Question

(25pts) 2. Design a four resister BJT (CE) bias circuit (using method 1 or 2) for the following specifications: loq=1.5 mA, VCEQ=5 V, Vcc=14 V. Find all resistor values. Assume a BJT transistor (npn) with B=180 and V Be=0.7V Draw the designed circuit with all values.

Answer 1

© It is the voltage dividon bias coquit. Giren IQ=0.70 VCE2=5V, Vcc=l4V , B=18 1.5mA VBE=0.70 for designing the conditions are 0 UCE < Vcc 2 ܘܐ 370. ③ VE = 10 cc to - Vic ③ BRE> 10R2 we have vce=50 and Vcc = 14 VCE SUCC so first condition is satisfied +Ver to vec (ofrom second condition) VE 1.4 V Emitter Resistance Re= 1.4 1,4 1.4 IE (+8) IB, (1+BA to (+180)/105m RE 928.176 37 lo X14 VE = 180 IB = Ic A It = (1+B) IB » Collector Resistance, Rc = Vec-VCE Ic Rc 14-5 (1.5X103) Rc-662 From the third condition, PRE = 10R2 R, < PRE 10 Rac 180 x 928.476 lo Ra < 16.707KA CS Scanned with CamScanner

R = VCCX R2 R2 Given that VBE = 0.70 Vg-VE = 0.7V Vo =0.7+VE → VB: = 0.7+164 = 2.1V = V8=2.1V Vs is the therewin voltage and it is 14x 16.707 X10 Rit R2 R. + (11,707x103) Ritr, VCC X R2 VB 3 VB = Vcc X R2 B Ri VB 14x16707) - (16707) 2.1 94,673 kr The BJT circuit is vec=141 & RC=okef, Ic RI 94,67312 + VCE B क VRE R28 RE= 928.176 - NIE 16.7o7te CS Scanned with CamScanner