2

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 enter image description here

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 enter image description here

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

enter image description here

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 enter image description here

RAN
  • 179
  • 1
  • 15
  • 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 Dec 06 '20 at 02:54
  • @AJN Please check the edit – RAN Dec 07 '20 at 19:49
  • 1
    @RAN Have you tried plotting that equation? Octave or (wx)Maxima can do just fine. – a concerned citizen Dec 07 '20 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 Dec 08 '20 at 13:58
  • @AJN That is because the \$ \omega_n^2 \$ term in the numerator and denominator does not match – RAN Dec 08 '20 at 16:05
  • It doesn't have to. "*Now to find the complex poles I would like to get the equation in the standard format which is*". Poles can be calculated by solving the quadratic expression in the denominator alone. No need for processing the numerator AFAIK. – AJN Dec 08 '20 at 16:11
  • But the ideal second order transfer function clearly states that \$ \omega_n^2 \$ should be in the numerator and the denominator. So \$ A_o/s^2 + 2 \zeta w_n s +\omega_n^2 \$ is not a standard form, am i correct? Besides in the equation which is in my edit posted, i cant find \$ \omega_n \$ anyway, what according to you is the \$ \omega_n \$ – RAN Dec 08 '20 at 16:16
  • @a concerned citizen, I tried plotting the equation and the gain is always 1 across all frequency points – RAN Dec 08 '20 at 17:14
  • @RAN Odd. For some values of the elements I see a 2nd order pole/zero transfer function i.e. complex conjugate poles and zeroes. For example with these random values: `r1=1k, r2=2k, gm1=3m, gm2=4m, c1=5p, c2=6p`. You may have to increase the maximum frequency to see the effect. I began writing an aswer, but what I get from a simple `gm` analysis is two 1st order filters in series. Tha can never have a complex conjugate root. What are your settings/schematic? I'm curious to see in LTspice. – a concerned citizen Dec 08 '20 at 20:35
  • If you want to rearrange your formula in the form of \$H(s)=H_0\frac{N(s)}{D(s)}\$ where \$N\$ and \$D\$ follow the form \$1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2\$ with different values for \$\omega\$ and \$Q\$ in \$N\$ and \$D\$ of course, then factor \$g_{m2}g_{m1}r_2r_1\$ in the numerator and \$g_{m2}g_{m1}r_2r_1+1\$ in the denominator. You then have \$H_0=\frac{g_{m2}g_{m1}r_2r_1}{g_{m2}g_{m1}r_2r_1+1}\$ as your dc gain and the rest are normalized polynomial forms with a RHPZ pair in the numerator if your whole formula makes sense of course. – Verbal Kint Dec 08 '20 at 21:21
  • @ a concerned citizen You are right that they are two low pass filters in series, if the feedback is disconnected, and the output is loaded with load capacitor, But in feedback, the input capacitance plays a role, which is the reason for complex conjugate poles. In my cases the values are: \$ r1 = 38.5k, r2=74.7k, gm1=1.5m, gm2=3.2m, c1=0.115p, cin=0.428p \$. Also I am using Cadence and not LT spice – RAN Dec 09 '20 at 08:42
  • @ a concerned citizen, I have added the small signal model, based on which I have derived the euqations – RAN Dec 09 '20 at 08:57

1 Answers1

1

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.

a concerned citizen
  • 21,167
  • 1
  • 20
  • 40
RAN
  • 179
  • 1
  • 15
  • 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 TAB to cycle between nicknames (there's no space), or type the first letter followed by TAB 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. – a concerned citizen Dec 11 '20 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 Dec 11 '20 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](https://en.wikipedia.org/wiki/Elliptic_filter), but with a smoother zero. – a concerned citizen Dec 12 '20 at 09:02
  • 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 Dec 14 '20 at 07:04
  • 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. – a concerned citizen Dec 14 '20 at 09:57