This set of Electronic Devices and Circuits Multiple Choice Questions & Answers (MCQs) focuses on “The Operating Point”.
1. Where should be the bias point set in order to make transistor work as an amplifier?
a) Cut off
d) Cut off and Saturation
Explanation: To operate transistor as an amplifier, it requires more current amplification factor and in cut off and saturation, the current amplification is less, therefore active region is better to fix the Q point.
2. Q point can be set to work on active region requires particular conditions. What are they?
a) BE reverse biased and BC forward biased
b) BE reverse biased and BC reverse biased
c) BE forward biased and BC reverse biased
d) BE forward biased and BC forward biased
Explanation: BJT requires the forward voltage nearly equal to 0.7v and the p-junction should be more positive in BE junction and n region should be more positive in BC junction. This will make the current to flow through emitter which is the sum of current through base and emitter.
3. The Q-point of a transistor is made to shift between Active and cut off Region, then how does the transistor behave?
Explanation: When the Q point lies in cut off, No current flows and hence it acts as a closed switch. When the Q point is shifted to saturation, Current flows through the circuit creating a closed switch. Thus the current flow makes the turn on and off of switch.
4. For a Fixed bias circuit having RC=2.2KΩ, RB=240Ω, VCC=12v and current amplification factor is 100 and the current flowing through the base is 20µA, the value if Collector current in saturation is_____________
Explanation: Vce = VCC – IC RC
For saturation, Vce = 0 ICsat = VCC/RC = 5.4mA.
5. The bias point of a transistor occurs when the supply voltage exceeds the breakdown voltage of a transistor.
Explanation: Bias point can be set on the basis of DC load line that is Q point can be found out only without applying any input. The DC load line is defined as the line drawn in response of collector current and base emitter voltage when no input is applied.
6. For a fixed bias circuit having RC=4.7KΩ and RB=1KΩ, VCC=10V, and base current at Bias point was found to be 0.2µA, Find β?
Explanation: ICQ = Vcc/Rc = 2.12mA
7. For a Voltage divider bias circuit, having R1=R2=10KΩ, RC=4.7 KΩ, RE=1 KΩ, What is the value of collector current at saturation if VCC=10V?
Explanation: By using Thevenin’s law Vth=VccR2/(R1+R2)=10V
8. For a Voltage divider circuit having RC=R1=R2=RE=1KΩ, if VCC=20V, find IC when Vce = VCC?
Explanation: when Vce = VCC,
VCC = Vce-IC (RC+RE) => VCC-Vce = 0 = IC.
9. Changes in the temperature will not affect the bias point.
Explanation: The temperature changes the β value of the Transistor which wills in turn shifts the Q-point of the Transistor. Once the temperature changes, it will increase the mobility of electrons resulting in a change of system current, hence temperature does affect the transistor parameter.
10. For the voltage divider circuit, if a diode is connected in reverse direction across the base Find the value of β? (Vd=1.2v and input is 1V)
d) cannot be determined
Explanation: Since the input voltage is less than the diode voltage, transistor never turns on during the positive half cycle, hence it is difficult to measure the β value. Also during a negative half cycle, the diode will be turned off and hence no current flows, so the value if beta cannot be determined, it can be 0 because the manufacturer will certainly provide the value.
Sanfoundry Global Education & Learning Series – Electronic Devices and Circuits.
To practice all areas of Electronic Devices and Circuits, here is complete set of 1000+ Multiple Choice Questions and Answers.