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EMF and Terminal Voltage

A depletion t...

Question

$A depletion type N-channel MOSFET is biased in linear region for use as a votlage controlled resistor. At V_{G S} = 2 V, the drain to source resistance (r_{d s}) found to be 500 Ω, if V_{G S} is increased to 3V, then the value of rds is______Ω (Assume, Threshold voltage, V_{T h} = - 0.5 V, the value of V_{D S} to be very small)$
357.14

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Solution

The correct option is A 357.14
$Drain to source current in linear region, I_{D S} = μ_{n} C_{o x} \frac{W}{L} [(V_{G S} - V_{T h}) V_{D S} - \frac{V_{D S}^{2}}{2}] I_{D S} = μ_{n} C_{o x} \frac{W}{L} [(V_{G S} - V_{T h}) V_{D S}] (Since V_{D S} is ver small Drain to source resistance, r_{d s} = \frac{V_{D S}}{I_{D S}} ∴ r_{d s} = \frac{1}{μ_{n} C_{o x} \frac{W}{L} (V_{G S} - V_{T h})} ∴ r_{d s} \propto \frac{1}{V_{G S} - V_{T h}} Given V_{G S 1} = 2 V a n d r_{d s 1} = 500 Ω V_{G S 2} = 3 V a n d r_{d s 2} = ? \frac{r_{d s 1}}{r_{d s 2}} = \frac{V_{G S 2} - V_{T h}}{V_{G S 1} - V_{T h}} \frac{500}{r_{d s 2}} = \frac{3 - (- 0.5)}{2 - (- 0.5)} ∴ r_{d s 2} = \frac{500 \times 2.5}{3.5} = 357.14 Ω$

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