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Given, Vgs is the gate-source voltage, Vds is the drain source voltage, and Vth is the threshold voltage of an enhancement type NMOS transistor, the conditions for transistor to be biased in saturation are
- a)Vgs < Vth; Vds ≥ Vgs - Vth
- b)Vgs > Vth; Vds ≥ Vgs – Vth
- c)Vgs > Vth; Vds ≤ Vgs – Vth
- d)Vgs < Vth; Vds ≤ Vgs – Vth
Correct answer is option 'B'. Can you explain this answer?
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Given, Vgs is the gate-source voltage, Vds is the drain source voltage...
At Vgs = 0, no current flows through the MOS transistors channel because the field effect around the gate is insufficient to create or open the n-type channel. Then the transistor is in its cut-off region acting as an open switch.
As we now gradually increase the positive gate-source voltage Vgs, the field effect begins to enhance the channel regions conductivity and there becomes a point where the channel starts to conduct. This point is known as the threshold voltage Vth. As we increase Vgs more positive, the conductive channel becomes wider (less resistance) with the amount of drain current (Id).
Therefore, the n-channel enhancement MOSFET will be in its cut-off mode when the gate-source voltage (VGS) is less than its threshold voltage level (Vth) and its channel conducts or saturates when Vgs is above this threshold level.
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Given, Vgs is the gate-source voltage, Vds is the drain source voltage, and Vth is the threshold voltage of an enhancement type NMOS transistor, the conditions for transistor to be biased in saturation area)Vgs < Vth; Vds ≥ Vgs - Vthb)Vgs > Vth; Vds ≥ Vgs – Vthc)Vgs > Vth; Vds ≤ Vgs – Vthd)Vgs < Vth; Vds ≤ Vgs – VthCorrect answer is option 'B'. Can you explain this answer?
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