Consider a small-signal model for MOSFET Which of the following

SOLUTION

Concept:

Transconductance (g_m) is the rate of change of drain current w.r.t V_GS, i.e.

\({g_m} = \frac{{\partial {I_{DS}}}}{{\partial {V_{GS}}}}\)

The current in a MOSFET in saturation regions is given by:

\({I_{DS}} = \frac{{{K_N}}}{2}{\left( {{V_{GS}} - {V_t}} \right)^2}\)   ---(1)

K_N = Transconductance parameter

\({K_N} = {\mu _n}{C_{ox}}\frac{W}{L}\)

∴ The transconductance will be:

\(\frac{{\partial {I_{DS}}}}{{\partial {V_{GS}}}} = {K_N}\left( {{V_{GS}} - {V_t}} \right)\)   ---(2)

Using Equation (1), we can write:

\({g_m} = \frac{{\partial {I_{DS}}}}{{\partial {V_{GS}}}} = {K_N}\sqrt {\frac{{2{I_{DS}}}}{{{K_N}}}} \)

\({g_m} = \sqrt {2{K_N}{I_{DS}}} \)

Dynamic resistance (r_ds): It is the reciprocal of the slope of drain characteristics.

Slope \( = \frac{{\partial {I_{DS}}}}{{\partial {V_{DS}}}} = \frac{{{I_{DS}} - 0}}{{({V_{DS}} - \left( {\frac{{ - 1}}{\lambda }} \right)}} \)

\(Slope= \frac{{{I_{DS}}}}{{{V_{DS}} + \frac{1}{\lambda }}}\)

\({r_{ds}} = \frac{1}{{slope}} = \frac{{\partial {V_{DS}}}}{{\partial {I_{DS}}}} = \frac{{{V_{DS}} + \frac{1}{\lambda }}}{{{I_{DS}}}}\)

\({r_{ds}} = \frac{{{V_{DS}} + \frac{1}{\lambda }}}{{{I_{DS}}}}\)

λ → channel length modulation parameter.

If \(\frac{1}{\lambda } \gg {V_{DS}}\) then:

\({r_{ds}} \cong \frac{1}{{\lambda {I_{dS}}}}\)

If the modulation does not have channel length modulation, then λ = 0 and r_ds = ∞

Option (a), (c) and (d) are true.