Question

352 Part 1 Semiconductor Devices and asic Applications sme Valon) assume Veacon) - 0.7 PROBLEMS (Note: In the following problems, unless otherwise stated, assum and cast) = 0.2 V for me transistors, and assume V Vec(sat) = 02 V for pup transistors region.aera . A Determine A and ig = 1.80 mA de. The colle 726 A and th 732 A. Delen di 2.961 is 110 P51 mine parande se current is 50 A Section 5.1 Basic Bipolar Junction Transistor 5.1 (a) In a bipolar transistor based in the forward-activere 28A rent is and the emitter current is the 325 a, and ic. (b) Repeat part (a) ifia - 20 mA and i 180 3.2 (a) A bipolar transistor is based is based in the forward-active mode for current is 726 A and the emitter current is mine , a, and is. (b) Repeat part(a)ific 2.902 mA and 53 a) The range of for a particular type of transistor is li Determine the corresponding range of or. (b) If the base determine the range of collector current 5.4 (a) A bipolar transistor is based in the forward-active mode. The parameters are 1.25 mA and P = 150. Determine (b) Repeat part (a) forie 4.52 mA and 8 = 80. (a) For the following values of common-base current gain a. den corresponding common-emitter current gain : ode. The met termine is. ic. and t gain a, determine the 0.990 0.995 0.9990 0.980 a 0.90 0.950 (b) For the following values of common-emitter current gain B. determin the corresponding common-base current gaina: B 20 50 100 150 220 400 5.6 An npn transistor with B = 80 is connected in a common-base configura tion as shown in Figure P5.6. (a) The emitter is driven by a constant-curren source with Tg = 1.2 mA. Determine 18. Ic., and Vc. (b) Repeat parti for lg = 0.80 mA. (c) Repeat parts (a) and (b) for B = 120. -- Rc=2k8 osonroolean reloald die w + IcSVALA Lidinions Figure P5.6

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Answer 1

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iB = 2.8 MA ie= 325mA When traneisfor is in achuc ngon I = ß IB diero I= IE - from groen data Ie from 0 e ① ß To= X IE 173 LAPPE - p= E —0 subehlahing gaen valves B= 325.2.8 = 115.0914 P= 115.07147 2.8 de 115 0714 -= 1+115.0914 0.99108 = 0.99137 Ice BIO 115.07144 2.8 322199 HA Ic= 322.198MA

"B= ZOMA ie= 1.8mA = 1800 MA B- Ic-iB Ba le 1800 - 20 -E89 150.20 = 89 = B= 89] IB-89 20 x= Bs 89_= 0.988 1 +D 1+89 = 220-9887 Ic= BIG - 89*20 - 1780HA | Ic= 1780 MA