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A common-source amplifier with a drain resistance, RD = 4.7 kΩ, is powered using a 10 V power supply. Assuming that the transconductance, gm, is 520 µA/V, the voltage gain of the amplifier is closest to:
  • a)
    1.22
  • b)
    -1.22
  • c)
    2.44
  • d)
    -2.44
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A common-source amplifier with a drain resistance, RD = 4.7 kΩ, ...
Given data: RD = 4.7 kΩ, gm = 520 μA/V
Voltage gain of CS amplifier
=–gmRD
= –520 μAV × 4.7 kΩ = –2.44
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Most Upvoted Answer
A common-source amplifier with a drain resistance, RD = 4.7 kΩ, ...
The common-source amplifier is a type of field-effect transistor (FET) amplifier that has the source terminal connected to the ground and the input and output signals are applied to the gate and drain terminals, respectively. The drain resistance, RD, is a component in the amplifier circuit that helps to set the gain and stability of the amplifier.

In this case, the drain resistance, RD, has a value of 4.7 kΩ (kilo-ohms). This means that the amplifier circuit includes a resistor with a resistance of 4.7 kilo-ohms connected to the drain terminal of the FET.

The function of the drain resistance, RD, in a common-source amplifier is to provide a load for the amplifier, which affects the gain and output impedance of the circuit. It helps to convert the small AC signal voltage at the drain terminal into a larger AC voltage swing at the output.

By adjusting the value of the drain resistance, RD, the gain of the amplifier can be controlled. Higher values of RD can result in higher gain, but it can also reduce the stability of the amplifier. Lower values of RD can decrease the gain but improve stability.

Overall, the drain resistance, RD, is an important component in a common-source amplifier that affects the gain and stability of the circuit. Its value can be adjusted to optimize the performance of the amplifier for a specific application.
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A common-source amplifier with a drain resistance, RD = 4.7 kΩ, is powered using a 10 V power supply. Assuming that the transconductance, gm, is 520 µA/V, the voltage gain of the amplifier is closest to:a)1.22b)-1.22c)2.44d)-2.44Correct answer is option 'D'. Can you explain this answer?
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A common-source amplifier with a drain resistance, RD = 4.7 kΩ, is powered using a 10 V power supply. Assuming that the transconductance, gm, is 520 µA/V, the voltage gain of the amplifier is closest to:a)1.22b)-1.22c)2.44d)-2.44Correct answer is option 'D'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A common-source amplifier with a drain resistance, RD = 4.7 kΩ, is powered using a 10 V power supply. Assuming that the transconductance, gm, is 520 µA/V, the voltage gain of the amplifier is closest to:a)1.22b)-1.22c)2.44d)-2.44Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A common-source amplifier with a drain resistance, RD = 4.7 kΩ, is powered using a 10 V power supply. Assuming that the transconductance, gm, is 520 µA/V, the voltage gain of the amplifier is closest to:a)1.22b)-1.22c)2.44d)-2.44Correct answer is option 'D'. Can you explain this answer?.
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