Test 1 - Analog Circuits (Electronics and Communication)

Description: Test 1 of Analog Circuits (Electronics and Communication)
Number of Questions: 19
Created by:
Tags: Analog Circuits GATE
Attempted 0/19 Correct 0 Score 0

In the silicon BJT circuit shown below, assume that the emitter area of transistor Q1 is half that of transistor Q2.

The value of current I0 is approximately

  1. 0.5 mA

  2. 2 mA

  3. 9.3 mA

  4. 15 mA


Correct Option: B
Explanation:

Generally, the gain of a transistor amplifier falls at high frequencies due to the

  1. internal capacitances of the device

  2. coupling capacitor at the input

  3. skin effect

  4. coupling capacitor at the output


Correct Option: A
Explanation:

The gain of amplifier is

$$A_i = \dfrac{-g_m}{g_o + j\omega C}$$

Thus the gain of a transistor amplifier falls at high frequencies due to the internal capacitance that are diffusion capacitance and transition capacitance.

Voltage series feedback (also called series-shunt feedback) results in

  1. increase in both input and output impedances

  2. decrease in both input and output impedances

  3. increase in input impedance and decrease in output impedance

  4. decrease in input impedance and increase in output impedance


Correct Option: C
Explanation:

In voltage series feedback amplifier, input impedance increases by factor $(1+A\beta)$ and output impedance decreases by the factor $(1+A\beta)$

$$R_{if} = R_i(1+A\beta)\\

R_{of} = \dfrac{R_0}{(1+A\beta)} $$

The OPAMP circuit shown above represents a

  1. high pass filter

  2. low pass filter

  3. band pass filter

  4. band reject filter


Correct Option: B
Explanation:

The diodes and capacitors in the circuit shown are ideal. The voltage v(t) across the diode D1 is

  1. – 1 + Cos ($\omega t$)

  2. sin ($\omega t$)

  3. 1 – cos ($\omega t$)

  4. 1 – sin ($\omega t$)


Correct Option: A
Explanation:

An n-channel depletion MOSFET has following two points on its ID -VGS curve: (i) VGS = 0 at ID = 12 mA and (ii) VGS = - 6 volts at ID = 0

Which of the following Q-points will give the highest trans-conductance gain for small signals?

  1. VGS = - 6 Volts

  2. VGS = - 3 Volts

  3. VGS = 0 Volts

  4. VGS = 3 Volts


Correct Option: B
Explanation:

 The transconductance for the junction FET is given by gm = (2 Idss/|Vp |) [1 - (Vgs/Vp)] Vp is the pinch off voltage. Max transconductance can occur between pinch off voltage and at origin. So it lies between - 6V and 0. So Q point = - 3 V

The effect of current shunt feedback in an amplifier is to

  1. increase the input resistance and decrease the output resistance

  2. increase both input and output resistances

  3. decrease both input and output resistances

  4. decrease the input resistance and increase the output resistance


Correct Option: D
Explanation:

The given solution is right.                    Here  Ri  = input resistance without feedback                              Rif  = input resistance with feedback                              R of  = output resistance with feedback                  

In the CMOS circuit shown, electron and hole mobilities are equal, and M1 and M2 are equally sized. The device M1 is in the linear region if

  1. Vin < 1.875 V

  2. Vin < 3.125 V

  3. Vin > 3.125 V

  4. 0 <Vin < 5 V


Correct Option: B
Explanation:

The Op-amp circuit shown in the figure below is a filter. The type of filter and its cut-off frequency are respectively

  1. high pass, 10000 rad/sec

  2. low pass, 1000 rad/sec

  3. high pass, 1000 rad/sec

  4. low pass, 10000 rad/sec


Correct Option: C
Explanation:

At low frequency capacitor is open circuit and voltage acr s non-inverting terminal is zero. At high frequency capacitor act as short circuit and all input voltage appear at non-inverting terminal. Thus, this is high pass circuit. The frequency is given by

$$\omega = \dfrac{1}{RC} = \dfrac{1}{1 \times 10^3 \times 1 \times 10^{-6}} = 1000 \ rad/sec$$

What is the frequency of radio waves radiated out by an oscillating circuit consisting of condenser of capacity 0.02 microfarad and inductance 8 microhenry?

  1. 3.98 x 105 Hz

  2. 4.56 x 105 Hz

  3. 6.21 x 104 Hz

  4. 7.12 x 104 Hz


Correct Option: A
Explanation:

C = 0.02 x 10-6 F L = 8 x 10-6 H $$v = \dfrac{1}{{2\pi}\sqrt{LC}}$$  After putting the values, we get the frequency = 3.98 x 105 Hz

Consider the common emitter amplifier shown below with the following circuit parameters.

