# Compensation of EA in current mode switch power

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#### huojinsi

##### Full Member level 5
Hi everyone:
In the current mode switch power, generally the compensation of EA is as the configuration of the picture. The purpose of EA compensation is to maximize low frquency gain and minimize high frequency gain. I have some doubts about the compensation of EA in current mode switch power:
1. Maximizing low frequency gain is obtained from the ratio of Rf/Ri, not from introducing a zero, is it ?
2. From some papers, I know the purpose of this pole Fp=1/(2πRfCf) is to cancel the zero of output filter capacitor ESR. How to minimize high frequency gain ?
Pls correct me, thks in advance!

#### Davood Amerion

your formula is in laplace domain. output/input will be:
H(s) = Vo(s)/Vi(s)=-(Rf || xCF) / Ri
H(s) = -(Rf*(1/CS))/Rf+(1/CS)) / Ri
H(s) = (-Rf/Ri) * (1/(1+sRfC)

in frequency domain this formula will be:
Vo/Vi = -Rf / (Ri *(RfCf * ω +1 )
in this formula when ω = 0 (DC gain)
Vo/Vi = -Rf/Ri
and for ω = (infinity) Vo/Vi = 0

I want to add accurate formula with presence of Rd and Vref is :
Vo = [ (-Zf/Ri) * Vi ] + [ ( (R1Zf + ZfRd + R1Rd) / R1Rd ) * Vref ]
where Zf is impedance of feedback path (Rf || Cf) = Rf/(SCRf+1) or Rf/(ωCRf+1)

saeidjabbari

### saeidjabbari

Points: 2

#### johnsmith101

##### Member level 3
I've seen a transconductance amplifier (Gm) used more often than a voltage amplifier for compensating current-mode converters. For an easy to follow guide on how to compensate using a Gm amplifier, you can use the following Maxim datasheet:

hxxp://pdfserv.maxim-ic.com/en/ds/MAX1566-MAX1567.pdf

See pg. 28 for a step-up converter and pg. 29 for a step-down converter.

I hope this helps.

-John

#### VVV

This type of error amplifier has never been my preference, even though some personalities advocate its use. I prefer the type 2 EA, that introduces a zero and helps boost the phase margin.

Anyway, to answer your question, look at the picture of the Bode plot and note that the resulting gain (in dB) is merely an addition of the top two. The gain above the crossover frequency is the "high-frequency" gain. As you can see, lowering it means bringing the Rf, Cf pole closer to the ESR zero. That will make the gain begin to drop faster after the crossover frequency and result in lower gain at higher frequencies.
Bringing the Rf, Cf pole close to the ESR zero effectively cancels that zero, but the actual goal of the pole is to make the high-frequency gain drop. If that pole did not exist, the gain would be flat from the ESR zero onwards. (Actually, another pole is sure to occur, at least from the opamp itself if not from other parasitics, but its frequency could be too high and uncontrollable.)

The low-frequency gain can only be improved by increasing the EA gain, that is increasing the Rf/Ri ratio, as can be seen from the picture. The flat portion of the EA characteristic is the EA gain and is only affected by Rf/Ri.

Note that I have not included any info on the phase relationships. That can be relatively easily added, just by following the definition of the phase shift at a certain frequency f, in the presence of a pole or zero at frequencies fp, fz respectively. If you do that, remember that the EA has a -180deg phase shift at low frequency, due to the inverting configuration.

### huojinsi

Points: 2

#### huojinsi

##### Full Member level 5
Thks VVV
From ur picture, i can clearly understand the effect of the pole" Fp=1/(2πRfCf)". I think this pole of EA cann't fastly decrease the gain of high frequency (above crossover frequency). No matter the pole is to cancel the ESR zero or not, the final gain curve always drops at "-1" slope.

Is my viewpoint right?

#### huojinsi

##### Full Member level 5
Hi VVV:
From ur uploaded picture, i found the ESR zero is small than the pole of EA. But i am not clear which of ESR zero and the pole of EA is large. If the ESR zero is large than the pole of EA, the resulting gain is drawn on the down picture, is it right?
Pls tell me! Thks in advance!

#### VVV

Yes, your picture is absolutely correct.
The EA pole will make the HF characterisitc drop off at a -1 slope, but if the pole is at a very high frequency, then you will not have much attenuation after crossover, until you approach the EA pole, because your slope is zero (due to the ESR zero) and not many dB below 0dB. So you would have a large portion of fairly high gain at high frequency, even though eventually the slope is -1.
By the way, the slope should be -1 only at crossover, not at high frequency. At HF it can be -2 for instance, which usually happens because there can be poles at frequencies higher than the EA pole.

### huojinsi

Points: 2
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