Q21GATE 2005MCQ

Assume that the N-channel MOSFET shown in the figure is ideal, and that its threshold voltage is +1.0 V. The voltage $V_{ab}$ between nodes a and b is:

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📖 Explanation

MOSFET is N_channel. Gate through souce is so connected that MOSFET will be in enhance mode and so conductivity of the channel will be increased very much and effectively ' $\beta$ ' terminal act as short circuited, So, $V_{ab}=0V$ .

Q22GATE 2005MCQ

Assume that the threshold voltage of the N-channel MOSFET shown in figure is + 0.75 V. The output characteristics of the MOSFET are also shown The voltage gain of the amplifier is

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📖 Explanation

Since, $r_d \gt \gt R$ All current will pass through R $V_{out}=-g_mV_{GS}R$ $\;\; =-1 \times 10^{-3}\times 2 \times 10^{-3}\times 10\times 10^{-3}$ $\; \; =-20\; mV$ Volate gain $=\frac{V_{out}}{V_{\in}} =\frac{-20\; mV}{2 \; mV}=-10$

Q23GATE 2004MCQ

Two perfectly matched silicon transistor are connected as shown in figure. The value of the current I is

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📖 Explanation

Both transistor are perfectly matched, hence $V_{BE2}=V_{BE1}$ therefore, $\frac{I_{c1}}{I_{c2}}=exp[(V_{BE1}-V_{BE2})/V_T]=1$ Also ' $\beta$ ' is same. $I_R=\frac{+5-0.7}{1K}=4.3\; mA$ Writing KCL at node B, $I_c+2I_b-I_R=0$ $I_c=\beta I_b$ $\therefore \; I_c=\left ( \frac{\beta }{\beta +2} \right )I_R$ $\approx I_R$ (Because $\beta$ is very large) Hence, $I_{c1}=I_{c2}=I_R=4.3\; mA$

Q24GATE 2004MCQ

The transconductance $g_{m}$ of the transistor shown in figure is 10 mS. The value of the input resistance $R_{IN}$ is

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📖 Explanation

Input Resistance= $\frac{h_{fe}}{g_m}=\frac{\beta }{g_m} =\frac{50}{10\times 10^{-3}}=5k\Omega$ But overall input resistance seen from source is $R_{IN}=10\left \| 10 \right \|5=2.5\;k\Omega$

Q25GATE 2004MCQ

The value of R for which the PMOS transistor in figure will be biased in linear region is

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📖 Explanation

Here $V_S=+4V$ $V_G=0V$ $V_T=-1V$ Therefore, $V_{SG}=4V$ $V_{SD}=V_S-V_D$ $=4-I_a R$ Now for linear region of operation $\;\; V_{SD} \lt (V_{SG}-V_T)$ $\Rightarrow 4-I_a R \lt (4-1)$ $\Rightarrow I_a (R) \gt 1$ $\Rightarrow 10^{-3}\times R\gt 1$ $\;\;\; R \gt 1000\Omega$

Q26GATE 2003MCQ

In the circuit of figure, assume that the transistor has $h_{FE}$ = 99 and $V_{BE}$ = 0.7 V. The value of collector current $I_{C}$ of the transistor is approximately

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📖 Explanation

Using KVL in Base-emitter Loop $4-33\times I_b-0.7-3.3\times100I_b=0(I_c=\beta I_b)$ $\therefore \; I_b=\frac{4-0.7}{(33+330)}$ $\therefore \; I_c=\frac{3.3\times 100}{33+330} =\frac{3.3}{(0.33+3.3)}mA$

Q27GATE 2003MCQ

For the n-channel enhancement MOSFET shown in figure, the threshold voltage $V_{th}$ = 2 V. The drain current $I_{D}$ of the MOSFET is 4 mA when the drain resistance $R_{D}$ is 1 k $\Omega$ . If the value of $R_{D}$ is increased to 4 k $\Omega$ , drain current $I_{D}$ will become

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📖 Explanation

$I_D=K(V_{GS}-V_{th})^2$ $4=K(6-2)^2$ $\therefore \; K=\frac{1}{4}mA/V^2$ $V_{GS}=10-4\times 1=6V$ when $R_D$ is increased to $4k\Omega$ $V_{GS}=10-4I_D$ $I_D=\frac{1}{4}(10-4I_D-2)^2$ $4I_D=16I^2_D+64-64I_D$ $16I^2_D-68I_D+64=0$ $I_D=2.84 \; mA,1.4 \; mA$ For MOSFET to be on, $V_{GS}$ must be greater than $V_{th}$ and this is possible only if $\;\;\; I_D=1.4\; mA$ $If \; I_D=2.84\; mA$ then $V_{GS}$ become -ve and less than $V_{th}$ so transistor will be off for this value which is not possible since $V_{DG}=0\geq -V_{th}$ .

Q28GATE 2003MCQ

The variation of drain current with gate-to-source voltage ( $I_{D}$ - $V_{GS}$ characteristic) of a MOSFET is shown in figure. The MOSFET is

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Q29GATE 2003MCQ

In the circuit shown in figure, the current gain ( $\beta$ ) of the ideal transistor is 10. The operating point of the transistor ( $V_{ce},I_C$ ) is

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Q30GATE 2001MCQ

An n-channel JFET, having a pinch-off voltage ( $V_{p}$ ) of -5 V, shows a transconductance ( $g_{m}$ ) of 1 mA/V when the applied gate-to-source voltage ( $V_{GS}$ ) is -3V. Its maximum transconductance (in mA/V) is

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📖 Explanation

$g_m=-\frac{2I_{DSS}}{V_p}\left ( 1-\frac{V_{GS}}{V_p} \right )$ $1 \; mA/V=-\frac{2I_{DSS}}{5} \left( 1-\frac{-3}{-5} \right ) =\frac{2}{5}\times \left( 1-\frac{3}{5} \right )I_{DSS}$ $\Rightarrow \; \; I_{DSS}=\left ( \frac{5}{2} \right )^2 \; mA$ $(g_{m})_{max}=-\frac{2I_{DSS}}{V_p}=-\frac{2 \times 25}{4 \times (-5)}=2.5 \; mA/V$