Continue to Site

# question on frequency compensation for 2-stage OTA

Status
Not open for further replies.

#### airboss

##### Member level 3
stage ota

Hi,

A question regarding the frequency compensation for a 2-stage OTA.
In Razavi's textbook, he assumed the dominant pole & the 1st non-dominant pole before compensation are the O/P pole of the 1st stage & the 2nd stage, respectively. He explained that to acquire each pole's location we actually need to do simulation. Also, he compared the pole frequencies by deriving the poles before and after compensation.

When I went to an interview a while ago, an interviewer told me the dominant pole is supposed to be at the O/P of the 2nd stage even w/o simulation. Another interviewer asked me to explain how Miller compensation works w/o going through the math; so basically, he asked me the intuition behind.

1) so, where's the dominant pole? where's the 1st non-dominant pole?
2) what're the pole frequencies after Miller compensation? how do i approach this problem w/o doing math?

airboss

dominant pole of ota

As I see, before the Miller compensation, the dominant pole is the pole of the second stage,because the load capacitor is usually large.After the compensation, the dominant pole is the first pole.

Hi airboss

Actually I have heard it before. Take a look at this paper, and consider the sentences below:

H. Yang and D. Allstot, "Considerations for Fast Settling Operational Amplifiers", TCAS 1990.

" Our results show that the previous techniques based on Miller-multiplied capacitance models are suitable only for those amplifiers in which the first-stage pole is dominant prior to compensation. In most SC circuits, the internal opamps are transconductance amplifiers whch drive on-chp capacitive loads, and thus for these opamps, the second-stage pole is dominant."

"As previously stated, on-chip CMOS and GaAs transconductance amplifiers are usually second-stage dominant-pole systems with R,'C,'< R2'C2' in the model of Fig. 6(a). Conventional application of the Miller-multiplied capacitance effect in developing the equivalent circuit of Fig. 6(b) predicts that the first-stage pole moves to a lower frequency for a nonzero value of cgd and will become dominant if C and/or g,, is large enough as indicated in Fig. 8(a). This commonly-held view (c.f. [13, p. 3771 is incorrect in terms of basic root locus theory wherein the poles of a two-pole system can never cross one another as pole-splitting negative feedback is applied [l]."

### airboss

Points: 2
compensate 2 stage ota

w/o=without

Status
Not open for further replies.