GATE 2018 Analog Electronics Questions and Solutions

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\begin{questions}

\question In the circuit shown below, the \((W/L)\) value for \(M_2\) is twice that for \(M_1\). The two nMOS transistors are otherwise identical. The threshold voltage \(V_T\) for both transistors is \(1.0\, V\). The voltage at \(V_2\) is (in volts, accurate to two decimal places):

[Image of MOSFET small signal model analysis]

Let \(K_n = \mu_n C_{ox} (W/L)\), \(K_2 = 2K_1\).
For \(M_1\): \(V_{GS1} - V_T = 2-1 = 1\,V\), in linear region.
For \(M_2\): \(V_{GS2} - V_T = 2-V_2-1 = 1-V_2\) and \(V_{DS2} = 3.3 - V_2 > (V_{GS2} - V_T)\) (saturation).
Set \(I_{D1} = I_{D2}\):

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Solve: \(2V_1 - V_1^2 = 2(1 + V_2^2 - 2V_2)\), which simplifies to \(3V_2^2 - 6V_2 + 2 = 0\).
Quadratic formula gives \(V_2 = 1.58\) or \(V_2 = 0.42\). Since \(V_{GS2} > V_T\), only \(V_2 = 0.42\)~V is valid.
\end{solution}

\question In the op-amp circuit shown, with ideal op-amp, what is the voltage (in volts, to one decimal place) at node A connected to negative input?

\question A DC current of \(26\,\mu\)A flows through the circuit shown. The diode (ideality 1) at quiescent point has junction capacitance \(0.5\) nF. Assume \(V_T=26\)mV. For \(\omega = 2\times10^6\) rad/s, small-signal input is \(5\sin(\omega t)\) mV across the diode. The amplitude of the small-signal component of diode current is (in \(\mu\)A, to one decimal place):

Small-signal resistance \(r_d = V_T/I_{DC} = 26\,mV / 26\,\mu A = 1k\Omega\)
At \(f=2\times 10^6/2\pi = 3.2 \times 10^5\) Hz, \(X_C = 1/(\omega C) = 1/(2 \times 10^6 \times 0.5 \times 10^{-9}) = 1k\Omega\)
Combined, \(Z = 1k\Omega\angle{-45^\circ}\); Input amplitude \(= 5\)mV.
Output current amplitude \(I=5mV / 1.414k\Omega = 3.5\,\mu\)A.
Correct answer: B \end{solution}

\question A regulated \(5\)V is to be maintained across a \(1k\Omega\) resistor by the circuit. The Zener reverse breakdown starts from \(I_{zmin}=2\)mA, max allowed current \(I_{zmax}\); input \(V_1\) can vary by 5\% from \(6\)V. Diode's breakdown resistance is negligible. The value of \(R\) and required minimum power dissipation of the diode are:

[Image of zener regulator circuit with variable input voltage]

\(V_1=6V\pm5\%=5.7\) to \(6.3\)V
Load current \(I_L=5V/1k=5\)mA. Minimum current through resistor for zener to engage: \(I_s(min)=5+2=7\)mA.
\(R=(V_1(min)-V_z)/I_s(min) = (5.7-5)/0.007 \approx 100\Omega\)
\(R=100\Omega\): at \(V_1=6.3V\): \(I_s=13\)mA, so \(I_z(max)=13-5=8\)mA.
So, \(P_{z}(min)=5V\times8\)mA = \(40\)mW.
Correct answer: B \end{solution}

\question[2] A good transimpedance amplifier has:
\begin{choices} \choice Low input, high output impedance \choice High input, high output impedance \choice High input, low output impedance \choice Low input, low output impedance \end{choices} \begin{solution}

[Image of transimpedance amplifier diagram]

Desirable: high input impedance (for current sensing), low output impedance (for load driving).
Correct answer: C \end{solution}

\question In the circuit shown below with ideal op-amp and unit step input \(v_{in}(t) = u(t)\), Zener voltage is \(2.5\)V, \(t=0\) capacitors are zero. At what time \(t\) (ms) does the output cross \(-10\)V?

\question Two identical nMOS transistors \(M_1\) and \(M_2\) are connected as shown. Input between \(G\) and \(S\), output between \(D\) and \(S\). The transconductance \(g_m\) and output resistance \(r_o\) for the combination are best estimated as:
\begin{choices} \choice \(g_m \approx g_{m1} \cdot g_{m2} \cdot r_{o2}\) and \(r_o \approx r_{o1} + r_{o2}\) \choice \(g_m \approx g_{m1} + g_{m2}\) and \(r_o \approx r_{o1} + r_{o2}\) \choice \(g_m \approx g_{m1}\) and \(r_o \approx r_{o1} \cdot g_{m2} \cdot r_{o2}\) \choice \(g_m \approx g_{m1}\) and \(r_o \approx r_{o2}\) \end{choices} \begin{solution}

For cascade amplifiers: overall \(g_m \approx g_{m1}g_{m2}r_{o2}\), \(r_o \approx r_{o1}+r_{o2}\).
Correct answer: A \end{solution}

\end{questions}