The NMOS transistor has VT = 0·7 V, µn COX = 100 µA/V2,

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The NMOS transistor has V_T = 0·7 V, µ_n C_OX = 100 µA/V², L = 1 µm, W = 32 µm. If I_D = 0·4 mA and V_D = 0·5 V. The values of R_D and R_S are respectively given by—

1. 5 Ω, 3·25 kΩ
2. 3·25 kΩ, 5 kΩ
3. 2·5 kΩ, 6·5 kΩ
4. None of these

Correct Option: C

The given circuit

Given, V_T = 0·7 V, U_nC_ox = 100 µA/V²
L = 1 µm, W = 32 µm, I_D = 0·4 mA, V_D = 0·5 V
From figure,

or R_D =	V_DD – V_D
	I_D

=	2·5 – 0·5		= 5 kΩ
	0·4 mA

Since, V_DS > V_GS, this means the NMOS transistor is operating in saturation region.

Now, I_D =		1	µn C_OX ×	W	(V_GS – V_T) ²
		2		L

V_GS – V_T = V_DS(Condition for saturation region)
or V_GS – V_T = 0·5 V
or V_GS = 0·5 + V_T = 1·2 V
or V_G – V_S = 1·2 V
or O – V_S = 1·2 V or V_S = 1·2 V

Now, R_S =		V_S – V_SS	=	– 1·2 V – (– 2·5 V)
		I_D		0·4 mA

or R_S =	1·3 V		= 3·25 kΩ
	0·4 mA

Hence alternative (A) is the correct choice.