Module 2 Unit 2 TRANSISTOR (BJT) Review Questions 1 Define current gains α and β. How are they related? 2 In transistor design, the base has most critical features as compared to emitter and collector. Discuss. 3 Why do we bias a transistor? What are the considerations in choosing an appropriate biasing scheme? 4 Draw I - V characteristics (Collector characteristics) for a transistor and discuss the salient features. 5 Explain ‘base width modulation’ and its influence on transistor characteristics. 6 What is collector feedback bias? How does this biasing provide stability to the circuit? 7 Give reasons for the wide use of ‘voltage divider bias’ in BJT amplifiers. 8 Discuss the flow of three currents IE, IB and IC in a forward biased emitter junction and reverse biased collectior junction. Problems 2.1 A transistor has current gain of 0.99 when used in common base (CB) configuration. How much will be the current gain of this transistor in common emitter (CE) configuration ? Solution :- The current gain in common base circuit is written as α, and it has been given equal to 0.99. α and current gain in common emitter configuration β are related as 1 Therefore, 0.99 = 99 1 0.99 or β = 99 2.2 Determine the minimum value of current gain β required to put the transistor in saturation when Vin = +5V. Assume, VBE(sat) = 0.8 V, VCE(sat) = 0.12 V +12V V 5k IB Vin 80k k VBE IC + VCE - Solution:- Let IB and IC be base current and collector current respectively (see fig. above). Taking +5V as input voltage Vin, the summation of voltages (Kirchhoff’s voltage law, KVL) in the base-emitter loop gives 8 X 103 X IB + VBE – 5V = 0 We assume that the transistor is in saturation, so that VBE = VBE(Sat) = 0.8 V Therefore , 80 X 103 X IB + 0.8 V – 5V = 0 or , IB 4.2V 0.0525mA 80 10 3 Summation of voltages in the collector loop gives, 5 X 103 X IC + VCE = 12 V or IC 12 0.2 2.36mA 5 103 And the condition for the transistor in saturation is that the minimum value of gain β , 2.36mA min I C I 45, B 0.525mA or βmin = 45. 2.3 The fixed bias circuit shown in figure uses a silicon transistor with VBE = 0.7V. (a) Find the collector current, IC, and voltage VCE, if β of transistor is 60. (b) Find IC and VCE if β changes to 80. What conclusions may be drawn? +VCC (9V) 60k k 0.5k VBE + VCE - Solution:(a) Summation of voltages in base-emitter loop results in VCC = RBIB + VBE or I B VCC VBE 9 0.7 0.138 mA RB 60 10 3 and, IC = βIB = 60 X 0.138 mA IC = 8.28 mA Summation of voltages in the collector circuit gives, RCIC +VCE = VCC Or VCE = VCC – ICRC = 9 - 8.28 X 10-3 X 0.5 X 103 = 9 – 4.14 Or VCE = 4.86 V Thus the operating point values are IC = 8.28 mA , and VCE = 4.86 V (b) Proceeding in the same way, and if the current gain β equals 80, then Since IB = 0.138 mA, therefore IC = βIB = 80 X 0.138 mA or IC = 12.42 mA And VCE = VCC- ICRC = 9 – 12.42 X 10-3 X 0.5 X 103 = 9 – 6.21 VCE = 2.79 V With β = 80, the operating point values are : IC = 12,42 mA, and VCE = 2.79 V CONCLUSIONS:Changes in collector current IC when β changes from 60 to 80 are from 8.28 mA to 12.42 mA. And VCE changes from 4.86V to 2.79V. These changes are drastic and operation may shift to regions to give distorted output. 