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schematic

simulate this circuit – Schematic created using CircuitLab

For Vin below threshold voltage, M1 isn't conducting, neither does M2 since Id=0. So what happens circuit wise, how come that Vdd=Vout ( we did this in class)?

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  • \$\begingroup\$ Think about it in very simple terms. PMOS conducts, while NMOS doesn't. So it's same as Vout is connected to VDD and disconnected from GND. \$\endgroup\$ – Gregory Kornblum Mar 21 '16 at 20:24
  • \$\begingroup\$ More importantly you have to understand, what is transistor. \$\endgroup\$ – Gregory Kornblum Mar 21 '16 at 20:24
  • \$\begingroup\$ You do NOT know the current through M2, unless you know what is connected to Vout. \$\endgroup\$ – Brian Drummond Mar 21 '16 at 22:56
  • \$\begingroup\$ If M2 is "on", consider what would happen if Vout was not equal to Vdd. \$\endgroup\$ – The Photon Mar 21 '16 at 23:45
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Note that M1 is N channel while M2 is P channel, although they both invert from gate to drain in this configuration.

With Vin at 0, M1 is off. With the gate of M2 at 0, M2 is presumably on. I say "presumably" since you have provided no specs on Vdd and the required gate voltage for M2 to be on. I'm assuming these have been arranged such that a P channel FET with source tied to Vdd is on when its gate is at 0 V.

From above, replace M1 with a open switch and M2 with a closed switch. Vout is now obvious. Draw it out as open/closed switches if that helps you see it.

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  • \$\begingroup\$ You presumed correctly. But I am having hard time figuring out why is M2 closed and M1 open, since both of its currents are 0. \$\endgroup\$ – rkcepelin Mar 21 '16 at 20:58
  • \$\begingroup\$ It's because this one is M2 is on, right? \$\endgroup\$ – rkcepelin Mar 21 '16 at 21:00
  • \$\begingroup\$ @rkc: FETs are voltage controlled not current controlled. Current doesn't matter toward open/closed switch behavior. It's the gate-source voltage that matters. \$\endgroup\$ – Olin Lathrop Mar 21 '16 at 21:03
  • \$\begingroup\$ @rkcepelin - But pretend, just for a moment, that there was a load on the output. Now consider that M1 has leakage current (which it does). What does this do to the output? \$\endgroup\$ – WhatRoughBeast Mar 21 '16 at 21:04
  • \$\begingroup\$ @rkcepelin, Even if you think about a mechanical switch, it can be closed and still have no current flowing through it if the external circuit doesn't provide any current. \$\endgroup\$ – The Photon Mar 21 '16 at 23:44

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