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Consider the ...

Question

Consider the circuit shown below

If $g_{m_{1}} a n d g_{m_{2}}$ are the transconductances of transistors $T_{1} a n d T_{2}$ respectively $A_{v} = \frac{V_{o u t}}{V_{i n}}$ of the circuit will be
(Assume $λ = 0$ for both the transistors)

A

A_{V} = - \frac{g_{m_{1}}}{g_{m_{2}}}

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B

A_{V} = - \frac{g_{m_{2}}}{g_{m_{1}}}

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C

A_{V} = \frac{g_{m_{2}}}{g_{m_{1}}}

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D

A_{V} = \frac{g_{m_{1}}}{g_{m_{2}}}

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Solution

The correct option is B $A_{V} = - \frac{g_{m_{2}}}{g_{m_{1}}}$
MOS $T_{2}$ serve as drain resistance for $T_{1}$

Calculating $R_{o u t} o f T_{2}$

Applying KCL at node, A

$I_{x} = g_{m} V_{g s}$

$V_{g s_{2}} = V_{x}$

Thus $\frac{V_{x}}{I_{x}} = \frac{1}{g_{m_{2}}} = R_{o u t}$

for transistor $T_{1}$

$V_{o u t} = - g_{m_{1}} V_{g s_{1}} R_{o u t}$

$V_{g s_{1}} = V_{i n}$

$\frac{V_{o u t}}{V_{i n}} = - g_{m_{1}} R_{o u t}$

$A_{V} = - \frac{g_{m_{1}}}{g_{m_{2}}}$

$∵$ $R_{o u t} = - \frac{1}{g_{m_{2}}}$

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