTORIAL 4 – Questions & Answers Q4 – Solution (b) (cont’d) iO 30 Ai 3704 iI 0.0081 TUTORIAL 4 – Questions & Answers Q4 – Solution (b) (cont’d) VBE 3 iE 3 VT ln iS 30 10 3 0.026 ln 13 5 10 VBE3 0.6453 V TUTORIAL 4 – Questions & Answers Q4 – Solution (b) (cont’d) iE 1 VEB1 VT ln iS 2.508 10 3 0.026 ln 13 5 10 VEB1 0.5807 V TUTORIAL 4 – Questions & Answers Q4 – Solution (b) (cont’d) vI vO VBE 3 VEB1 6 0.6453 0.5807 6.0646 V Av vO vI 6 6.0646 0.989