Test 3 - Analog Circuits (Electronics and Communication)

Description: Test 3 of Analog Circuits (Electronics and Communication)
Number of Questions: 20
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Tags: Analog Circuits GATE
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In the following transistor circuit, VBE = 0.7 V, r3 = 25 mV / IE, and $\beta$ and all the capacitances are very large

The mid-band voltage gain of the amplifier is approximately

    • 180
    • 120
    • 90
    • 60

Correct Option: D
Explanation:

In the circuit below, the diode is ideal. The voltage V is given by

  1. min (Vi, 1)

  2. max (Vi, 1)

  3. min (-Vi, 1)

  4. max (-Vi, 1)


Correct Option: A
Explanation:

In the circuit shown below, for the MOS transistors, $\mu_n \ C_{ox}$ = 100$\mu A / V^2$and the threshold voltage VT = 1V. The voltage Vx at the source of the upper transistor is

  1. 1V

  2. 2V

  3. 3V

  4. 3.67V


Correct Option: C
Explanation:

In the following transistor circuit, VBE = 0.7 V, r3 = 25 mV / IE and $\beta$ and all the capacitances are very large.

The value of DC current IE is

  1. 1 mA

  2. 2 mA

  3. 5 mA

  4. 10 mA


Correct Option: A
Explanation:

Consider the Op-Amp circuit shown in the figure.

The transfer function V0 (s) / Vi (s) is

  1. $ \dfrac {1-sRC}{1+sRC}$

  2. $ \dfrac {1-sRC}{1+sRC}$

  3. $ \dfrac {1}{1-sRC}$

  4. $ \dfrac {1}{1+sRC}$


Correct Option: A
Explanation:

The input impedance (Zi) and the output impedance (Z0) of an ideal trans - conductance (voltage controlled current source) amplifier are

  1. Zi = 0, Z0 = 0

  2. Zi = 0, Z0 = $\infty$

  3. Zi = $\infty$, Z0 = 0

  4. Zi = $\infty$, Z0 = $\infty$


Correct Option: D
Explanation:

In the transconductance amplifier, it is desirable to have large input impedance and large output impedance.

Assuming the OP-AMP to be ideal, the voltage gain of the amplifier shown below is

  1. $ - \dfrac{R_2}{R_1}$

  2. $ - \dfrac{R_3}{R_1}$

  3. $ - \dfrac{R_2 || R_3}{R_1}$

  4. $ - \dfrac{R_2 || R_3}{R_1}$


Correct Option: A
Explanation:

Consider the Op-Amp circuit shown in the figure.

If Vi = V1 sin ($\omega t$) and V0 = V2 sin ($\omega t$+ $\phi$), the minimum and maximum values of $\phi$(in radians) are respectively

  1. $ \dfrac{-\pi}{2} and \dfrac{\pi}{2}$

  2. 0 and $\dfrac{\pi}{2}$

  3. -$\pi$ and 0

  4. $ \dfrac{-\pi}{2}$ and 0


Correct Option: C
Explanation:

A bipolar transistor is operating in the active region with a collector current of 1 mA. Assuming that the $\beta$ of the transistor is 100 and the thermal voltage (VT) is 25 mV, the transconductance (gm) and the input resistance ($r_x$)the transistor in the common emitter configuration, are

  1. gm = 25 mA/V and $r_x$= 15.625 k$\Omega$

  2. gm = 40 mA/V and $r_x$= 4.0 k$\Omega$

  3. gm = 25 mA/V and $r_x$= 2.5 k$\Omega$

  4. gm = 40 mA/V and $r_x$= 2.5 k$\Omega$


Correct Option: D
Explanation:

If the input to the ideal comparator shown in figure is a sinusoidal signal of 8V (peak to peak) without any DC component, then the output of the comparator has a duty cycle of

  1. $\dfrac{1}{2}$

  2. $\dfrac{1}{3}$

  3. $\dfrac{1}{6}$

  4. $\dfrac{9}{12}$


Correct Option: B
Explanation:

Consider the following circuit using an ideal OPAMP. The I - V characteristic of the diode is described by the relation I = I0$e^{\left (\frac{V}{V1}-1 \right)}$where VT = 25 mV, I0 = 1$\mu$A and V is the voltage across the diode (taken as positive for forward bias). For an input voltage Vi = - 1 V, the output voltage V0 is

  1. 0 V

  2. 0.1 V

  3. 0.7 V

  4. 1.1 V


Correct Option: B
Explanation:

Given rd = 20 k$\Omega$, IDSS = 10 mA, Vp = -8 V.

