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In our final exam we had a question about the following MOSFET push pull configuration:

schematic

simulate this circuit – Schematic created using CircuitLab

where $$V_{TN}=V_{TP} = 0$$ $$K_N=K_P=4 mA/V^2$$

The question was to plot \$V_{in}\$ vs \$V_{out}\$ for \$-10 < V_{in} < 10\$. The answer is given as enter image description here

But I think this is not correct. I wrote the following equation to find the relationship between input and output when \$10 > V_{in} > 0\$ $$ I_D = K_N(V_{GS} - V_{TN})^2=4 \times 10^{-3} (V_{in}-V_{out})^2 $$ and $$ V_{out} = I_D \times 50 = 0.2(V_{in} - V_{out})^2 $$ Similarly for \$0 > V_{in} > -10\$ $$ I_D = K_P(V_{SG} + V_{TP})^2=4 \times 10^{-3} (V_{out}-V_{in})^2 $$ and $$ V_{out} = - I_D \times 50 = -0.2(V_{in} - V_{out})^2 $$

Plot of these two equations are

enter image description here which is substantially different from the one given in the answer.

Am I doing something wrong?

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  • \$\begingroup\$ Upside down power supplies, always a good way to throw away your students in the wrong direction. \$\endgroup\$
    – winny
    Jan 22 '17 at 21:40
  • \$\begingroup\$ sorry, that is my mistake in drawing the schematic. Original question has V2=10V. \$\endgroup\$
    – srhat
    Jan 22 '17 at 21:50
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The teacher musta never touched a MOSFET with a 10-foot pole in their lives.

You guys forgot the currents have a sign...

Let I be the current flowing in resistor R1, from top to bottom:

I = Vout/R

If only the top NMOS is conducting, then:

I = K(Vin - Vout)^2

However the bottom PMOS is upside down, which introduces a nagging minus sign... If only the PMOS is conducting, then:

I = - K(Vin - Vout)^2

...and when Vin=0, then Vout=0 and both FETs are OFF.

Fortunately, since both FETs are identical save the polarity, the whole thing is symmetrical around zero, so we only need to study what happens in one polarity. Say, Vin>0.

Therefore, Vout = RK(Vin-Vout)^2 as we saw.

To solve this, simply solve for Vin! And stitching both polarities together, we get:

Vin = Vout + sign(Vout) * sqrt( abs(Vout)/RK )

...and this looks quite like the curve posted originally.

From this it is easy to get G = Vout/Vin.

Cross-check:

This gives Vin=17.071067811865476, Vout=10.0 (sorry I didn't limit it to +/- 10V)

Now, Vout=10V so I=0.2

K(Vin-Vout)^2 = 0.2 also

Bingo. Teacher wrong.

Also, anyone who has handled a push pull MOSFET stage in their life knows that the curve calculated by the OP looks correct. This is how they behave. If anyone can get a push pull to make a straight line like the teachers'.... run to the patent office at once! You gonna make billions!

Another demonstration in case you're incredulous

Let Gm (transconductance) be :

gm = dI / d(Vin-Vout)

For Vin>0, I = K(Vin - Vout)^2 then gm=2K(Vin-Vout)

For Vin=OV, both FETs are fully OFF, I=0, Vout=0, and Gm=0.

However, since gm is also the derivative of the gain curve divided by R............

...and the teacher's gain curve is a straight line with a constant derivative...

Then, nope. Still no workie.

Note:

These ain't perfect FETs because their Vt is zero! They're normal square-law FETs with a zero Vt. We can build NMOS with positive or negative Vt, it's easy, you can grab'em at gigikey for a nickel, check DN2540...

Now, if anyone can find me a PMOS with positive threshold voltage... I'd be interested.

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the clue is

VTN=VTP=0

Both have zero threshold voltages so will act as ideal N and P MOSFET source followers with no cross over region. The answer given is correct.

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  • \$\begingroup\$ But the problem is not about VTN and VTP. The problem is having a resistor connected to the source terminal. The more current we have on R, the less voltage difference we have between Gate and Source, which acts as a negative feedback and reduces the gain. Normally I would expect gain to be only slightly less than 1 but when I do the calculations I find gain to be much less than 1 as shown on the plot. \$\endgroup\$
    – srhat
    Jan 22 '17 at 21:18
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your calculation is wrong: think about what Vtn / Vtp is.

source followers have close 1x of voltage gain, putting aside their threshold voltage.

edit: to help you a little bit more.

your equation can be rewritten as

Vout / R = K(Vin - Vout)^2.

divide both side by Vout and denote G = Vout / Vin, you have 1 / R = K (1 / G - 1)^2.

take a square root on both side and you get

G = Vout / Vin = 1/ (1 + x) < 1 -> you can figure out what x is.

it shows you that the gain is slightly less than 1, and independent of Vin, just as we would have expected.

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  • \$\begingroup\$ Are you sure that gain is only slightly less than one? I solved the expression Vout / R = K(Vin - Vout)^2 on wolfram alpha for R = 50 and \$K = 0.4 * 10^{-3}\$ and \$V_{in} = 10\$. The answer is Vout = 5. This means gain is 0.5 for Vin = 10. \$\endgroup\$
    – srhat
    Jan 22 '17 at 21:11

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