Analog Electronic Circuits

Limits of Operation

Question 1
Marks : +2 | -2
Pass Ratio : 100%
What is the region on the output characteristics for VCE < VCEsat called?
Active region
Cutoff region
Saturation region
Active & Cutoff region
Explanation:
The region below VCE < VCEsat is called the saturation region.
Question 2
Marks : +2 | -2
Pass Ratio : 100%
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?
npn
pnp
faulty
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.
Question 3
Marks : +2 | -2
Pass Ratio : 100%
Given that the collector power dissipation is 300 mW, what is the value of collector to emitter voltage for collector current = 50 mA?
6 V
3 V
0 V
2 v
Explanation:
P = VCE.IC = > 300mW = VCE(50 mA) = > VCE = 300/50 = 6 V.
Question 4
Marks : +2 | -2
Pass Ratio : 100%
For the given transistor, what is the correct sequence of the pins from left to right?
ECB
BCE
CEB
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.
Question 5
Marks : +2 | -2
Pass Ratio : 100%
What is the region on the output characteristics below IC = ICEO line called?
Active region
Cutoff region
Saturation region
Active & Saturation region
Explanation:
The region below IC = ICEO is called the cutoff region.
Question 6
Marks : +2 | -2
Pass Ratio : 100%
From the given curve tracer response, what is the value of β for IC = 7 mA and VCE = 5 V?
150
180
250
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.
Question 7
Marks : +2 | -2
Pass Ratio : 100%
For a BJT, what is typically the shape of the power dissipation curve, if it’s plotted on the output characteristics?
Parabola
Linear
Hyperbola
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.
Question 8
Marks : +2 | -2
Pass Ratio : 100%
Given that the collector power dissipation is 300 mW, what is the value of collector current for the collector to emitter voltage = 12 V?
50 mA
0 mA
25 mA
100 mA
Explanation:
P = VCE.IC = > 300mW = (12V)IC = > IC=300/12 mA = 25 mA.
Question 9
Marks : +2 | -2
Pass Ratio : 100%
From the given characteristics, the value of VCEsat is closest to which of the following values?
0.3 V
1 V
5 V
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.
Question 10
Marks : +2 | -2
Pass Ratio : 100%
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?
NPN
PNP
Faulty transistor
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.