This set of Tricky Linear Integrated Circuit Questions and Answers focuses on “Multiplier and Divider – 2”.

1. Determine the relationship between log-antilog method.

a) lnV_{x}×lnV_{y} = ln(V_{x}+V_{y})

b) lnV_{x} / lnV_{y} = ln(V_{x}-V_{y})

c) lnV_{x} -lnV_{y} = ln(V_{x}/V_{y})

d) lnV_{x}+ lnV_{y} = ln(V_{x}×V_{y})

View Answer

Explanation: The log-anilog method relies on the mathematical relationship that is the sum of the logarithm of the product of those numbers.

=> lnV

_{x}+lnV

_{y}= ln(V

_{x}×V

_{y}).

2. Which circuit allows to double the frequency?

a) Frequency doubler

b) Square doubler

c) Double multiplier

d) All of the mentioned

View Answer

Explanation: The multiplication of two sine wave of same frequency with different amplitude and phase allows in doubling a frequency.

3. Compute the output of frequency doubler. If the inputs V_{x} = Vsinωt and V_{y} = Vsin(ωt+θ) are applied to a four quadrant multiplier?

a) V_{o}= { V_{x}×V_{y}× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/ V_{ref}

b) V_{o}= { V_{x}×V_{y}× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/ 2

c) V_{o}= { V_{x}×V_{y}× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/(2×V_{ref})

d) V_{o}= – { V_{x}×V_{y}× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/(2×V_{ref})

View Answer

Explanation: The output of frequency doubler, V

_{o}= (V

_{x}×V

_{y})/V

_{ref}= [(Vsinωt +Vsin(ωt+θ)]/ V

_{ref}

= {V

_{x}×V

_{y}×[(sin

^{2}ωt×cosθ)+(sinθ×sinωt×cosωt)]}/ V

_{ref}

=> V

_{o}= { V

_{x}×V

_{y}× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/(2×V

_{ref}).

4. How to remove the dc term produced along with the output in frequency doubler?

a) Use a capacitor between load and output terminal

b) Use a resistor between load and output terminal

c) Use a Inductor between load and output terminal

d) Use a potentiometer between load and output terminal

View Answer

Explanation: The output of the frequency doubler contains a dc term and wave of double frequency. The dc term can be easily removed by using a 1µF coupling capacitor between load and output terminal of the doubler circuit.

5. Find the voltage range at which the multiplier can be used as a squarer circuit?

a) 0 – V_{in}

b) V_{ref} – V_{in}

c) 0 – V_{ref}

d) All of the mentioned

View Answer

Explanation: The basic multiplier can be used to square any positive or negative number provided the number can be represented by a voltage between 0 to V

_{ref}.

6. Calculate the output voltage of a squarer circuit, if it input voltage is 3.5v. Assume V_{ref} = 9.67v.

a) 2.86v

b) 1.27v

c) 10v

d) 4.3v

View Answer

Explanation: The output voltage of a squarer circuit, V

_{o}= V

_{i}

^{2}/ V

_{ref}= (3.5)

^{2}/ 9.67v = 1.27v.

7. Which circuit can be used to take square root of a signal?

a) Divider circuit

b) Multiplier circuit

c) Squarer circuit

d) None of the mentioned

View Answer

Explanation: A divider circuit can be used to find square roots by connecting both the inputs of the multiplier (in the divider) to the output of an op-amp.

8. Find the output voltage of squarer circuit?

a) V_{o} = √( V_{ref}/|V_{in}|)

b) V_{o} = √( V_{ref}×|V_{in}|)

c) V_{o} = √( V_{ref}^{2}×|V_{in}|)

d) V_{o} = √( V_{ref}×|V_{in}|^{2})

View Answer

Explanation: The output of the square root circuit is proportional to the square root of the magnitude of the input voltage.

9. Find the current, I_{L} flowing in the circuit given below

a) I_{L} = 1.777mA

b) I_{L} = 1.048mA

c) I_{L} = 1.871mA

d) None of the mentioned

View Answer

Explanation: Voltage, V

_{L}= V

_{o}

^{2}/ V

_{ref}=13

^{2}/10 = 16.9v.

∴ I

_{L}= V

_{L}/R = 16.9/12kΩ = 1.408mA.

10. A square root circuit build from multiplier is given an input voltage of 11.5v. Find its corresponding output voltage?

a) 11v

b) 15v

c) 13v

d) Cannot be determined

View Answer

Explanation: The input voltage must be negative or else the op-amp will saturate and lie between the ranges of -1 to -10v.

**Sanfoundry Global Education & Learning Series – Linear Integrated Circuits.**

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