Question

ASAP!

Question 1 [Soalan 1] (a) Describe the condition when a npn BJT transistor operates in saturation condition and what are the terminal currents and voltages conditions during saturation. [Terangkan keadaan bila satu transistor BJT npn beroperasi dalam keadaan tepu dan apakah keadaan arus dan voltan terminal semasa tepu. ] (20 Marks/Markah) (b) Consider the BJT transistor circuit in Figure 1. If Bpc = 100 and VBE = 0.65V, calculate: [Pertimbangkan litar transistor BJT dalam Rajah 1. Jika Bpc = 100 dan Vaz = 0.65V, kirakan:) i Base (IB), collector (Ic) and emitter (le) currents [Arus dasar (13), pengumpul (IC) dan pemancar (IE)] (25 Marks / Markah) ii. The voltages VB, VC, VE, Vce and VCB [Voltan-voltan V3. VC. VE Vc and Vcs] (40 Marks / Markah) 111. If the power dissipation derating factor for the transistor is 2mW/°C and the PD(max) is 0.5 W at 25°C. What is PD(max) at 100°C? [Jika faktor nyahkadaran pelesapan kuasa untuk transistor tersebut ialah 2mWPC dan PD mars talah 0.5 W pada 25°C. Apakah Pdmar) pada 100°C?] (15 Marks / Markah)

Answer 1

Given circuit, 5. Ic No 207 2012 B IB ABEE VBB=5v Bose- Emitter loop, KVL across the a) Apply -5+ 281201) + VBE = 0 Is(20k) = 5-0.65 IB = 435 = 0.2175X10 20 k so IB- 0.2175mA Given Bor 100; anapotele 100% 0.2175x103 Ic - B Tg = le : 21.75X10A to Collector Emitter loop, Apply K&L across the LE=IB + T = 0.215m + 21.75 m IE * 21.9 675mA

- 20+ Ic (900) + VCE = 0 Vce = 20- (21.75x103500) • 20 - (21.75X0.5) 20-10.875 VCE 29.125/ VCEL VCE(sat). The ondition for the circuit to be in saturation is VCE(sat) = 0.2 V; Vce = 9.125 V and thus the circuit is in The condition is not satisfied active region. that the circuit is in saturation, then It we assume NCE = 0.2V 19.8 20-0.2 - 39.6x103 20-VCE 500 1. 500 I c = 39.6mA in value of 'To not be There will any change that is 18 = 0.2175m A IE= IB + 1 = 0.2175m + 39.6m TE = 39.8175 mA

Question ) are ( () and (i) Questions already done in (mi). Given deo QmWloc dT Por-Po, am who le, - 1, -25°c) Given Po, = 0.5W at at I2 = 100°c; Pos? Por-0.5 3 E 2 X 10 Here 100 - 25 Pp2 -0.5 = 2x75x10 3 Pa -0.5 = 0.15 Pez = 0.15+0.5 Pex=0 -0.65W at To=100C