For the transistor in circuit shown below, β ≫ 1, V CE(sat) = 0.2

SOLUTION

For the given circuit with V_B = 2V, let us assume that the transistor is working in the active region.

The circuit is redrawn as:

Applying KVL, we can write:

V_B – 0.7 = V_E

2 – 0.7 = V_E

V_E = 1.3 V

Applying KVL at the emitter terminal, we get:

V_E - I_E R_E = 0

1.3 - I_E (1 k) = 0

I_E = 1.3 mA

Since β is very large, we can write:

I_C = I_E = 1.3 mA

Now, the voltage at the collector terminal will be:

6 - I_C R_E = V_C

V_C = 6 – (1.3 m) (10 k)

V_C = 6 – 13

V_C = -7 V

V_CE = -7 – 1.3

V_CE = - 8.3 V

Comparing this with V_CE(sat), we observe that:

V_CE < V_CE(sat)

This situation is not possible and hence our assumption of the transistor in the active region is incorrect.

∴ The transistor is operating in the saturation region.

Now, the equivalent circuit for saturation is drawn as:

We can write:

VC – VCE(sat) = 1.3

V_C – 0.2 = 1.3

V_C = 0.2 + 1.3

V_C = 1.5 V