# How would the pole and zero frequency expression look like for this transfer function I derived?

This is a practice problem in deriving the transition frequency function for mos and part of the set of questions is to determine the pole and zero frequency expression but this transfer function i derived is kinda different from what i'm used to from the previous lectures. I usually see this form: But my derived transfer function looks a little different. Could somebody guide me how to derive them?

• By reading your i_0/i_i expression I though that gap between C_g d and C_g s meant they were different variables/parameters and read it as $G(s) = \frac{g_m -s C_g d}{s(C_g s+C_g d)}$
– jDAQ
Dec 12, 2020 at 6:26
• Factor out the coefficients of $s$. For example, $C_g$ in the denominator.
– AJN
Dec 12, 2020 at 7:05
• If you're referring to the s term in the denominator, let $P_1=0$
– Chu
Dec 12, 2020 at 7:59
• @jDAQ sorry about that, my handwriting does suck
– user266967
Dec 12, 2020 at 8:52

If you look closely in the generic transfer function, the $$\s\$$ operator is single, everywhere, and there is a gain parameter. So try to isolate those from your t.f.:

\begin{align} \dfrac{g_m-sC_{gd}}{s(C_{gs}+C_{gd})}=\dfrac{-C_{gd}\left(s-\dfrac{g_m}{C_{gd}}\right)}{(C_{gd}+C_{gs})\left(s-\dfrac{0}{C_{gd}+C_{gs}}\right)}=-\dfrac{C_{gd}}{C_{gd}+C_{gs}}\dfrac{s-\dfrac{g_m}{C_{gd}}}{s-0} \end{align}

Which is a pure integrator -- that makes sense, since you have no resistive elements, only capacitors and a current source. Notice the denominator now shows the zero valued pole (as mentioned by Chu).

• whats a zero valued pole and how does it relate to an integrator? Is that the same as a pole at the origin?
– user266967
Dec 12, 2020 at 9:54
• Yes. Pole at origin.
– AJN
Dec 12, 2020 at 11:25
• Is it right that the zero frequency = -Cgd/gm and the pole frequency = Cgd + Cgs? Because we had a graded exercise based on this and when i used zero freq to be -Cgd/gm, i got it wrong, also pole freq = 1/Cgd+Cgs is also wring which i dont understand.
– user266967
Dec 12, 2020 at 13:03
• @Rein Try to compare the last part of the formula above with your generic transfer function: $\small{A}\frac{s-z}{s-p}$. What similarities do you see? Which one can be $z$, which one $p$, and which one $A$? Dec 13, 2020 at 9:17
• Thanks Chu! I actually figured things out and got perfect scores!
– user266967
Dec 14, 2020 at 5:17