The parameters for the transistor in circuit of fig. below

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The parameters for the transistor in circuit of fig. below are V_TN = 2 V and K_n = 0.2 mA/V². The power dissipated in the transistor is:

1. 5.84 mW
2. 2.34 mW
3. 0.26 mW
4. 58.4 mW

Correct Option: B

From fig. since gate is connected to the drain. Hence transistor will always in saturation.

I_D =	10 – V_GS	= K_n (V_GS – V_Tn)²
	10 kΩ

10 – V_GS = 0.2 × 10^–3 × 10 × 10³ (V_GS – 2)²
or 10 – V_GS = 2 (V_GS – 2)²
After solving quadratic equation, we get
V_GS = – 0.27 V, 3.77 V V_GS = – 0.27 is not possible since gate terminal is at 10 V. So, V_GS = 3.77 V is taken
i.e. V_GS = V_DS = 3.77 V

I_D =	10 – 3·77	= 0.623 mA
	10 kΩ

Power = I_D V_DS = 0.623 × 10^–3 × 3.77 = 2.35 mW