# Solving for Vout of CMOS Inverter with parasitic resistance

Discussion in 'Homework Help' started by Mary Kate, Apr 14, 2015.

1. ### Mary Kate Thread Starter New Member

Apr 14, 2015
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0
Due to a manufacturing error, a parasitic resistance (Rp=9kohms) has appeared in the CMOS inverter of Figure 1. Find the output voltage when Vin=0V, verifying the region of operation of the device.
Given: Vdd=1.8V; kn'=100uA/V^2; kp'= 50uA/V^2; lambda=0 for both devices
Mp: Vt=-0.5; W/L=5/0.18
Mn: Vt=0.4; W/L=3/0.18

Attempts: I know NMOS must be in cutoff and that Imp = Imn = 0. PMOS is conducting because Vgs<Vt. I believe that PMOS is in saturation because Vds (which should be equal to zero) > Vgs-Vt (which is -1.8-(-0.5)=-1.3). I know that if the resistor was not there, Vout=Vdd, but I am very confused about how the resistance will change Vout. When I simulated the circuit it gave Vout a value of 1.2V, but I do not know how to solve for that.
Any help would be very appreciated!

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2. ### WBahn Moderator

Mar 31, 2012
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Hint: Voltage divider.

3. ### Mary Kate Thread Starter New Member

Apr 14, 2015
2
0
Sorry I'm still confused. I only have one resistor so I'm not sure how to do voltage division.

4. ### WBahn Moderator

Mar 31, 2012
20,227
5,755
Don't your FETs have a channel resistance when they are turned on?

5. ### crutschow Expert

Mar 14, 2008
16,528
4,453
You said the device had a parasitic 9k resistance, so you have that resistor and the output load resistor.

6. ### WBahn Moderator

Mar 31, 2012
20,227
5,755
The output load resistor is unspecified. I thought it was the resistor shown in the schematic, too. But it turns out that that IS the parasitic resistance, Rp, that the problem was talking about.

7. ### crutschow Expert

Mar 14, 2008
16,528
4,453
You're right, I missed that.
So the output ON resistance has to be calculated from the given transistor parameters.

8. ### WBahn Moderator

Mar 31, 2012
20,227
5,755
That's my take, too.

@Mary Kate : Do you have what you need to go forth?