For the transistor shown β = 25. The range of V 1 such that 1.0 ≤

SOLUTION

We redraw the given circuit for dc analysis as

Now, we consider the following two cases for the circuit.

CASE I:

For V_CE = 4.5 V

Since, emitter is connected to the ground, i.e. V_E = 0. So, we have

V_C = V_CE = 4.5 V

Therefore, the collector current is

So, the base current is

= 20 μA

And the base voltage is

V_B = V_E + V_BE = 0 + 0.7 = 0.7 V

Also, the current through 100 kΩ resistor is

= 57 μA

Applying KCL at the base,

I₁ = I₂ + I_B

= 57 + 20 = 77 μA

So, V₁ = V_B + I₁ × 15k = 0.7 + 77μ × 15k = 1.855 V

CASE II:

For V_CE = 1.0 V, we have

V_C = V_E + V_CE = 0 + 1 = 1V,

So, collector and base currents are

Therefore, we obtain

I₂ = 0.057 mA

I₁ = I₂ + I_B = 0.057 + 0.16 = 0.217 mA

So, V₁ = (0.217 m) (15k) + 0.7 = 3.96 V

Thus, from the results obtained for two cases, we get the range of V₁ as

1.855 ≤ V ≤ 3.96 V