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CH9 FET Amplifiers And Switching Circuits Solved Examination 2024/2025 $15.99
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CH9 FET Amplifiers And Switching Circuits Solved Examination 2024/2025
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CH9 FET Amplifiers And Switching Circuits Solved Examination 2024/2025

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  • August 28, 2024
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  • 2024/2025
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CH9 FET Amplifiers And Switching
Circuits Solved Examination 2024/2025

The formula for the voltage gain of a common-source amplifier is RD/gm. Answer: F



Load resistance added to the output of an amplifier increases the voltage gain. Answer: F



The addition of a source bypass capacitor will increase the voltage gain. Answer: T



FETs are superior to BJTs in almost all switching applications. Answer: T



FET amplifiers generally have lower distortion than BJT amplifiers. Answer: F



Cascode amplifiers are used primarily in radio frequency applications. Answer: T



Most of the gain in a JFET-based cascode amplifier is provided by the CS amplifier. Answer: F



Class D amplifiers usually employ JFETs. Answer: F



The 3 stages in a class D amplifier are the pulse-width-modulator, the switching amplifier, and the low-
pass filter. Answer: T



A pulse-width-modulator employs a square-wave generator and a comparator. Answer: F



A MOSFET switch is turned on and off by changing the source voltage. Answer: F

, An analog MOSFET switch is turned on when the gate receives a positive voltage pulse. Answer: T



CMOS combines n-channel and p-channel D-MOSFETS. Answer: F



Power MOSFETs have a negative temperature coefficient and therefore are less prone to thermal
runaway. Answer: F



In an amplifier using a JFET, the gate current is approximately 0. Answer: T



A common-source amplifier has a ________ phase shift between the input and the output. Answer:
180°



Refer to the figure above. Assuming midpoint biasing, if VGS = -4 V, the value of RS that will provide this
value is Answer: 800 ohms



Refer to the figure above. If Vin = 50 mVp-p, the output voltage is Answer: 440 mVp-p.



Refer to the figure above. If the measured value of Vout were below normal, the problem

might be that Answer: C2 is open.



Refer to the figure above. If Vin = 1 Vp-p, the output voltage Vout would be Answer: undistorted.



Refer to the figure above. If ID = 6 mA, the value of VGS is Answer: -9 V.



Refer to the figure above. If gm = 6500 S and an input signal of 125 mVp-p is applied to the gate, the
output voltage Vout is Answer: 2.68 Vp-p.

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