Test 2 - Analog Circuits (Electronics and Communication)

Description: Second test of Analog Circuits (Electronics and Communication)
Number of Questions: 20
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Tags: Analog Circuits
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In the following limiter circuit, an input voltage Vi = 10 sin 100 $\pi$t is applied. Assume that the diode drop is 0.7 V when it is forward biased. The zener breakdown voltage is 6.8 V. The maximum and minimum values of the output voltage respectively are

  1. 6.1 V, - 0.7 V

  2. 0.7 V, - 7.5 V

  3. 7.5 V, - 0.7 V

  4. 7.5 V, - 7.5 V


Correct Option: C
Explanation:

For the positive half of $V_i,$ the diode $D_1$ is forward bias, $D_2$ is reverse bias and the zener diode is in breakdown state because state because $V_i > 6.8$

Thus output voltage is $V_0 = 0.7 + 6.8 = 7.5V$

For the negative half of $V_i$, $D_2$ is forward bias thus

Then $V_0 = -0.7 V$

If the differential voltage gain and the common mode voltage gain of a differential amplifier are 48 dB and 2 dB respectively, then its common mode rejection ratio is

  1. 23 dB

  2. 25 dB

  3. 46 dB

  4. 50 dB


Correct Option: C
Explanation:

$$CMMR = \dfrac{A_d}{A_c}$$

OR

$$20\ log\ CMMR = 20\ log\ A_d - 20\ log\ A_c$$

OR

$$ = 48-2 = 46 dB$$

Where $A_d \rightarrow$ Differential Voltage Gain and $A_c \rightarrow$ Common Mode Voltage Gain

The circuit below implements a filter between the input current ii and the output voltage vo. Assume that the opamp is ideal. The filter implemented is a

  1. low pass filter

  2. band pass filter

  3. band stop filter

  4. high pass filter


Correct Option: D
Explanation:

Assume that the $\beta$ of transistor is extremely large and VBE = 0.7V, IC and VCE in the circuit are shown in the figure.

  1. IC = 1 mA, VCE = 4.7 V

  2. IC = 0.5 mA, VCE = 3.75 V

  3. IC = 1 mA, VCE = 2.5 V

  4. IC = 0.5 mA, VCE = 3.9 V


Correct Option: C
Explanation:

Identify the circuit in the following figure?

  1. Low-pass filter

  2. High-pass filter

  3. Band-pass filter

  4. Band-reject filter


Correct Option: A
Explanation:

$$ \text{This is a Low pass filter, because}\\ \begin{array}{ll} At\quad \omega = \infty\qquad & \dfrac{V_o}{V_{in}} = 0 \\ and\ at\ \omega = 0 \qquad & \dfrac{V_o}{V_in} = 1 \end{array} $$

For an npn transistor connected as shown in figure, VBE = 0.7 Volts. Given that reverse saturation current of the junction at room temperature 300°K is 10-13 A, the emitter current is

  1. 30 mA

  2. 39 mA

  3. 49 mA

  4. 20 mA


Correct Option: C
Explanation:

$$I_E = I_s\left(\dfrac{V_{BE}}{enV_r} - 1\right) = 10^{-13}\left(\dfrac{0.7}{e^{1 \times 26 \times 10^{-3}}}-1\right) = 49\ m A $$

In the Op-Amp circuit shown, assume that the diode current follows the equation I = Is exp (V/VT). For Vi = 2V, V0 = V01, and for Vi = 4 V, V0 = V02. The relationship between V01 and V02 is

  1. V02 = $\sqrt2$V01

  2. V02 = e2 V01

  3. V02 = V01 In 2

  4. V01 - V02 = VT In 2


Correct Option: D
Explanation:

Here the inverting terminal is at virtual ground and the current in resistor and diode current is equal i.e.

$$ \begin{aligned} I_R = & \quad I_D \\ or \qquad \dfrac{V_i}{R} = & \quad I_s e^{\frac{V_0}{V_r}} \\ or \qquad V_D = &\quad V_T\ ln \dfrac{V_i}{I_sR} \\ \text{For the first condition}\\ V_D = 0 - V_{o1} = V_T \ ln \ \dfrac{2}{I_sR}\\ \text{For the first condition}\\ V_D = 0 - V_{o1} = V_T \ ln \ \dfrac{4}{I_sR}\\ \text{Subtracting the above equation}\\ V_{o1} - V_{o2} = &\quad V_T\ ln\dfrac{4}{I_sR} - V_T\ ln \dfrac{2}{I_sR} \\ or \quad V_{o1} - V_{o2} = &\quad V_T\ ln\dfrac{4}{2} = V_T\ ln 2 \end{aligned} $$

If the op-amp in figure is ideal, the output voltage Vout will be equal to

  1. 1 V

  2. 6 V

  3. 14 V

  4. 17 V


Correct Option: B
Explanation:

 

In the following a stable multivibrator circuit, which properties of v0 (t) depend on R2?

