Question

ECT2601/101/3/2019 TRANSISTOR BIAS CIRCUITS Question 2 Find lo. Vcn qnd V for the pnp transistor in figure 2 using the Thevenins Theorem applied o voltage divider Bias. (12) [12) VCC 12V IC Rc 3kQ R1 vc 04 2N5684G BDC 150 VB RE R2 30k0 1

Answer 1

Answer:

When we apply Thevenin's theorem to Base-emitter circuit consisting R1, R2 and Vcc, we get

Rth = R1 || R2 = R1*R2/(R1+R2) = 50*30/80 = 18.75K Ohm

and Vth = Vcc * R2/(R1+R2) = -12*30/80 = -4.5V

The reflected value of RE from emitter to base circuit is Beta*RE = 150*2 = 300K

As this an order of magnitude higher than Rth, therefore for all practical purposes VB = -4.5V

IB is given by

IB = (Vth-Vbe)/(Rth+(1+beta)RE)

IB =( -4.5+0.7)/(18.75+151*2) = -0.01185mA = -11.85microamps

Also we know that IE = (1+beta)IB= -151*11.85 = -1789microamps = -1.79mA

This is also nearly equal to Ic

Therefore Ic = -1.79mA

Also VE = -1.79*2= -3.58V

and the voltage drop across Rc = Vrc= -1.79*3 = -5.37V

To find out Vce we apply KVL

VCC = Vrc+Vce + VE

Therefore

Vce = -12+3.58+5.37 = -3.05volts, or in others words, the magnitude of Vce is 3.05volts.

Hope this helps.