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1.	What is IC 723?(a) A voltage regulator(b) A full-wave rectifier(c) A half-wave rectifier(d) A clipperI have been asked this question in a job interview.Query is from Voltage Regulators topic in chapter Regulators of Analog Circuits
Answer» RIGHT answer is (a) A voltage REGULATOR Explanation: The IC 723 is a voltage regulator, which can act as both a low voltage regulator as WELL as a high voltage regulator. Output can be set between 7-37 volts. 7 volts is the reference starting voltage.

2.	For a Zener diode shunt regulator, the source current is IS, the Zener diode current is IZ and the load current is IL. The source voltage is VS, Zener voltage is VZ and load voltage is VL. The load resistance is RL. What is the correct option for the safe operation of the diode?(a) IS = IZ + IL(b) IS =< IZmax + IL(c) IS =< IZmin + IL(d) VL = VZThis question was posed to me in a job interview.The above asked question is from Voltage Regulators topic in portion Regulators of Analog Circuits
Answer» The correct choice is (b) IS =< IZmax + IL To explain I would say: For proper OPERATION, the CURRENT through RS should be at LEAST equal to the sum of IZMIN a specified load current. IS >= IZmin + IL For safe operation of the diode, IS =< IZmax + IL.

3.	What is line regulation?(a) The process of keeping Zener diode voltage constant inspite of changes in AC supply(b) The process of keeping load voltage constant irrespective of the fluctuation in AC supply or the line voltage(c) The process of keeping load voltage constant irrespective of fluctuation in load current(d) The process of keeping Zener current constant irrespective of fluctuation in AC supplyI had been asked this question by my college professor while I was bunking the class.Question is from Voltage Regulators topic in portion Regulators of Analog Circuits
Answer» The correct choice is (b) The process of KEEPING load voltage constant IRRESPECTIVE of the fluctuation in AC supply or the line voltage Explanation: Line regulation is the process of keeping the load voltage constant, irrespective of fluctuation in AC supply or the line voltage. In line regulation, the load CURRENT is considered constant.

4.	Find the power rating of the diode in the given circuit. The breakdown voltage of the diode is 5V.(a) 200 mW(b) 125 mW(c) 250 mW(d) 300 mWThe question was posed to me in class test.This interesting question is from Voltage Regulators topic in chapter Regulators of Analog Circuits
Answer» The correct option is (C) 250 mW To explain I would say: SOURCE current IS = 15-5/200 = 0.05A = 50 mA The power rating of the diode is the maximum power it can dissipate which occurs when the LOAD is DISCONNECTED because then the whole current flows into the diode. Hence maximum power rating is P = VZxIZ = 5×50 = 250 mW.

5.	What is load regulation?(a) The process of keeping the load voltage constant irrespective of any change in AC supply(b) The process of keeping the load voltage constant irrespective of variations in load current(c) The process of keeping load voltage constant irrespective of variations in source current(d) The process of keeping load current constant irrespective of variations in AC supplyThe question was posed to me by my college professor while I was bunking the class.The doubt is from Voltage Regulators topic in section Regulators of Analog Circuits
Answer» The CORRECT option is (b) The PROCESS of keeping the load VOLTAGE constant irrespective of variations in load current For explanation: Load REGULATION is the process of keeping VO constant irrespective of variations in load current. The line voltage is taken to be constant during load regulation.

6.	In a power supply, the output voltage can vary due to multiple reasons. Which of these is not true if it is found that the output voltage is constant?(a) The voltage stability factor is very high(b) The output resistance is zero(c) The temperature coefficient is zero(d) The voltage stability factor is very smallI got this question by my school principal while I was bunking the class.Question is taken from Voltage Regulators topic in portion Regulators of Analog Circuits
Answer» CORRECT option is (a) The voltage stability factor is very high To elaborate: In a POWER supply, the output voltage can VARY DUE to changes in AC supply, LOAD current and the temperature. VO = f(VS, IL,T) ΔVO = SV.ΔVS + ROΔIL + STΔT.

7.	Which of these is a not drawback of Zener diode shunt regulator?(a) The output voltage is fixed(b) The output voltage can vary with temperature(c) Variation in load current needs to be minimal(d) It is difficult to designI had been asked this question in a national level competition.My enquiry is from Voltage Regulators topic in chapter Regulators of Analog Circuits
Answer» The CORRECT choice is (d) It is difficult to design The best explanation: The Zener diode shunt regulator is a simple VOLTAGE regulator circuit that provides a fixed output voltage. However, due to the presence of the Zener diode, the output is temperature DEPENDENT because the breakdown voltage of Zener diode depends on temperature. When load current VARIES a LOT, then Zener current also varies which causes a change in output voltage.

