Question

Design a universal bias for class -A CE-amplifier using NPN transistor for the following parameters values;Vcc=20V, Vв=4V, Iс=10mA, Vве=0.6V,Vсе=10V and

Answer 1

See the diagram for common emitter configuration of npn transistor, in a Universal biasing circuit.

We will not worry about the capacitors. We will worry about the DC biasing and the resistors RL, RE, R1 and R2.

Given Ic = 10 mA
Assuming the transistor has a current amplification factor  = β = 100.
So IB = base current ≈ Ic / β = 10 mA /50 = 0.2 mA

Given Vb = 4 V,   Vce = 10 V,  Vbe = 0.6 V  
=> Vcb = 10 - 0.6  = 9.4V
=> Vc = Vb+ Vcb = 13.4 V
=> RL = (Vcc - Vc)/Ic = (20 - 13.4)/10 mA = 660 Ω

Ve = Vb - Vbe = 3.4 V
Ie = Ic + Ib = 10 mA + 0.2 mA = 10.2 mA
Re = Ve/Ie = 3.4 / 10.2 = 333.3 Ω

We know in the design of R1 and R2, I1 , it is ensured that I1 > 10 * Ib for stability in I1 and not to vary due to changes in Ib.

Let   I1 = 10 * Ib = 2 mA
=> R1 = (Vcc - Vb) / I1 = 16/2 = 8 kΩ
=>  R2 = Vb / [I1 - Ib] = 4 / [2 - 0.2]  = 2.22 kΩ

So we have all the resistors known. So biasing done, except for capacitors.