# Linear Integrated Circuit Questions and Answers – IC Voltage Regulator

This set of Linear Integrated Circuit Multiple Choice Questions & Answers (MCQs) focuses on “IC Voltage Regulator”.

1. Which is not considered as a linear voltage regulator?
a) Fixed output voltage regulator
c) Switching regulator
d) Special regulator

Explanation: In linear regulator’s the impedance of active element may be continuously varied to supply a desired current to the load. But in the switching regulator, a switch is turned on and off.

2. What is the dropout voltage in a three terminal IC regulator?
a) |Vin| ≥ |Vo|+2v
b) |Vin| < |Vo|-2v
c) |V in| = |Vo|
d) |Vin| ≤ |Vo|

Explanation: The unregulated input voltage must be atleast 2V more than the regulated output voltage. For example, if Vo=5V, then Vin=7V.

3. To get a maximum output current, IC regulation are provided with
b) Heat sink
c) Peak detector
d) None of the mentioned

Explanation: The load current may vary from 0 to rated maximum output current. To maintain this condition, the IC regulator is usually provided with a heat sink; otherwise it may not provide the rated maximum output current.

4. For the given circuit, let VEB(ON)=1v, ß= 15 and IO=2mA. Calculate the load current a) IL = 23.45A
b) IL = 46.32A
c) IL = 56.87A
d) IL = 30.75A

Explanation: The equation for load current, IL = [(ß+1)IO]-[ß×(VEB(ON)/R1)]=[(15+1)×2]–[15×(1v/12 Ω)] =32-1.25 =30.75A.

5. Which type of regulator is considered more efficient?
a) All of the mentioned
b) Special regulator
c) Fixed output regulator
d) Switching regulator

Explanation: The switching element dissipates negligible power in either on or off state. Therefore, the switching regulator is more efficient than the linear regulators.

6. State the reason for thermal shutdown of IC regulator?
a) Spikes in temperature
b) Decrease in temperature
c) Fluctuation in temperature
d) Increase in temperature

Explanation: The IC regulator has a temperature sensor (built-in) which turn off the IC, when it becomes too hot (usually 125oC-150oC). The output current will drop and remains there until the IC has cooled significantly.

7. Find the difference between output current having a load of 100Ω and 120Ω for 7805 IC regulator. Consider the following specification: Voltage across the load = 5v; Voltage across the internal resistor= 350mv.
a) 8.4mA
b) 7mA
c) 9mA
d) 3.4mA

Explanation: Given the voltage across the internal resistor to be 350mv, which is less than 0.7v. Hence the transistor in 7805 is off.
When load = 100Ω, IL= IO= Ii= 5v/100 Ω = 50mA
When load=120Ω, IO= 5v/120 Ω = 41.6mA.
So, the difference between the output voltage = 50-41.6mA = 8.4mA.

8. The change in output voltage for the corresponding change in load current in a 7805 IC regulator is defined as
a) All of the mentioned
b) Line regulation
d) Input regulation

Explanation: Load regulation is defined as the change in output voltage for a change in load current and is also expressed in millivolts or as a percentage of output voltage.

9. An IC 7840 regulator has an output current =180mA and internal resistor =10Ω. Find the collector current in the output using the transistor specification: ß=15 and VEB(ON) =1.5v.
a) 270mA
b) 450mA
c) 100mA
d) 50mA

Explanation: The collector current from transistor, IC= ßIB
Where, IB= IO-(VEB(ON)/R1) = 180mA-(1.5v-10Ω) = 0.03A.
Therefore, IC= 15×0.03 = 0.45A = 450mA.

10. How the average temperature coefficient of output voltage expressed in fixed voltage regulator?
a) miilivolts/oC
b) miilivoltsoC
c) None of the mentioned
d) oC/ miilivolts

Explanation: The temperature stability or average temperature coefficient of output voltage, is the change in the output voltage per unit change in temperature and expressed in miilivolts/oC.

11. In the circuit given below, let VEB(ON)=0.8v and ß=16. Calculate the output current coming from 7805 IC and collector current coming from transistor Q1 for a load of 5Ω. a) IO =111mA, IC= 808mA
b) IO =111mA, IC= 829mA
c) IO =111mA, IC= 881mA
d) IO =111mA, IC= 889mA

Explanation: When load = 5Ω, IL= 5v/5Ω =1A. The voltage across R1 is 7Ω × 1A=7v. Since, IL is more than 100mA, the transistor Q1 turns on and supplies the extra current required.
Therefore, IL =(ß+1)IO-[ß×(VEB(ON)/R1)
IO = [IL/(ß+1)]+ [ß×(VEB(ON)/R1) = [1/(16+1)]+[16×(0.8/2Ω)] ≅111mA.
=> IC=IL-IO=1A-111mA =889mA.

12. Calculate the output voltage for LM314 regulator. The current IADJ is very small in the order of 100µA. (Assume VREF=1.25v) a) 17.17v
b) 34.25v
c) 89.34v
d) 23.12v

Explanation: The output voltage, VO =VREF[1=(R2/R1)]+(IADJ×R2)=1.25Vin× [1+(3kΩ/240Ω)] +( 100µA×3kΩ )= 16.875 +0.3.
=> VO=17.17v.

13. Compute the input voltage of 7805c voltage regulator with a current source that will deliver a 0.725A current to 65Ω, 10w load. (Assume reference voltage =5v)
a) Vin = 84v
b) Vin = 34v
c) Vin = 54v
d) Vin = 64v

Explanation: VO=VREF+VL =VREF+(IL×RL) = 5v+(0.725A×65Ω) = 52.125v
=> Input voltage, Vin = VO + dropout voltage = 52.125v+2v.
=> Vin ≅54v.

14. Which of the following is not a characteristic of adjustable voltage regulators?
a) Non-versatile
b) Better performance
c) Increased reliability
d) None of the mentioned

Explanation: Adjustable voltage regulators are versatile; it has improved over-load protection allowing greater output current over operating temperature range.

Sanfoundry Global Education & Learning Series – Linear Integrated Circuits.

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