Consider a long-channel MOSFET with a channel length 1 - m and width 10 - m- The device parameters are acceptor concentration NA - 5- 1016 cm-3 - electron mobility -n - 800 cm2-V-s - oxide capacitance-area Cox - 3-45 - 10-7 F-cm2 - threshold voltage VT - 0-7 V - The drain saturation current IDsat for a gate voltage of 5 V is mA rounded off to two decimal places- - 0 - 8-854 - 10-14 F-cm- - Si - 11-9-

I_D_sat = 12μ _n C_oxWL (V_GS V_T)^2 = 12× 800× 3.45× 10^ 7×101(5 0.7)^2)A = 25.5162 mA≃ 25.52 mA ...

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Standard XII

Physics

Solenoid

Consider a lo...

Question

Consider a long-channel MOSFET with a channel length $1 μ m$ and width $10 μ m$ . The device parameters are acceptor concentration $N_{A} = 5 \times 10^{16} c m^{- 3}$ , electron mobility $μ_{n} = 800 c m^{2} / V - s$ , oxide capacitance/area $C_{o x} = 3.45 \times 10^{- 7} F / c m^{2}$ , threshold voltage $V_{T} = 0.7 V$ . The drain saturation current $(I_{D s a t})$ for a gate voltage of $5 V$ is mA (rounded off to two decimal places). $ϵ_{0} = 8.854 \times 10^{- 14} F / c m, ϵ_{S i} = 11.9$ ]

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Solution

$I_{D_{s a t}} = \frac{1}{2} μ_{n} C_{o x} \frac{W}{L} (V_{G S} - V_{T})^{2}$
$= \frac{1}{2} \times 800 \times 3.45 \times 10^{- 7} \times \frac{10}{1} (5 - 0.7)^{2}) A$
$= 25.5162 m A ≃ 25.52 m A$

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IDsat=12μnCoxWL(VGS−VT)2 =12×800×3.45×10−7×101(5−0.7)2)A =25.5162 mA≃25.52 mA

$I_{D_{s a t}} = \frac{1}{2} μ_{n} C_{o x} \frac{W}{L} (V_{G S} - V_{T})^{2}$
$= \frac{1}{2} \times 800 \times 3.45 \times 10^{- 7} \times \frac{10}{1} (5 - 0.7)^{2}) A$
$= 25.5162 m A ≃ 25.52 m A$