Question

The current in an enhancement mode NMOS transistor biased in saturation mode was measured to be 1 mA at a drain-source voltage of 5 V. When the drain-source voltage was increased to 6 V while keeping gate-source voltage same, the drain current increased to 1.02 mA. Assume that drain to source saturation voltage is much smaller than the applied drain-source voltage. The channel length modulation parameter λ (in V^-1) is _______.

Answer 1

Concept:

Saturation Current in an enhancement mode N-MOS is given as:

\({{\rm{I}}_{{\rm{DS}}}} = {\rm{\;}}\frac{1}{2}{\rm{\;}}{{\rm{\mu }}_{\rm{n}}}{{\rm{C}}_{{\rm{ox}}}} \times \frac{{\rm{W}}}{{\rm{L}}}{\rm{\;}}{\left[ {{{\rm{V}}_{{\rm{GS}}}} - {{\rm{V}}_{\rm{T}}}} \right]^2}\left( {1 + {\rm{\lambda }}.{{\rm{V}}_{{\rm{DS}}}}} \right)\) 

Calculation:

Case-I:

V_DS = 5V, I_DS = 1 mA

\(1{\rm{\;mA}} = {\rm{\;}}\frac{1}{2}{{\rm{\mu }}_{\rm{n}}}{{\rm{C}}_{{\rm{ox}}}}\frac{{\rm{W}}}{{\rm{L}}}{\rm{\;}}{\left( {{{\rm{V}}_{{\rm{GS}}}} - {{\rm{V}}_{\rm{T}}}} \right)^2}\left( {1 + {\rm{\lambda }}.5} \right)\) --- (1)

Case-II:

V_DS = 6V, I_DS = 1.02 mA

\(1.02 = {\rm{\;}}\frac{1}{2}{{\rm{\mu }}_{\rm{n}}}{{\rm{C}}_{{\rm{ox}}}}\frac{{\rm{W}}}{{\rm{L}}}{\rm{\;}}{\left( {{{\rm{V}}_{{\rm{GS}}}} - {{\rm{V}}_{\rm{T}}}} \right)^2}{\rm{\;}}\left( {1 + {\rm{\lambda }}.6} \right)\) --- (2)

Dividing Equation (1) by Equation (2), we get,

\(\frac{1}{{1.02}} = \frac{{1 + 5{\rm{\lambda }}}}{{1 + 6{\rm{\lambda }}}} \Rightarrow 1 + 6{\rm{\lambda }} = 1.02 + 5.1{\rm{\lambda }}\)  

⇒ 0.9 λ  = 0.02

\({\rm{\lambda }} = \frac{2}{{90}} \approx 0.0222\)