Question

Refer to the following circuit (Q1 is a generic transistor), assuming β = 200 when the transistor is in the active region.
(a) If R1 is open, calculate VB, VE, VC.
(b) If R1 is short, calculate IB, IE and VC.
(c) If R2 is short, calculate IE and VC.
(d) If R3 is short, calculate VB, IC, and IB.

R1 47ΚΩ R3 3.3kΩ 12V Vcc Q1 R2 12kΩ R4 1.0kΩ

Answer 1

(a) If R is open R3 Vo 12V Rh Case In this Voltage at Bage is zero so the transistor Qi is in ait off region. when But in cut off IB=0, Ic=0, TE Ve=12V VB-OV, VE-OV, shooted b) I f R, is Wir R3 12V Yв VBE VE 2 R2 :R4

Vo = 12 V VE VE - - VB VBE 12-07 VE - 11.3V Vc = Vcc-Ic Ro VE 11.3m 4 IES - 11.3 16 RE Ica IE Vc = 12- 11.3 mx 33K Vc -25.29v the BJT in Saturation since Vio regionon -0.2V. VCE (sat) VCE Hence TVB =12v VE - 11.3V Ve= VE + VCE - 11.3 +0.2 Vc 11.5 V

11:3 IE= 11.3mA VE RE IB = IE B+ 11.3mA 201 = 0.0565mA VCE 2 11.5V R3 Ri 12V VB VBE Ry since is in cut off region VB-0 the BJT IE=0 Vc = Vec- Ic Rc = 12 -0.R3 Vc - 12V

If Rg is short R 120 VB Ru Rule by Voltage devision Vcc. R2 VB2 12 x 12k 4th+12k BitR2 VB = 2.44 v] bias devider Base current of Voltage circuit is IB = Vigo - VBE RB + B+R4 R, R2 u7x12 k where - RB = Rilla 47+12 RitR2 RB = 9.56k.

. 2.44-0.7. IB = 9.56k + 201816 1074 210.56k Is - 0.008264 mA IB - 8.264 AA Ic = BIB = 200x 8-264M Ic = 1.653 mA