8.	Consider the following circuit. Find the power dissipation of the transistor given that the diode breakdown voltage is 5V, R1=20kΩ, R2=100kΩ, R3=200kΩ, R4=10Ω. The source voltage VS=20V.(a) 12.12W(b) 9.375W(c) 9.27W(d) 10.575WI have been asked this question in a job interview.This key question is from Voltage Regulators in chapter Regulators of Analog Circuits
Answer» CORRECT ANSWER is (b) 9.375W For explanation I would say: Power DISSIPATION = VCE x IC VCE = VC – VE = VS – VOUT = 20-5(1+R2/R3) = 20-5×3/2 = 20-7.5 = 12.5V IC ≈ IE = load current, since R2 and R3 are greater than the load, so most current flows through the load. IL = VOUT/10 = 0.75A P = 12.5 x 0.75 = 9.375W.

9.	The following is a shunt regulator. Find maximum power dissipation of the Zener diode and transistor, given that the source voltage varies from 20-40V, the Zener breakdown voltage is 5V, the output voltage is 10V, the resistance R1=50Ω, R2=20kΩ, β=99, VBE=0.5V.(a) PTransistor = 5.12 W, PZener = 0.051 W(b) PTransistor = 0.41 W, PZener = 0.57 W(c) PTransistor = 5.94 W, PZener = 0.057 W(d) PTransistor = 6.22 W, PZener = 5.66 WThe question was posed to me during an interview for a job.The question is from Voltage Regulators topic in portion Regulators of Analog Circuits
Answer» Correct ANSWER is (c) PTransistor = 5.94 W, PZener = 0.057 W Explanation: Output voltage = 10V = VCE Current ACROSS R1 = VS-VO/R1 For maximum power dissipation, supply is maximum I = 40-10/50 = 0.6A Current I = IZ + IC = IB + βIB = 100IB IB = 6mA IC = 0.594 A PTransistor = 100.594 = 5.94 W PZener = 10-0.5 6mA = 57mA = 0.057 W.

10.	What is not related to a transistorized series regulator?(a) The output can be varied by using a variable resistor(b) The output is independent of temperature(c) The overload and short circuit protection is not required(d) The circuit has negative feedback responsible for regulationI got this question in an interview for internship.This intriguing question originated from Voltage Regulators in portion Regulators of Analog Circuits
Answer» Correct choice is (c) The overload and short circuit protection is not required To explain I would say: In a transistorized series regulator, the Zener diode maintains emitter voltage. An increase in output is canceled by a decrease in the output. The circuit has negative feedback for this regulation. The output MAY be varied by using a variable resistor and change in output due to temperature due to the Zener diode is canceled by the change in VBE of the transistor. However, when the LOAD is REDUCED, or an accidental short-circuit occurs, overload and a short circuit occurs, and protection is needed to prevent that.

11.	In a transistorized series regulator, how is the overload and short-circuit protection provided?(a) By the use of a thermistor(b) By using two additional diodes and a current sensing resistor to protect the series transistor(c) By using a diode and an additional resistor to protect the transistor(d) By using a diode along with a capacitor of a small capacitance value in seriesThis question was addressed to me during an interview.I'm obligated to ask this question of Voltage Regulators topic in section Regulators of Analog Circuits
Answer» Correct choice is (b) By using two additional DIODES and a CURRENT sensing resistor to protect the series TRANSISTOR For explanation: Series transistor can be protected by connecting two additional diodes and a current sensing resistor to the circuit. When the load current is smaller, the diode is off and load current is through the transistor and the new resistor. When the load current increases, the voltage DROP across the new resistor increases and soon the diodes start CONDUCTION. A limiting current flows through the transistor.

12.	What is the output of the IC 7924?(a) 12V(b) -12V(c) 24V(d) -24VI had been asked this question during an interview for a job.This intriguing question comes from Voltage Regulators topic in chapter Regulators of Analog Circuits
Answer» RIGHT option is (d) -24V Easiest explanation: The IC of series 78xx and 79xx are fixed voltage regulators, wherein the 78 represents those with a positive OUTPUT and 79 is for those with a negative output. The xx VALUE represents the magnitude of the output voltage being achieved. For proper operation, the input voltage should be at least 2V GREATER than the output voltage.

13.	In the IC 7805, what is the minimum input voltage for proper functioning?(a) 5V(b) 6V(c) 7V(d) 8VI have been asked this question in an online quiz.I'm obligated to ask this question of Voltage Regulators in division Regulators of Analog Circuits
Answer» Right answer is (C) 7V The EXPLANATION is: For a FIXED voltage IC regulator, the input voltage should be at least 2V greater than the output voltage. A MINIMUM voltage of 2V should be allowed to drop in the internal circuit of the IC.