Question

Match the elements of the list given below:

List I	List II
a) P-N junction diode	\(\left( i \right)~{{\varphi }_{m}}-\left( x+\frac{{{E}_{g}}}{2}-{{\psi }_{E}} \right)=0\)
b) MOSFET	\((ii)\frac{z}{L}{{\mu }_{0}}{{C}_{ox}}~\left( {{V}_{G}}-{{V}_{T}} \right){{V}_{D}}\)
c) JFET	\(\left( iii \right)~I={{I}_{0}}\left( e\frac{qV}{kT}-1 \right)\)
d) Ideal MIS diode	\(\left( iv \right)~{{I}_{DS}}={{I}_{DSS}}{{\left[ 1-\frac{{{V}_{GS}}}{{{V}_{p}}} \right]}^{2}}\)

 

The correct option is
1. (a) – (iii), (b) - (ii), (c) - (iv), (d) – (i)
2. (a) – (i), (b) - (ii), (c) - (iii), (d) – (iv)
3. (a) – (ii), (b) - (iii), (c) - (iv), (d) – (i)
4. (a) – (iv), (b) - (i), (c) - (ii), (d) – (iii)

Answer 1

Correct Answer - Option 1 : (a) – (iii), (b) - (ii), (c) - (iv), (d) – (i)

1) For a p-n junction diode, the current equation is given by:

\(I={{I}_{o}}\left( {{e}^{\frac{qV}{kT}}}-1 \right)\)

I₀ = Reverse Saturation Current has a strong dependence on temperature.

V = Applied voltage

For a forward applied voltage, the current will increase exponentially.

2) In the saturation region, the current through the JFET is given by:

\(I={{I}_{DSS}}{{\left[ 1-\frac{{{V}_{GS}}}{{{V}_{p}}} \right]}^{2}}\)

I_DSS = Maximum Drain current, when V_GS = 0

V_GS = Applied Gate to Source voltage

V_p = Pinch-off voltage

3) For a MOSFET operating in the triode or active region, the drain current is given by:

\({{I}_{D}}=\frac{Z}{~2L}{{\mu }_{o}}{{C}_{ox}}\left[ 2\left( {{V}_{G}}-{{V}_{T}} \right){{V}_{D}}-\frac{V_{DD}^{2}}{2} \right]\)

Z and L are the dimensions of the insulating layer

V_T = Threshold voltage

V_G = Applied Gate to Source voltage

For small values of V_D(called a deep active region), the higher power can be neglected, and the current variation approximates a linear variation w.r.t. V_D, i.e.

\({{I}_{d}}=\frac{Z}{L}{{\mu }_{o}}{{C}_{ox}}\left( {{V}_{G}}-{{V}_{T}} \right){{V}_{D}}\)

4) Ideal MIS diode, the difference in the work function (ϕ_ms) of the metal and semiconductor is 0, i.e.

ϕ_m = ϕ_s

Where ϕ is the amount of work done to move the particle from the Fermi-level to vacuum.

The semiconductor Work function (ϕ_s) is given by:

\(\phi_s=\chi+\frac{E_g}{2}-\psi_E\)

The difference in work function will be, therefore:

\({{ϕ }_{m}}-\left( \chi +\frac{{{E}_{g}}}{2}-{{\psi }_{E}} \right)=0\)