# Differential Amplifier MCQ [Free PDF] – Objective Question Answer for Differential Amplifier Quiz

1. A Differential Amplifier should have the collector resistor’s value (RC1 & RC2) as

A. 5kΩ, 5kΩ
B. 5Ω, 10kΩ
C. 5Ω, 5kΩ
D. 5kΩ, 10kΩ

The values of collector current will be equal in the differential amplifier (RC1=RC2).

2. A Differential Amplifier amplifies

A. Input signal with higher voltage
B. Input voltage with smaller voltage
C. Sum of the input voltage
D. None of the Mentioned

The purpose of the differential amplifier is to amplify the difference between two signals.

3. The value of emitter resistance in Emitter Biased circuit are RE1=25kΩ & RE2=16kΩ. Find RE

A. 9.756kΩ
B. 41kΩ
C. 9.723kΩ
D. 10kΩ

In emitter biased circuit, RE1 & RE2 are connected in parallel combination.

⇒ RE = RE1 II RE2

= (RE1× RE2)/(RE1+RE2)

= (25kΩ × 16kΩ)/(25kΩ+16kΩ) = 9.7561kΩ.

4. If the output is measured between two collectors of transistors, then the Differential amplifier with two input signals is said to be configured as

A. Dual Input Balanced Output
B. Dual Input Unbalanced Output
C. Single Input Balanced Output
D. Dual Input Unbalanced Output

When two input signals are applied to the base of the transistor, it is said to be Dual Input. When both collectors are at the same DC potential with respect to ground, then it is said to be a Balance Output.

5. A differential amplifier is capable of amplifying

A. DC input signal only
B. AC input signal only
C. AC & DC input signal
D. None of the Mentioned

Direct connection between stages removes the lower cut-off frequency imposed by the coupling capacitor; therefore it can amplify both AC and DC signals.

6. In an ideal Differential Amplifier, if the same signal is given to both inputs, then the output will be

A. Same as input
B. Double the input
C. Not equal to zero
D. Zero

In an ideal amplifier, the Output voltage

⇒ Vout = Vin1 −Vin2

7. An emitter bias Dual Input Balanced Output differential amplifier has VCC=20v, β=100, VBE=0.7v, RE=1.3kΩ. Find IE

A. 7.42mA
B. 9.8mA
C. 10mA
D. 8.6mA

Emitter current can be found out by substituting the values in the equation,

⇒ IE = (VEE − VBE)/(2RE)

= (20v − 07v)/(2 × 1.3kΩ) = 7.42mA.

8. Find IC, given VCE=0.77v, VCC=10v, VBE=0.37v and RC=2.4kΩ in Dual Input Balanced Output differential amplifier

A. 0.4mA
B. 0.4A
C. 4mA
D. 4A

Substitute the values in the collector to the emitter voltage equation,

VCE= VCC+ VBE-RC IC

⇒IC = (VCC-VCE+VBE)/RC

= (10v-0.77v+0.37v)/2.4kΩ = 4mA

9. Find the correct match

Configuration Voltage gain and Input resistance
1. Single Input Unbalanced Output i. Ad = Rc/re , Ri1 Ri2 = 2βacRE
2. Dual Input Balanced Output ii. Ad= Rc/2re , Ri1 Ri2 = 2βacRE
3. Single Input Balanced Output iii. Ad= Rc/re , Ri = 2βacRE
4. Dual Input Unbalanced Output iv. Ad = Rc/2re , Ri = 2βacRE

A. 1-i , 2-iii, 3-iv, 4-ii
B. 1-iv, 2-ii, 3-iii, 4-i
C. 1-ii, 2-iv, 3-i , 4-iii
D. 1-iii, 2-i, 3-ii, 4-iv

The correct match is

Configuration Voltage gain and Input resistance
1. Single Input Unbalanced Output Ad= Rc/re , Ri = 2βacRE
2. Dual Input Balanced Output Ad = Rc/re , Ri1 Ri2 = 2βacRE
3. Single Input Balanced Output Ad= Rc/2re , Ri1 Ri2 = 2βacRE
4. Dual Input Unbalanced Output Ad = Rc/2re , Ri = 2βacRE

10. Obtain the collector voltage, for the collector resistor (RC. =5.6kΩ, IE=1.664mA, and VCC=10v for single input unbalanced output differential amplifier

A. 0.987v
B. 0.682v
C. 0.555v
D. None of the mentioned

Substitute the given values in the collector voltage equation,
VC= VCC – RC×IC

⇒ VC= 10v – 5.6kΩ×1.664mA (∵ IC ≅ IE )

⇒ VC= 0.682v.