  1. Only the frequency

  2. Only the amplitude

  3. Both the amplitude and the frequency

  4. Neither the amplitude nor the frequency


Correct Option: A
Explanation:

The $R_2$ decide only the frequecy

In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7 V. The thermal voltage Vt = kT/q = 25 mV and small signal input vi = Vp cos $\omega t$ where Vp = 100 mV.

The ac output voltage vac is

  1. 0.25cos $\omega t$ mV

  2. 1cos $\omega t$ mV

  3. 2cos $\omega t$ mV

  4. 22cos $\omega t$ mV


Correct Option: B
Explanation:

The forward resistance of each diode is

$$ \begin{aligned} r & = \dfrac{V_T}{I_C} \dfrac{25 \ mV}{1 \ mA } = 25\Omega \\ Thus \quad V_{ac} \quad & = V_i \times \left( \dfrac{4(r)}{4(r) + 9900} \right) \\ & = 100 \ mV \ cos(\omega t)0.01 = 1\ cos(\omega t) \ mV \end{aligned} $$

In the amplifier circuit shown in figure, the values of R1 and R2 are such that the transistor is operating at VCE = 3V and IC = 1.5mA when its $\beta$is 150. For a transistor with $\beta$of 200, the operating point (VCE, IC) is

  1. (2V, 2 mA)

  2. (3V, 2 mA)

  3. (4V, 2 mA)

  4. (4V, 1 mA)


Correct Option: A
Explanation:

 

In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7 V. The thermal voltage Vt = kT / q = 25mV. The small signal input vi = Vp cos $\omega t$ where Vp = 100 mV.

The bias current IDC through the diodes is

  1. 1 mA

  2. 1.28 mA

  3. 1.5 mA

  4. 2 mA


Correct Option: A
Explanation:

The current flows in the circuit if all the diodes are forward biased. In forward biased there will be 0.7 drop across each diode. Thus $ I_{DC} = \dfrac{12.7-4(0.7)}{9900} = 1\ mA$

An ideal op-amp is an ideal

  1. voltage controlled current source

  2. voltage controlled voltage source

  3. current controlled current source

  4. current controlled voltage source


Correct Option: B
Explanation:

An ideal OPAMP is an ideal voltage controlled voltage source.

In the circuit shown below, capacitors C1 and C2 are very large and are shorts at the input frequency. vi is a small signal input. The gain magnitude $\left | \dfrac{V_0}{V_1} \right |$ at 10 M

  1. maximum

  2. minimum

  3. unity

  4. zero


Correct Option: A
Explanation:

The correct full wave rectifier circuit is


Correct Option: C
Explanation:

The circuit shown in figure is best described as a

  1. bridge rectifier

  2. ring modulator

  3. frequency discriminatory

  4. voltage doubler


Correct Option: D
Explanation:

This circuit having two diode and capacitor pair in parallel, works as voltage doubler.

The amplifier circuit shown below uses a silicon transistor. The capacitors CC and CE can be assumed to be short at signal frequency and the effect of output resistance R0 can be ignored. If CE is disconnected from the circuit, which one of the following statements is TRUE?

  1. The input resistance Ri increases and the magnitude of voltage gain AV decreases.

  2. The input resistance Ri decreases and the magnitude of voltage gain AV decreases.

  3. Both input resistance Ri and the magnitude of voltage gain AV decrease.

  4. Both input resistance Ri and the magnitude of voltage gain AV increase.


Correct Option: A
Explanation:

The circuit shown in a

  1. low pass filter with f3dB = $\dfrac{1}{(R_1 + R_2)C}rad/s$

  2. high pass filter with f3dB = $\dfrac{1}{(R_1)C}rad/s$

  3. low pass filter with f3dB = $\dfrac{1}{(R_1)C}rad/s$

  4. high pas filter with f3dB = $\dfrac{1}{(R_1 + R_2)C}rad/s$


Correct Option: B
Explanation:

The cascode amplifier is a multistage configuration of

  1. CC-CB

  2. CE-CB

  3. CB-CC

  4. CE-CC


Correct Option: B
Explanation:

The CE configuration has high voltage gain as well as high current gain. It performs basic functions of amplifications. The CB configuration has lowest $R_i$ and highest $R_o$. It is used as last step to match a very low impedance source and to drain a high impedance load. Thus cascade amplifier is a multistage configuration of CE-CB

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