# Complex conjugate poles for OTA unity gain

So I am designing an OTA, exactly like Miller OTA, but I tried to see the behaviour without Miller compensation. The circuit diagram is shown below

As seen there is no compensation network, neither is the output loaded. If connected in unty gain configuration i.e Vout shorted to Vin- then I get a frequency response with gain peaking as shown below

This suggest that there is complex conjugate pole. So I went ahead with the small signal analysis. The input signal is connected in Vin+ hence $$\frac{Vo}{Vin} = \frac{1}{1+\frac{1+sr1C1}{gm2\cdot gm6\cdot r1\cdot r2 - (sCin \cdot r2)(1+sr1C1)}}$$ where $$\ r1=ro2 \parallel ro4 \$$ ;$$\ r2=ro7 \parallel ro6 \$$ $$\ Cin \$$ is capacitance between gate and source of M1 and C1 is capacitance between gate and source of M6. This equation checks out because for DC, the gain is $$\ \frac{Vo}{Vin} = \frac{1}{1+\frac{1}{gm2\cdot gm6\cdot r1\cdot r2}} \$$ and assuming $$\ gm2\cdot gm6\cdot r1\cdot r2 >> 1\$$ the gain is roughly equal to one. Now to find the complex poles I would like to get the equation in the standard format which is

And this is where I am stuck. So my questions are as follows:

1. Is the gain equation correct?
2. Can anyone show me how to get this in standard form? I am guessing there will be a zero as well with a value r1C1.

EDIT: I end up with the following equation which does not make sense to me: $$\frac{Vo}{Vin} = \frac{gm2\cdot gm1\cdot r2\cdot r1 -s\cdot Cin\cdot r2 + s^2 \cdot r1\cdot r2\cdot C1\cdot Cin}{s^2 \cdot r1\cdot r2\cdot C1\cdot Cin +s\cdot (r1c1-r2cin) +gm2\cdot gm1\cdot r2\cdot r1 +1}$$

EDIT: Added small signal model for reference

• Simplify the denominator. Try to get just the denominator into the standard form. Let the numerator be what ever it turns out to be. To find poles, you need just the denominator.
– AJN
Commented Dec 6, 2020 at 2:54
• @AJN Please check the edit
– RAN
Commented Dec 7, 2020 at 19:49
• @RAN Have you tried plotting that equation? Octave or (wx)Maxima can do just fine. Commented Dec 7, 2020 at 20:21
• @RAN Why doesn't the equation make sense ? It has, in the denominator, an $s^2$ term, an $s$ term and a constant. Can't you work out the pole locations after substituting the compoenent values ?
– AJN
Commented Dec 8, 2020 at 13:58
• @AJN That is because the $\omega_n^2$ term in the numerator and denominator does not match
– RAN
Commented Dec 8, 2020 at 16:05

After doing all the calculations juiciously and going back to the basics of manipulating equations (also from the suggestion from Verbal Kint) I arrived at the following equation $$\dfrac{V_o}{V_{in}} = A_0 \cdot \dfrac{s^2 + \dfrac{s}{r_1c_1} + \dfrac{g_{m6} g_{m2}}{c_{in} c_1}}{s^2 + \dfrac{(r_1c_1 + c_{in} r_2)}{r_1r_2c_1c_{in}}s + \dfrac{g_{m6} g_{m2}}{c_{in} c_1}}$$ where $$\ A_0 = \dfrac{g_{m1}g_{m2}r_1r_2}{g_{m1}g_{m2}r_1r_2+1} \$$, $$\\omega_n= \sqrt{\dfrac{g_{m6}g_{m2}}{c_{in} c_1}}\$$, $$\ \zeta = \dfrac{ (r_1c_1 + c_{in} r_2)}{2 r_1r_2 \sqrt{c_1c_{in}g_{m6}g_{m2}}} \$$. Looks like $$\ \zeta \$$ will always be less than one and hence there will always be a complex conjugate root present.

a concerned citizen, AJN, Verbal Kint, please can you comment on this answer.

• The small signal circuit you show can never have a complex conjugate root, but the I/O capacitance will add a zero (damped by r1). Unless I am missing something, but anything's possible now. But your calculated transfer function should make much more sense now, even only at a glance. When using the ping operator, @, use <kbd>TAB</kbd> to cycle between nicknames (there's no space), or type the first letter followed by <kbd>TAB</kbd> to make it easier, otherwise the person will never get a notification. If you get them it's because this is your question. Also, you can select your answer. Commented Dec 11, 2020 at 22:36
• @aconcernedcitizen I hope it looks better now. The small signal circuit model, according to me, will always have a complex conjugate pole, because $\zeta$ will always be less than one. Please can you elaborate on your comment?
– RAN
Commented Dec 11, 2020 at 22:46
• I was referring to the last edit in the OP, the small signal schematic, but now (that it's morning) I see I missed the $V_{in_+}$ and $V_{in_-}$ notations. Your transfer function makes more sense now, complex conjugate roots and all. But you should try plotting beyond the 1 GHz limit, to see the zero, too. The response will be that of a 2nd order Cauer/elliptic filter, but with a smoother zero. Commented Dec 12, 2020 at 9:02
• Yes there will two zeros as well, I have not focused on them, but generally speaking, two cascaded amplifiers in a feedback configuration, like in this post, will always have a complex conjugate pole?
– RAN
Commented Dec 14, 2020 at 7:04
• Not always, it depends on the parameters. If you look closely, $\zeta$ has $r_1$ and $r_2$ as terms, whereas $\omega_n$ doesn't, which means that the damping will be a factor of those two resistances. Try increasing their values (even if unreasonable) to see the effects. Commented Dec 14, 2020 at 9:57