Question

A depletion type N-channel MOSFET is biased in its linear region for use as a voltage controlled resistor. Assume threshold voltage V_TH = - 0.5 V, V_GS = 2.0 V, V_DS = 5 V, W/L = 100, C_OX = 10^-8 F/cm² and μ_n = 800 cm²/V-s. The value of the resistance of the voltage controlled resistor (in Ω) is _________.

Answer 1

Concept:

For a field-effect transistor (FET) under certain operating conditions, the resistance of the drain-source channel is a function of the gate-source voltage alone and the JFET will behave as an almost pure ohmic resistor

For a MOSFET in the saturation region:

VGS > Vth and

VDS > VGS – Vth

The current equation for a MOSFET in the saturation region is given by:

\({{I}_{D}}=\mu C_{ox} \frac{W}{2L}\left\{\left( {{V}_{GS}}-{{V}_{th}} \right)^2\right\}\)

Taking the derivative of the above w.r.t. V_DS, we get:

\(\frac{\partial I_D}{\partial V_{DS}}=\mu C_{ox}\left( {\frac{W}{L}} \right)\left( {{V_{GS}} - {V_{TH}}} \right)\)

Voltage-controlled Resistor (r_DS) is defined as:

\({r_{DS}} = \frac{{{V_{DS}}}}{{{I_{DS}}}} = \frac{1}{{{\mu _n}{C_{ox}}\left( {\frac{W}{L}} \right)\left( {{V_{GS}} - {V_{TH}}} \right)}}\)

Calculation:

Given:

V_GS = 2.0 V,

V_TH = -0.5 V,

\(\frac{W}{L} = 100\)

C_ox = 10^-8 f/cm²

μ_n = 800 cm²/V-s,

V_DS = 5 V

Putting on the respective values in Equation (1), we get:

\({r_{DS}} = \frac{1}{{800 \times {{10}^{ - 8}} \times 100 \times \left( {2 - \left( { - 0.5} \right)} \right)}}\)

\({r_{DS}} = 500\;{\rm{\Omega }}\)