B = 100, gm = 0.3861 A/V, r0 = $\infty$, rp = 259 W, RS = 1 kW, RB = 93 kW, RC = 250 W, RL = 1 kW, C1 = $\infty$ and C2 = 4.7 mF.

The lower cut-off frequency due to C2 is

  1. 33.9 Hz

  2. 27.1 Hz

  3. 13.6 Hz

  4. 16.9 Hz


Correct Option: B
Explanation:

In a transconductance amplifier, it is desirable to have

  1. a large input resistance and a large output resistance

  2. a large input resistance and a small output resistance

  3. a small input resistance and a large output resistance

  4. a small input resistance and a small output resistance


Correct Option: A
Explanation:

In the transconductance amplifier, it is desirable to have large input resistance and large output resistance. ideally, Id = Gm.Vs Id = Gm.Vi(Rd/Rd+RL) if Rd>>RL, Id = Gm Vi Vi = Vs if Ri >> Rs

Consider the common emitter amplifier shown below with the following circuit parameters.

B = 100, gm = 0.3861 A/V, r0 = $\infty$, rp = 259 W, RS = 1k W, RB = 93kW, RC = 250 W, RL = 1k W, C1 = $\infty$ and C2 = 4.7mF.

The resistance seen by the source Vs is

  1. 258$\Omega$

  2. 1252$\Omega$

  3. 93 K$\Omega$

  4. $\infty$


Correct Option: B
Explanation:

By using given solution figure and solution                            Rin = 1000 + [(9300 * 259)/(9300 + 259)] = 1252 om                                                      

The input resistance Ri of the amplifier shown in the figure is

  1. $\dfrac{30}{4}K\Omega$

  2. 10 k$\Omega$

  3. 40 k$\Omega$

  4. infinite


Correct Option: B
Explanation:

Since the inverting terminal is at virtual ground, the current flowing through the voltage source is $$ I_s = \dfrac{V_s}{10K} \\ or \quad \dfrac{V_s}{I_s} = 10 \ K\Omega = R_{in} $$

The current ib through the base of a silicon npn transistor is 1 + 0.1 cos(10000 $\pi$t) mA. At 300 K, the r$\pi$ in the small signal model of the transistor is

  1. 250$\Omega$

  2. 27.5$\Omega$

  3. 25$\Omega$

  4. 22.2$\Omega$


Correct Option: C
Explanation:

Choose the correct match for input resistance of various amplifier configurations shown below.||| |---|---| | Configuration| Input resistance| | CB: Common Base| O: Low| | CC: Common Collector| MO: Moderate| | CE: Common Emitter| HI: High|

  1. CB-LO, CC-MO, CE-HI

  2. CB-LO, CC-HI, CE-MO

  3. CB-MO, CC-HI, CE-LO

  4. CB-HI, CC-LO, CE-MO


Correct Option: B
Explanation:

The voltage gain Av of the circuit shown below is

  1. |Av| $\approx$ 200

  2. |Av| $\approx$ 100

  3. |Av| $\approx$ 20

  4. |Av| $\approx$ 10


Correct Option: D
Explanation:

Directions: Consider the CMOS circuit shown below, where the gate voltage v0 of the n-MOSFET is increased from zero, while the gate voltage of the p −MOSFET is kept constant at 3 V. Assume that, for both transistors, the magnitude of the threshold voltage is 1 V and the product of the trans-conductance parameter is 1 mAV-2.

For small increase in VG beyond 1 V, which of the following gives the correct description of the region of operation of each MOSFET?

  1. Both the MOSFETs are in saturation region.

  2. Both the MOSFETs are in triode region.

  3. n -MOSFET is in triode and p −MOSFET is in saturation region.

  4. n -MOSFET is in saturation and p −MOSFET is in triode region.


Correct Option: D
Explanation:

For small increase in $v_G$ beyond 1V the n-channel MOSFET goes into saturation as $V_{GS} \rightarrow +ive \ and\ p-MOSFET$ is always in active region or triode region.

Directions: Consider the CMOS circuit shown, where the gate voltage v0 of the n-MOSFET is increased from zero, while the gate voltage of the p−MOSFET is kept constant at 3 V. Assume that, for both transistors, the magnitude of the threshold voltage is 1 V and the product of the trans-conductance parameter is 1 mA V-2.

Estimate the output voltage V0 for VG = 15 V.

  1. 4 - $\dfrac{1}{\sqrt2}$

  2. 4 + $\dfrac{1}{\sqrt2}$

  3. 4 - $\dfrac{\sqrt3}{2}$

  4. 4 + $\dfrac{\sqrt3}{2}$


Correct Option: D
Explanation:

Vo = 4 + √3/2 = 4 + 0.866 = 4.866 V This voltage is near to supply voltage, while when it increases to more than 15 V, then the output will decrease.

- Hide questions