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In the circui...

Question

In the circuit shown below, the MOSFET has threshold voltage of 1 V and $k_{n} = \frac{{k^{'}}_{n}}{2} (\frac{W}{L}) = 20 μ A / V^{2} .$ The body terminal of the device is shorted to the source terminal and the channel length modulation parameter $λ$ is zero.

It is is required to keep the MOSFET always in the saturation region, then the maximum value of the gate voltage $(V_{g})$ that can be applied will be _________ V.

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Solution

Given $k_{n} = 20 m A / V^{2}, R_{D} = 5 k Ω$
$V_{D S} = 1.1 - I_{D} R_{D}$

$I_{D} = k_{n} {(V_{g} - V_{T})}^{2}$

$= k_{n} {(V_{g} - 1)}^{2}$
for saturation region of operation,

$V_{D S} \geq V_{g s} - V_{T}$

$V_{D S} \geq (V_{g s} - 1 V)$

$(1.1 - I_{D} R_{D}) \geq (V_{g s} - 1)$

$1.1 - R_{D} k_{n} {(V_{g s} - 1)}^{2} \geq (V_{g s} - 1)$

$1.1 - (0.1) {(V_{g s} - 1)}^{2} \geq (V_{g s} - 1)$

$1.1 - (0.1) (V_{g}^{2} - 2 V_{g} + 1) \geq (V_{g s} - 1)$

$1.1 - \frac{V_{g}^{2}}{10} + \frac{2 V_{g}}{10} - 0.1 \geq (V_{g s} - 1)$

$- \frac{V_{g}^{2}}{10} + \frac{2 V g}{10} - V_{g} + 2 \geq 0$

$- V_{g}^{2} - 8 V_{g} + 20 \geq 0$

at the edge of saturation, $v_{g} = v_{g max}$

so, $V_{g max}^{2} - 8 V_{g max} + 20 = 0$

$V_{g max}^{2} - 8 V_{g max} - 20 = 0$

$V_{g max} = \frac{- 8 \pm \sqrt{64 + 80}}{2} V$

$= \frac{- 8 \pm 12}{2} V = - 10 V, 2 V$

Valid $V_{g m a x} = 2 V$

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