# Analog Electronic Circuits Questions and Answers – Limits of Operation

This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Limits of Operation”.

1. For a BJT, what is typically the shape of the power dissipation curve, if it’s plotted on the output characteristics?
a) Parabola
b) Linear
c) Hyperbola
d) Circular

Explanation: Power Dissipation in a BJT is given by P=VCE.IC. This is in the form of k=xy which is the equation of a hyperbola.

2. What is the region on the output characteristics below IC = ICEO line called?
a) Active region
b) Cutoff region
c) Saturation region
d) Active & Saturation region

Explanation: The region below IC = ICEO is called the cutoff region.

3. What is the region on the output characteristics for VCE < VCEsat called?
a) Active region
b) Cutoff region
c) Saturation region
d) Active & Cutoff region

Explanation: The region below VCE < VCEsat is called the saturation region.

4. From the given characteristics, the value of VCEsat is closest to which of the following values?

a) 0.3 V
b) 1 V
c) 5 V
d) 20V

Explanation: From the given characteristics, the saturation voltage is obtained through the VCE vs IC graph where in the approximate saturation region is the area where the dotted vertical line near to the origin is present. Hence we can estimate the value of VSat to be that of 0.3V.

5. Given that the collector power dissipation is 300 mW, what is the value of collector current for the collector to emitter voltage = 12 V?
a) 50 mA
b) 0 mA
c) 25 mA
d) 100 mA

Explanation: P = VCE.IC = > 300mW = (12V)IC = > IC=300/12 mA = 25 mA.

6. Given that the collector power dissipation is 300 mW, what is the value of collector to emitter voltage for collector current = 50 mA?
a) 6 V
b) 3 V
c) 0 V
d) 2 v

Explanation: P = VCE.IC = > 300mW = VCE(50 mA) = > VCE = 300/50 = 6 V.

7. From the given curve tracer response, what is the value of β for IC = 7 mA and VCE = 5 V?

a) 150
b) 180
c) 250
d) 120

Explanation: From the curve, we get change in IC = (8.2-6.4) mA and change in IB = 10 uA. Hence, IC/beta; = (1.8/0.01) = 180.

8. If the positive lead of a DMM, with the mode set to ohmmeter is connected to the base and the negative lead to the emitter and a low resistance reading is obtained, then what is the type of transistor that is being tested?
a) NPN
b) PNP
c) Faulty transistor
d) Not a transistor, it is a FET

Explanation: There are multiple ways to test the BJT to be npn or pnp using a DMM based upon which terminals are connected to which lead of the DMM. If positive is to base and negative to the emitter of the BJT, and if a low reading is obtained and not a over limit, then the transistor is NPN.

9. If the positive lead of a DMM, with the mode set to ohmmeter is connected to the base and the negative lead to the emitter and a high resistance reading is obtained, then what is the type of transistor that is being tested?
a) npn
b) pnp
c) faulty
d) not a transistor, it is a FET

Explanation: If the positive lead of a DMM, with the mode set to ohmmeter is connected to the base and the negative lead to the emitter and a low resistance reading is obtained, then what is the type of transistor that is being tested is npn.

10. For the given transistor, what is the correct sequence of the pins from left to right?

a) ECB
b) BCE
c) CEB
d) CBE

Explanation: With the curved side facing us, the answer can either be collector-base-emistter, left to right, or emitter-base-collector. Hence the correct option is CBE, and that applies for an NPN transistor.

Sanfoundry Global Education & Learning Series – Analog Circuits.

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