Question 1 |

The line integral of \int _{C} F\left ( r \right ).dr is

\frac{1}{2} | |

\frac{1}{4} | |

\frac{1}{6} | |

\frac{1}{3} |

Question 2 |

\frac{dy}{dx}+\frac{x}{1-x^{2}}y=x\sqrt{y}

The integrating factor of the differential equation is

\left ( 1-x^{2} \right )^{-3/4} | |

\left ( 1-x^{2} \right )^{-1/4} | |

\left ( 1-x^{2} \right )^{-3/2} | |

\left ( 1-x^{2} \right )^{-1/2} |

Question 3 |

Y=2X+3

Let \sigma ^{2}_{X} and \sigma ^{2}_{Y} denote the variances of X and Y, respectively. The variances are related as

\sigma ^{2}_{Y}=2 \sigma ^{2}_{X} | |

\sigma ^{2}_{Y}=4 \sigma ^{2}_{X} | |

\sigma ^{2}_{Y}=5 \sigma ^{2}_{X} | |

\sigma ^{2}_{Y}=25 \sigma ^{2}_{X} |

Question 4 |

\text{15 kHz} | |

\text{30 kHz} | |

\text{60 kHz} | |

\text{20 kHz} |

\mathrm{NR}=2 \times f_{\mathrm{max}}=2 \times 15=30 \mathrm{kHz}

Question 5 |

k=0,1,2,,15 | |

k=0 | |

k=15 | |

k=0 and k=15 |

then

y(k)=z(k) at k=N-1

where, y(n)= Linear convolution of x(n) and h(n)

z(n)= Circular convolution of x(n) and h(n)

Since, N=16 (Given)

Therefore, \quad y(k)=z(k) at k=N-1=15

Question 6 |

10^{20} \text{cm}^{-6} | |

2 \times 10^{20} \text{cm}^{-6} | |

10^{32} \text{cm}^{-6} | |

2 \times 10^{32} \text{cm}^{-6} |

\begin{aligned} N_{A} &=10^{16} \mathrm{~cm}^{-3} \\ g_{0 p} &=1020 \mathrm{~cm}^{-3} \mathrm{~s}^{-1} \\ \tau &=100 \mu \mathrm{s} \\ n_{i} &=10^{10} \mathrm{~cm}^{-3} \end{aligned}

Product of steady state electron-hole concentration =?

At thermal equilibrium (before shining light)

\begin{array}{ll} \text { Hole concentration, } & p_{o} \simeq N_{A}=10^{16} \mathrm{~cm}^{-3} \\ \text { Electron concentration, } & n_{0}=\frac{n_{i}^{2}}{p_{0}}=\frac{10^{20}}{10^{16}}=10^{4} \mathrm{~cm}^{-3} \end{array}

After, illumination of light,

Hole concentration, p=p_{o}+\delta p

Electron concentration, \quad n=n_{o}+\delta n

Due to shining light, excess carrier concentration,

\begin{aligned} \delta p &=\delta n=g_{o p} \cdot \tau=10^{20} \times 100 \times 10^{-6}=10^{16} \mathrm{~cm}^{-3} \\ \therefore \qquad p &=10^{16}+10^{16}=2 \times 10^{16} \mathrm{~cm}^{-3}\\ n&=10^{4}+10^{16} \simeq 10^{16} \mathrm{~cm}^{-3} \end{aligned}

So, product of steady state electron-hole concentration

\begin{aligned} &=n p=10^{16} \times 2 \times 10^{16} \\ &=2 \times 10^{32} \mathrm{~cm}^{-6} \end{aligned}

Question 7 |

If the charge of an electron is q, then the magnitude of the electric field developed inside this semiconductor bar is

\frac{\Delta }{qL} | |

\frac{2\Delta }{qL} | |

\frac{\Delta }{2qL} | |

\frac{3\Delta }{2qL} |

\begin{aligned} E &=\frac{1}{q}\frac{ d E_{v}}{d x} \\ &=\frac{1}{q} \frac{\Delta}{L} \\ &=\frac{\Delta}{q L} \end{aligned}

Question 8 |

Ignoring the body effect, the ac small signal voltage gain \left ( \partial V_{out}/\partial V_{in} \right ) of the circuit is

-g_{m2}\left ( r_{01}\left | \right |r_{02}\right ) | |

-g_{m2}\left ( \frac{1}{g_{m1}}\left | \right |r_{02} \right ) | |

-g_{m1}\left ( \frac{1}{g_{m2}}\left | \right |r_{01}\left | \right |r_{02} \right ) | |

-g_{m2}\left ( \frac{1}{g_{m1}}\left | \right |r_{01}\left | \right |r_{02} \right ) |

Drain to gate connected MOSFET M_1 acts as load.

