Consider the circuit shown in figure below. If the β of

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Consider the circuit shown in figure below. If the β of the transistor is 30 and I_CBO = 20 nA and the input voltage is + 5 V, the transistor would be operating in—

1. Saturation region
2. Active region
3. Breakdown region
4. Cut-off region

Correct Option: B

Given, β = 30, I_CBO = 20 nA, V_i = 5V

From above figure
5 – I₁ × 15 kΩ – VBE = 0

or I₁ =	(5 – 0·7)V	= 0·2867 mA
	15 kΩ

Again from figure
– 12 + I₂ × 100 kΩ – V_BE = 0

or I₂ =	12·7 V	= 0·127 mA
	100 kΩ

And, I_B = I₁ – I₂ = 0·2867 mA – 0·127 mA
= 0·158 mA
We know that
I_C = βI_B + (1 + β) I_CBO
or I_C = 30 × 0·158 + (1 + 30) × 20 × 10^–9
or I_C = 4·74 mA + 0·00062 mA
or I_C = 4·74062 mA
Now, V_CE = V_CC – I_CR_C
or V_CE = 12 – 4·7062 × 10^–3 × 2·2 × 10^{3^{or V_CE = 12 – 10·385 = 1·6146V
Since V_CE > 0·2V. It means transistor is operating in Active region
Hence alternative (B) is the correct choice.}}