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If \$ V_{GS} >V_t\$, then the MOS transistor is in the triode zone, therefore isn't working as an differential amplifier.I understand this.Only when the transistor is in the saturation zone, MOS works as an amplifier,a differential pair amplifier.I want to understand what happens when \$ V_{GS} <V_t\$ ? I mean,what exactly makes and a MOS transistor work as an amplifier?

Also,bonus question,when we generally analyze amplifiers,why do we take the current source as ideal and with an infinite output resistance? What does it mean when a current source has an infinite output resistance?

nidhin
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user3543012
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  • Answer to your current source question: Input/output resistance of a device is defined as change of the voltage on it per change the current through, that is \$\frac{\Delta V}{\Delta I}\$. The voltage on a current source can freely change according with the load, but change of its current is always zero. Therefore, it has infinite internal resistance. – hkBattousai Jun 01 '14 at 11:18
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    Your question is too difficult to answer because you made a bunch of statements instead of just asking a question. These statements would need to be de-bunked first. – Olin Lathrop Jun 01 '14 at 11:28

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What happens when \$V_{GS}<V_t\$ ?

Ideally drain current is zero.

What exactly makes and a MOS transistor work as an amplifier?

Biasing into saturation region makes a MOS transistor as an amplifier.

Why do we take the current source as ideal and with an infinite output resistance?

See answers to this question: Why use only oversimplified/ideal equivalent circuits?

What does it mean when a current source has an infinite output resistance?

An ideal current source can develop any amount of voltage across its terminals so as to supply a constant current. or \$\Delta I = 0\$ $$R = \frac{\Delta V}{\Delta I} = \infty$$

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nidhin
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