Zi and Z0 of the circuit are respectively

  1. 2 M$\Omega$ and 2 k$\Omega$

  2. $2\ M\Omega \quad and \quad \dfrac{20}{11} K\Omega$

  3. Infinity and 2 k$\Omega$

  4. $Infinity \quad and \quad \dfrac{20}{11} K\Omega$


Correct Option: B
Explanation:

In the op-amp circuit given in the figure, the load current iL is

  1. $ - \dfrac{V_s}{R_2}$

  2. $ \dfrac{V_s}{R_2}$

  3. $- \dfrac{V_s}{R_L}$

  4. $ \dfrac{V_s}{R_1}$


Correct Option: A
Explanation:

For the BJT QL in the circuit shown below, $\beta = \infty$, $V_{BE_{on}} = 0.7, V_{CE_{sat}}$0.7V. The switch is initially closed. At time t = 0, the switch is opened. The time t at which Q1 leaves the active region is

  1. 10 ms

  2. 25 ms

  3. 50 ms

  4. 100 ms


Correct Option: C
Explanation:

Consider the Schmidt trigger circuit shown below: A triangular wave which goes from - 12 V to 12 V is applied to the inverting input of OPMAP. Assume that the output of the OPAMP swings from + 15 V to - 15 V. The voltage at the non-inverting input switches between

  1. − 12 V to + 12 V

    • 7.5 V to 7.5 V
    • 5 V to + 5 V
  2. 0 V and 5 V


Correct Option: C
Explanation:

A small signal source Vi = A cos 20 t + B sin 106 t is applied to a transistor amplifier as shown below. The transistor has $\beta$= 150 and hie = 3 $\Omega$. Which expression best approximate V0 (t)?

  1. V0 (t) = - 1500 (A cos 20 t + B sin 106 t)

  2. V0 (t) = - 150 (A cos 20 t + B sin 106 t)

  3. V0 (t) = - 1500 B sin 106 t

  4. V0 (t) = - 150 B sin 106 t


Correct Option: D
Explanation:

For a BJT the common base current gain $\alpha$ = 0.98 and the collector base junction reverse bias saturation current ICO 0.6$\mu A$. This BJT is connected in the common emitter mode and operated in the active region with a base drive current IB = 20XA. The collector current IC for this mode of operation is

  1. 0.98mA

  2. 0.99mA

  3. 1.0mA

  4. 1.01mA


Correct Option: D
Explanation:

An ideal sawtooth voltage waveform of frequency 500 Hz and amplitude 3 V is generated by charging a capacitor of 2 $\mu$F in every cycle. The charging requires

  1. constant voltage source of 3 V for 1 ms

  2. constant voltage source of 3 V for 2 ms

  3. constant current source of 3 mA for 1 ms

  4. constant current source of 3 mA for 2 ms


Correct Option: D
Explanation:

Given, rd = 20 k$\Omega$, IDSS = 10 mA, Vp = -8 V

Transconductance in milli-Siemens (mS) and voltage gain of the amplifier are respectively

  1. 1.875 mS and 3.41

  2. 1.875 mS and - 3.41

  3. 3.3 mS and -6

  4. 3.3 mS and 6


Correct Option: B
Explanation:

Given rd = 20 k$\Omega$, IDSS = 10 mA, Vp = -8 V.

ID and IDS under DC conditions are respectively

  1. 5.625 mA and 8.75 V

  2. 4.500 mA and 11.00 V

  3. 7.500 mA and 5.00 V

  4. 6.250 mA and 7.50 V


Correct Option: A
Explanation:

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