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When the input voltage is high, Q1 is off and Q2 is on. In this case, the shorted Q2 pulls the output voltage down to ground. On the other hand, when the input voltage is low, Q1 is on and Q2 is off. Now, the shorted Q1 pulls the output voltage up to VDD.

BUT I read that in CMOS, the high is +VDD and the low is 0. How does Q1 turns on when input voltage is zero(won't it only turn on when Vin is negative) ? enter image description here

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Q1 is a p-channel MOSFET (aka "PMOS").

It turns "on" when the potential difference between its gate and source (\$V_{gs}\$) is negative.

But where is its source connected? It's not connected to ground (in fact, none of its terminals are connected to ground). It's connected to Vdd.

So what turns Q1 on is having the input voltage (its gate voltage) brought substantially below Vdd.

Since it has no connection to ground it doesn't "care" what the gate voltage is in relation to ground (i.e. whether it's positive or negative).

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Q1 is a PMOS, so the "turn-on" condition is Vgs (gate-source voltage) < -Vt.

Vgs here is defined as vin - Vdd. So when Vin = 0, the gate-source voltage is sufficiently negative for turn-on.

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MOSFETs turn on and off based on the voltage DIFFERENCE between the gate and source terminal. That's all they care about. They don't care about GND or anything else.

When you say:

(won't it only turn on when Vin is negative) ?

You are saying when Vin is negative relative to GND. Just saying "the voltage at the gate" implies you referencing to GND.

But as we just said, the MOSFET doesn't care about GND. In fact, it cannot know what GND is since it has no pins connected to GND.

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