If there is a CMOS inverter such that the gate of the PMOS transistor is always attached to the ground and the input voltage is only applied to the gate of NMOS, then how would the inverter behave, as in: Will it be similar to a NMOS inverter with a resistor connected between its source and Vdd supply?

I need to calculate the Vout for Vin =0 and Vin=2.5 volts and the switching voltage ( where Vin = Vout); thus I would like to find the equations characterizing the behaviour of the inverter.

So, for any Vin, the PMOS is always saturated and it can be replaced by an resistor of resistance same as ON resistance of the PMOS and the current flowing throught it would always be the saturated current. Please tell me if my approach is correct or if not, how should I tackle this problem?

To clarify, the devices are all short channel and channel length modulation is ignored.

Here is a schematic:

Inverter Schematic

  • \$\begingroup\$ If you provide a link to the schematic we can add it in for you. \$\endgroup\$
    – Oli Glaser
    Commented Sep 9, 2012 at 8:29
  • \$\begingroup\$ I put the schematic on google plus .... heres the link : plus.google.com/u/0/photos/104893953340544861875/albums/… thanks a lot for your help \$\endgroup\$
    – anshu
    Commented Sep 9, 2012 at 8:33
  • \$\begingroup\$ any help guys .... \$\endgroup\$
    – anshu
    Commented Sep 9, 2012 at 11:33
  • \$\begingroup\$ Back in the days before CMOS logic became common, NMOS chips were designed in which the load for each logic gate was a depletion-mode PFET acting as a current source. But in that case, the PFET's gate was connected to its source. I'm having trouble imagining why you'd want to use an enhancement-mode PFET with its gate grounded, which would needlessly complicate the physical layout. \$\endgroup\$
    – Dave Tweed
    Commented Sep 9, 2012 at 14:15
  • \$\begingroup\$ Minor nit: I don't think NMOS chips used PFETs. I think you mean depletion-mode NFETs. I would expect that in CMOS unless one uses extra processing steps to control depletion-mode FET behavior a long and skinny PFET would make a better passive pull-up than would an NFET. \$\endgroup\$
    – supercat
    Commented Nov 25, 2013 at 20:08

2 Answers 2


M2 is essentially acting like a pullup resistor in this case. Real resistors are difficult to make on silicon chips, so a PFET in on-state is good enough for this purpose.

The chip designer can vary parameters like the channel length, width, and possibly doping level. Depending on the characteristics of the transistor, it could act more like a current source than a resistor at the operating point. Sometimes a "long tail FET" is used to make a rough current source. Without knowing the parameters of M2, we don't know if it is more like a resistor or more like a current source, although in this application that wouldn't make much of a difference. Ideally you'd want a current source for a pullup, but lots and lots of places you see resistors doing that job well enough.

  • \$\begingroup\$ thanka a lot for exaplaining this .... i also believe that the PMOS should act as a pull-up resistor...... \$\endgroup\$
    – anshu
    Commented Sep 9, 2012 at 14:42

If the circuit that you show is for a conventional CMOS type inverter you will probably not want to analyze the lower FET transistor in the ON state when the upper FET is also in the forced ON state. The typically available commercial CMOS type parts can often source as much current via the upper FET as the lower FET can sink in the high and low states of the output respectively. Forcing both FETs on at once in the analysis will make the output go to some level toward the median between the VDD and GND as opposed to pulling the output toward GND. On some logic parts this type of operation could end up damaging the device. Note that this very thing can happen on a real circuit board if two logic signal drives are shorted together with one driver trying to pull the shorted net high whilst the other is trying to pull it low.

You can note the characteristics of a buffer part such as a TI 74AHC125 part from its data sheet at: http://www.ti.com/lit/ds/symlink/sn74ahc125.pdf Note in the data sheet that the part has symmetric source/sink capability of +/-8mA in normal operation at VDD=5V. Also note that the abs max current ratings for the output are also specified symmetrically at +/-25mA. This indicates that the conduction characteristics of the output upper and lower FETs are nearly the same.

When considering that the upper P-FET could be acting as a pullup or current source it is necessary that the conduction characteristics of the FET be adjusted so that the lower N-FET will be able to easily sink all of the current from the P-FET as well as any external load so that the output can pull down to a valid low voltage level. Without that the output would, as indicated before, go to some level in the no-mans land between a valid low level and valid high level.


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