For given circuit, AC equivalent is as shown.

Replace M_2 with small signal model

\begin{aligned} \frac{V_{\text {out }}}{V_{\text {in }}} &=\frac{-g_{m 2} V_{g s}\left(r_{\infty} \| R_{\text {eq }}\right)}{V_{g s}} \\ A_{V} &=-g_{m 2}\left( \frac{1}{g_{m 1}}|| r_{o1} || r_{o 2} \right) \end{aligned}

Question 9 |

If V_{\text{OUT}}=1 volt for V_{\text{IN}}-0.1 volt and V_{\text{OUT}}=6 volt for V_{\text{IN}}=1 volt, where V_{\text{OUT}} is measured across R_{L} connected at the output of this OPAMP, the value of R_{F}/R_{\text{IN}} is

3.28 | |

2.86 | |

3.82 | |

5.55 |

**MARKS TO ALL AS PER IIT ANSWER KEY**

\begin{aligned} V &=V^{+} \\ \frac{V_{\text {out }} R_{\text {in }}+V_{\text {in }} R_{F}}{R_{\text {in }}+R_{F}} &=\frac{V_{\text {ref }} R_{2}}{R_{1}+R_{2}} \\ \frac{1 \times R_{\text {in }}+0.1 \times R_{F}}{R_{\text {in }}+R_{F}} &=\frac{V_{\text {ref }} R_{2}}{R_{1}+R_{2}} &\ldots(i)\\ \frac{6 R_{\text {in }}+1 \times R_{F}}{R_{\text {in }}+R_{F}} &=\frac{V_{\text {ref }} R_{2}}{R_{1}+R_{2}} &\ldots(ii) \end{aligned}

Equate equation (i) and (ii),

\begin{aligned} 1 \times R_{\text {in }}+0.1 \times R_{F} &=6 \times R_{\text {in }}+1 \times R_{F} \\ -5 R_{\text {in }} &=0.9 R_{F} \\ \therefore \quad \frac{R_{F}}{R_{\text {in }}} &=\frac{-5}{0.9}=-5.55 \end{aligned}

(According to the given data magnitude is taken)

Question 10 |

Assuming \left | V_{\text{IN}} \right |\ll \left | V_{\text{CC}} \right | and \left | V_{\text{REF}} \right |\ll \left | V_{\text{CC}} \right | , the condition at which V_{\text{OUT}} equals to zero is

V_{\text{IN}}\:=\:V_{\text{REF}} | |

V_{\text{IN}}\:=\:0.5\:V_{\text{REF}} | |

V_{\text{IN}}\:=\:2\:V_{\text{REF}} | |

V_{\text{IN}}\:=\:2\:+\:V_{\text{REF}} |

KCL at node \mathrm{V}^{-}:

\begin{aligned} \frac{V_{\text {IN}}-0}{R}+\frac{\left(-V_{\text {REF }}-0\right)}{R}+\frac{V_{\text {OUT }}-0}{R_{F}} &=0 \\ \frac{V_{\text {OUT }}}{R_{F}} &=\frac{1}{R}\left(V_{\text {REF }}-V_{\text {IN }}\right) \\ V_{\text {OUT }} &=\frac{R_{F}}{R}\left(V_{\text {REF }}-V_{\text {IN }}\right)\\ \text { We want, }\qquad V_{\text {OUT }}&=0 \\ \Rightarrow\qquad V_{\text {REF }}-V_{\text {IN }}&=0 \\ \Rightarrow\qquad V_{\text {IN }}&=V_{\text {REF }} \end{aligned}