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A depletion type NMOS is operated in enhancement mode. Vp = - 4 volts. For VGS = + 3 volts as VDS is increased, ID becomes nearly constant when Vps equals
- a)1 volt
- b)3 volts
- c)4 volt
- d)7 volts
Correct answer is option 'D'. Can you explain this answer?
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A depletion type NMOS is operated in enhancement mode. Vp = - 4 volts....
Depletion type NMOS in Enhancement Mode
Depletion type NMOS is a type of NMOS transistor that is designed to operate in depletion mode. However, when it is operated in enhancement mode, the threshold voltage (Vp) is negative.
Constant ID for Vps = 7 volts
When the gate-to-source voltage (VGS) is 3 volts, the transistor is in the saturation region. As VDS is increased, the drain current (ID) increases until it reaches a maximum value. However, when Vps equals 7 volts, ID becomes nearly constant.
Explanation
This is because, in enhancement mode, the gate voltage must be higher than the threshold voltage to turn on the transistor. When VGS is 3 volts, it is higher than Vp, and the transistor is turned on. As VDS is increased, the length of the channel decreases, and the voltage drop across the channel increases. This causes the drain current to increase until it reaches a maximum value.
However, when Vps equals 7 volts, the voltage drop across the channel is so high that the channel is completely pinched off, and the drain current becomes nearly constant. This is because the voltage drop across the channel is equal to VGS - Vp, which is only 4 volts. When VDS is increased beyond this point, the voltage drop across the channel remains constant, and the drain current does not increase further.
Conclusion
Therefore, the correct answer is option 'D' - 7 volts. At this voltage, the depletion region extends from the drain to the source, and the channel is completely pinched off. As a result, the drain current becomes nearly constant.
Depletion type NMOS is a type of NMOS transistor that is designed to operate in depletion mode. However, when it is operated in enhancement mode, the threshold voltage (Vp) is negative.
Constant ID for Vps = 7 volts
When the gate-to-source voltage (VGS) is 3 volts, the transistor is in the saturation region. As VDS is increased, the drain current (ID) increases until it reaches a maximum value. However, when Vps equals 7 volts, ID becomes nearly constant.
Explanation
This is because, in enhancement mode, the gate voltage must be higher than the threshold voltage to turn on the transistor. When VGS is 3 volts, it is higher than Vp, and the transistor is turned on. As VDS is increased, the length of the channel decreases, and the voltage drop across the channel increases. This causes the drain current to increase until it reaches a maximum value.
However, when Vps equals 7 volts, the voltage drop across the channel is so high that the channel is completely pinched off, and the drain current becomes nearly constant. This is because the voltage drop across the channel is equal to VGS - Vp, which is only 4 volts. When VDS is increased beyond this point, the voltage drop across the channel remains constant, and the drain current does not increase further.
Conclusion
Therefore, the correct answer is option 'D' - 7 volts. At this voltage, the depletion region extends from the drain to the source, and the channel is completely pinched off. As a result, the drain current becomes nearly constant.
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A depletion type NMOS is operated in enhancement mode. Vp = - 4 volts. For VGS = + 3 volts as VDS is increased, ID becomes nearly constant when Vps equalsa)1 voltb)3 voltsc)4 voltd)7 voltsCorrect answer is option 'D'. Can you explain this answer?
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