11. For the circuit shown below, determine the Output voltage (Assume β=5, differential input resistance=12 kΩ)

A. 4.33v
B. 2.33v
C. 3.33v
D. 1.33v

From the circuit dig, RC=10kΩ, Vin1= 1.3v and Vin2=0.5v,
Differential input resistance = 2 βre,
⇒ 12kΩ = 2×5×Re
⇒ Re = 1.2 kΩ

Output voltage Vo = RC/2Re(Vin1-Vin2)
⇒ Vo = 10kΩ/(2 ×1.2kΩ) × (1.3v-0.5v)
⇒ Vo = 3.33v.

12. In a Single Input Balanced Output Differential amplifier, given VCC=15v, RE = 3.9kΩ, VCE=2.4 v and re=250Ω. Determine Voltage gain

A. 26
B. 56
C. 38
D. 61

In single Input Balance Output amplifier,

⇒ IE = (VEE-VBE)/2RE

=(15v-0.7v)/(2×3.9kom)

= 1.83mA (∵VCC=VEE)

From the equation, VCE = VCC +VBE-RC×IC

⇒ RC = (14.3v – 2.4v)/1.83mA = 6.5kΩ

The voltage gain, Vo
⇒ Vo = RC/re

= 6.5kΩ/250Ω = 26(no units).

13. Which is not the internal circuit of an operational amplifier?

A. Differential amplifier
B. Level translator
C. Output driver
D. Clamper

Clamper is an external circuit connected to the output of the Operational amplifier, which clamps the output to the desired DC level.

14. The purpose of the level shifter in the Op-amp internal circuit is to

B. Increase impedance
C. Provide high gain
D. Decrease input resistance

The gain stages in Op-amp are directly coupled. So, a level shifter is used for the adjustment of the DC level.

15. How a symmetrical swing is obtained at the output of Op-amp?

A. Providing amplifier with negative supply voltage
B. Providing amplifier with positive voltage
C. Providing amplifier with positive& negative voltage
D. None of the mentioned

For example, consider a single voltage supply +15v. During a positive half cycle, the output will be +5v and -10v during the negative half cycle.

Therefore, the maximum peak to peak output swing, -5v (-10v) = -15v (Asymmetrical swing).

So, to get a symmetrical swing both positive and negative supply voltage with bias point fixed suitably is required.

16. What is the purpose of the differential amplifier stage in the internal circuit of Op-amp?

A. Low gain to the differential mode signal
B. Cancel difference mode signal
C. Low gain to the common-mode signal
D. Cancel common-mode signal

Any undesired noise, common to both of the input terminals is suppressed by a differential amplifier.

17. Which of the following is not preferred for the input stage of Op-amp?

A. Dual Input Balanced Output
B. Differential Input Single-ended Output
D. Single Input Differential Output

Cascaded DC amplifier suffers from a major problem of drift of the operating point, due to temperature dependency of the transistor.

18. What will be the emitter current in a differential amplifier, where both the transistor are biased and matched? (Assume current to be IQ)

A. IE = IQ/2
B. IE = IQ
C. IE = (IQ)2/2
D. IE = (IQ)2

Due to the symmetry of the differential amplifier circuit, current IQ divides equally through both transistors.

19. From the circuit, determine the output voltage (Assume αF=1) A. VO1=3.9v , VO2=12v
B. VO1=12v , VO2=3.9v
C. VO1=12v , VO2=0v
D. VO1=3.9v , VO2=-3.9v

The voltage at the common emitter ‘E’ will be -0.7v, which makes Q1 off and the entire current will flow through Q2.

⇒ VO1 = VCC VO2 = VCC-αF×IQ×RC,

⇒ VO1 = 12v , VO2 = 12v-1×3mA × 2.7k = 3.9v.

20. At what condition does differential amplifier function as a switch

A. 4VT < Vd < -4VT
B. -2VT ≤ Vd ≤ 2VT
C. 0 ≤ Vd < -4VT
D. 0 ≤ Vd ≤ 2VT

For Vd > 4VT, the output voltage is VO1 = VCC, VO2= VCC-αF IQRC. Therefore, a transistor Q1 will be ON and Q2 will be OFF. Similarly for Vd> -4VT, both transistors Q2 & Q1 will be ON.

21. For Vd > ±4VT, the function of the differential amplifier will be

A. Switch
B. Limiter
C. Automatic gain control
D. Linear Amplifier

At this condition, the input voltage of the amplifier is greater than ±100mv and thus acts as a limiter.

22. Change in value of common-mode input signal in differential pair amplifier make

A. Change in voltage across the collector
B. Slight change in collector voltage
C. Collector voltage decreases to zero
D. None of the mentioned