\section*{1. GATE EE 2008} \subsection*{Analog Circuits}
In the following transistor circuit \(V_{BE}=0.7\,\mathrm{V}\), \(r_e = 25\,\mathrm{mV}/I_E\), and \(\beta\) and all the capacitances are very large. The mid-band voltage gain of the amplifier is approximately:
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Solution:
The mid-band voltage gain is given by:
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To find \(r_e\), we first need \(I_E\). Assuming \(I_E \approx 1\,\mathrm{mA}\) (calculated in the next question):
@@EQ1@@
@@EQ2@@
@@EQ3@@
Correct answer: D.
\section*{2. GATE EE 2008} \subsection*{Analog Circuits}
In the following transistor circuit \(V_{BE}=0.7\,\mathrm{V}\), \(r_e = 25\,\mathrm{mV}/I_E\), and \(\beta\) and all the capacitances are very large. The value of DC current \(I_E\) is:
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Solution:
To find the DC current \(I_E\), we perform a DC analysis. The capacitors are open circuits.
We find the Thevenin equivalent for the base-biasing network (\(20k\Omega\), \(10k\Omega\), and \(9V\)).
Thevenin Voltage (\(V_{Th}\)):
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Thevenin Resistance (\(R_{Th}\)):
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Apply KVL to the base-emitter loop. Since \(\beta\) is large, \(I_B\) is negligible (\(I_B \approx 0\)).
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@@EQ7@@
@@EQ8@@
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Correct answer: A.
\section*{3. GATE EE 2008} \subsection*{Analog Circuits}
Consider the Schmidt trigger circuit shown below. A triangular wave which goes from \(-12\,\mathrm{V}\) to \(12\,\mathrm{V}\) is applied to the inverting input of OPAMP. Assume that the output of the OPAMP swings from \(+15\,\mathrm{V}\) to \(-15\,\mathrm{V}\). The voltage at the non-inverting input switches between:
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Solution:
This is an inverting Schmitt trigger. The voltage at the non-inverting input (\(V_+\)) sets the trigger points.
Based on the circuit configuration (assuming standard feedback resistors), the relationship is derived using KCL at the non-inverting node.
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Since \(V_0\) swings from \(-15\,\mathrm{V}\) to \(+15\,\mathrm{V}\):
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The voltage at the non-inverting input switches between \(-5\,\mathrm{V}\) and \(+5\,\mathrm{V}\).
Correct answer: C.
\section*{4. GATE EE 2008} \subsection*{Analog Circuits}
An astable multivibrator circuit using IC 555 timer is shown below. Assume that the circuit is oscillating steadily. The voltage \(V_C\) across the capacitor varies between:
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Solution:
In a standard 555 timer astable configuration, the capacitor voltage \(V_C\) oscillates between the lower and upper threshold voltages.
These internal thresholds are set by the 555's internal voltage divider:
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Given \(V_{CC} = 9\,\mathrm{V}\):
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The capacitor voltage \(V_C\) varies between \(3\,\mathrm{V}\) and \(6\,\mathrm{V}\).
Correct answer: B.
\section*{5. GATE EE 2008} \subsection*{Analog Circuits}
Two identical NMOS transistors M1 and M2 are connected as shown below. \(V_{bias}\) is chosen so that both transistors are in saturation. The equivalent \(g_m\) of the pair is defined to be \(\frac{\partial I_{out}}{\partial V_i}\) at constant \(V_{out}\). The equivalent \(g_m\) of the pair is:
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Solution:
This configuration is a cascode stage (M1) on top of a common-source stage (M2).
The input voltage \(V_i\) is the gate voltage of M2 (\(V_{GS2}\)).
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Since M1 is in series with M2, \(I_{out} = I_{D1} = I_{D2}\). The definition of the equivalent \(g_m\) is:
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By definition, the transconductance of M2 is \(g_{m2} = \frac{\partial I_{D2}}{\partial V_{GS2}}\). Therefore, \(g_m = g_{m2}\). Since M1 and M2 are identical, \(g_{m1} = g_{m2}\). Thus, the equivalent \(g_m\) is nearly equal to the \(g_m\) of M1 (and M2).
Correct answer: C.
\section*{6. GATE EE 2008} \subsection*{Analog Circuits}
The OPAMP circuit shown above represents a:
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Solution:
This is an active inductor circuit. The transfer function \(H(s) = V_o/V_i\) is derived as:
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Where \(Z_{in} = R_1 + Ls\) and \(Z_f = R_2 || \frac{1}{Cs}\).
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This is a standard second-order low-pass filter transfer function, which has a constant numerator and a second-order polynomial in \(s\) in the denominator.
Correct answer: B.
\section*{7. GATE EE 2008} \subsection*{Analog Circuits}
Consider the following circuit using an ideal OPAMP. The I-V characteristic of the diode is described by the relation \(I = I_0(e^{V/V_T} - 1)\) where \(V_T = 25\,\mathrm{mV}\), \(I_0 = 1\,\mathrm{\mu A}\) and V is the voltage across the diode (taken as positive for forward bias). For an input voltage \(V_i = -1\,\mathrm{V}\), the output voltage \(V_O\) is:
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Solution:
1. Since the op-amp is ideal, \(V_{-} = V_{+} = 0\,\mathrm{V}\).
2. The input current \(I_{in}\) is:
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3. The feedback current \(I_f\) flows from \(V_o\) to \(V_{-}\). To balance the node, \(I_f = 10\,\mathrm{\mu A}\). This current flows through the diode (forward biased). 4. Using the diode equation with \(I_D = I_f = 10\,\mathrm{\mu A}\) and \(I_0 = 1\,\mathrm{\mu A}\):
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5. The output voltage \(V_o\) is the sum of the voltage drop across the \(4k\Omega\) resistor and the diode \(V_D\):
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Correct answer: B.
\section*{8. GATE EE 2008} \subsection*{Analog Circuits}
For the circuit shown in the following figure, transistor M1 and M2 are identical NMOS transistors. Assume the M2 is in saturation and the output is unloaded. The current \(I_x\) is related to \(I_{bias}\) as:
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Solution:
This is a standard NMOS current mirror circuit.
The current \(I_{bias}\) flows into the drain of M1. M1 is "diode-connected" (\(V_{DS} = V_{GS}\)), ensuring it is in saturation.
Since the gate of M2 is connected to the gate of M1, and their sources are both connected to ground, \(V_{GS2} = V_{GS1}\).
Because M1 and M2 are identical transistors and have the same \(V_{GS}\), their drain currents must be equal:
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Correct answer: B.
\section*{9. GATE EE 2008} \subsection*{Analog Circuits}
In the following limiter circuit, an input voltage \(V_i = 10\sin(100\pi t)\) is applied. Assume that the diode drop is \(0.7\,\mathrm{V}\) when it is forward biased. When it is forward biased, the zener breakdown voltage is \(6.8\,\mathrm{V}\). The maximum and minimum values of the output voltage respectively are:
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Solution:
This circuit clips (limits) the output voltage \(V_o\).
For the positive half of \(V_i\) (\(V_i > 0\)):
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For the negative half of \(V_i\) (\(V_i < 0\)):
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The maximum and minimum values are \(7.5\,\mathrm{V}\) and \(-0.7\,\mathrm{V}\).
Correct answer: C.