2.4 Find out the operating point current ICQ, and voltage VCEQ in the circuit shown. (VBE = 0.7 V, β of transistor is 200). +9V(VCC) 2.5k 50k k RE RC IC + VCE 4.3k -9V(VEE) Solution:Considering the potential difference across the resistor RE, the emitter current IE (≈IC) may be written as, I E IC VEE VBE 9 0.7 RE 4.310 3 or IC = ICQ = 1.93 mA Voltage summation in the collector circuit leads to, VCC = RCIC + VCE or VCE = VCC – RCIC = 9 – 2.5 X 103 X 1.93 X 10-3 = 9 – 4.83 or VCE = VCEQ = 4.17V 2.5 Calculate the approximate value for the base resistor RB which will forward bias the emitter junction of silicon transistor (β = 100) in the circuit. Collectior – emitter voltage VCE of 2.5 V reverse biases the collector. (VBE = 0.7V) +5V RB IC VBE 1.5k + VCE IE 1k Solution:We assume that the emitter junction is in forward bias and VBE = 0.7V. Summation of voltages in the base-emitter circuit results in, RBIB +VBE + REIE = VCC Since IC = IE, and I B or I B IC IE Then above equation yields, RB . IE VBE RE I E VCC R or I E B RE VCC VBE 5.0 0.7 R or I E B RE 4.3 ( A) Now, summation of voltages in the collector – emitter circuit gives, RCIC + VCE + REIE = VCC Because IC = IE And substituting for other parameters IE(RC + RE) = VCC – VCE = 5.0 – 2.5 = 2.5 or I E 2.5 2.5 1 mA RC RE 1.5 1 10 3 Substituting IE = 1mA in eqn (A) R 110 3 B 110 3 4.3 100 RB = 330 X 103Ω or RB = 330 kΩ 2.6 The operating point values of current IC(=ICQ) and voltage VCE(=VCEQ) in the circuit have magnitudes of 0.9 mA and 3.72 V respectively when the current gain β for the transistor is 100. The transistor in the circuit is replaced by another one with β = 200. Calculate the new values of ICQ and VCEQ. What do you infer? +6V (VCC) R1 R2 4k 2k RC 0.7V 1k IC + VCE 1.3k Solution:Thevenized voltage Vth across R2 is VTH 2 10 3 VCC 2 10 3 4 10 3 2 6 6 2V That is VTH = 2V And, as we have derived the relation earlier, I E I CQ VTH VBE RE RTH Where Thevenized resistance RTH is RTH =R1 R2 = 4k 2k = 1.33 kΩ, Then with β= 200, 2 0.7 1.310 0.006 10 3 0.995 mA I E I C I cQ or I CQ 3 Further, summation of voltage in the collector circuit under the condition IC = IE leads to, VCC = IC (RC + RE) + VCE Or, VCE = VCEQ = VCC-IC (RC + RE) = 6V – 0.995 mA (1k + 1.3k) Or VCEQ = 3.71V Inference: The circuit is highly stable as the change in collector current is 0.995 – 0.99 = 0.005 mA only (change is ≈0.5%) which is negligible. Similarly VCE changes only 0.3% which is also negligible when β changes by 100% i.e from 100 to 200. (2.7) Calculate the value of resistor RB in the circuit shown to put VCE at 3.0V (VBE = 0.7V and β = 80) +9V RB IB RC 2k IC + IE VBE RE 3k Solution:Summation of voltages in the base-emitter circuit gives, RBIB + VBE + REIE = VCC Now, VBE = 0.7 V, and IC = IE. Also, I B IC IC 80 Substituting these in above equation, and taking resistances in kΩ, IC 3k I C VCC VBE 80 9 0.7 8.3 I or RB . C 3k . I C 8.3 ( A) 80 RB . Summation of voltages in the collector circuit results in, ICRC + VCE + IE X 3k = 9V or IC X 2k + 3.0V + IC X 3k = 9V or IC (2k +3k) = 9 - 3 = 6V or I C 6V 1.2mA 5k Substituting the value of IC in eqn (A) 1.2 3k 1.2 mA 8.3V 80 4.7 or RB 313.3 k 0.015 mA RB 313 k RB .