$C o n s i d e r t h e c i r c u i t s h o w n i n t h e f i g u r e b e l o w .$ $(λ = 0) V a l u e s o f V_{1} = 6 V a n d V_{2} = 2 V . V a l u e o f I_{D}$

$The two transistors Q_{1} and Q_{2} are exactly matched with each other. \frac{μ_{n} C_{o x} W}{L} = 10 {mA/V}^{2}, V_{t} = 0.5 V$
( $λ = 0) V a l u e s o f V_{1} = 6 V a n d V_{2} = 2 V . V a l u e o f I_{D} i n m A i s$

11.25

Open in App

Solution

The correct option is A 11.25
$Since Q_{1} and Q_{2} are matched and the same drain current flows through both of them, thus V_{G S_{1}} = V_{G S_{2}} V_{D_{1}} = V_{D_{2}} .$
$(∴ Gate and drain of the two transistors are shorted).$
$V - D S = V_{G S_{1}} = (V_{1} - V_{2}) / 2 = 2$

$Thus, I_{D} = \frac{μ_{n} C_{o x}}{2} \frac{W}{L} (V_{G S_{1}} - V_{t})^{2} = \frac{10 \times 10^{- 3}}{2} (2 - 0.5)^{2}$
$I_{D} = 11.25 mA$

Suggest Corrections

Similar questions

I_{D} = 0.4 m A

V_{D} = 0.5 V .

V_{t} = 0.7 V, μ_{n} C_{\propto} = 100 μ A / V^{2}, L = 1 μ m

ω = 32 μ m .

λ = 0.

R_{s} i s

Consider the circuit shown in the figure below; The circuit represents a

V_{T} = 1 V

λ = 0

\frac{μ_{n} C_{o x} W}{L} = 2 m A / V^{2}

T_{2} (V_{D S 2})

V_{T} = 1 V a n d μ_{n} C_{o x} (\frac{ω}{L}) = 2 m A / V^{2} .

(V_{D S})

If an n-channel enhancement MOSFET is biased in common source mode such that its gate to source voltage V_{G S} = 2 V and V_{T} = 1 V . the value of trans conductance is______m ℧ (Neglecting channel length modulation ).

V_{D S} = 3 V, μ_{n} C_{o x} (\frac{W}{L}) = 0.5 m A / V^{2}

Join BYJU'S Learning Program