Electronic Devices and Circuits Questions and Answers – Biasing the FET

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This set of Electronic Devices and Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Biasing the FET”.

1. Which of the following relation is true about gate current?
a) IG=ID+IS
b) ID=IG
c) IS= IG
d) IG=0
View Answer

Answer: d
Explanation: The FET physical structure which contains silicon dioxide provides infinite resistance. Hence no current will flow through the gate terminal.
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2. Which of the following equations gives the relation between ID and Vgs?
a) ID=IDSS (1-Vgs/Vp)2
b) ID=IDSS (1-Vgs/Vp)1
c) ID=IDSS (1-Vgs/Vp)3
d) ID=IDSS (1-Vgs/Vp)4
View Answer

Answer: a
Explanation: The above equation called as Shockley’s equation depicts the relation between ID and Vgs. When Vgs becomes equal toVp, the current will become zero, which clearly satisfies the physical nature of FET.

3. For a fixed bias circuit the drain current was 1mA, what is the value of source current?
a) 0mA
b) 1mA
c) 2mA
d) 3mA
View Answer

Answer: c
Explanation: We know that for an FET same current flows through the gate and source terminal, Hence source current=1mA.

4. For a fixed bias circuit the drain current was 1mA, VDD=12V, determine drain resistance required if VDS=10V?
a) 1KΩ
b) 1.5KΩ
c) 2KΩ
d) 4KΩ
View Answer

Answer: c
Explanation: VDS=VDD-ID RD
=>10=12-RD×1mA
=>RD=2/1mA=2 KΩ.

5. Which of the following equation brings the relation between gate to source voltage and drain current in Self Bias?
a) Vgs=VDD
b) Vgs=-ID Rs
c) Vgs=0
d) Vgs=1+ID Rs
View Answer

Answer: b
Explanation: VRs=ID Rs
ButVRs+Vgs=0
Vgs=-ID Rs.
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6. For a self-bias circuit, find drain to source voltage if VDD=12V, ID=1mA, Rs=RD=1KΩ?
a) 1V
b) 2V
c) 10V
d) 5V
View Answer

Answer: c
Explanation: VDS=VDD-ID (RD+Rs)
=>VDS=12-1mA(1KΩ+1KΩ)
=>VDS=10V.

7. Find the gate voltage for voltage divider having R1=R2=1KΩ and VDD=5V?
a) 1V
b) 5V
c) 3V
d) 2.5V
View Answer

Answer: d
Explanation: VG = R2×VDD/R1+R2
=>VG=1×5/2
=> VG= 2.5V.

8. Find the gate to source voltage for voltage divider having R1=R2=2KΩ and VDD=12V, ID=1mA and RS=4KΩ?
a) 3V
b) 2V
c) 0V
d) 1V
View Answer

Answer: b
Explanation: VG = R2×VDD/R1+R2
=>VG=2×12/4
=>VG=6V
=>VGS=VG-ID Rs
=>VGS=2V.

9. What will happen if values of Rs increase?
a) Vgs Increases
b) Vgs Decreases
c) Vgs Remains the same
d) Vgs=0
View Answer

Answer: b
Explanation: Increasing values of Rs result in lower quiescent values of ID and more negative values of Vgs.
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10. What is the current flowing through the R1 resistor for voltage divider (R1=R2=1KΩ, VDD=10V)?
a) 5mA
b) 3mA
c) 1mA
d) 2mA
View Answer

Answer: a
Explanation: IR1=IR2 =VDD/R1+R2
=>IR1 = 10/2KΩ
=>IR1 = 5mA.

Sanfoundry Global Education & Learning Series – Electronic Devices and Circuits.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn