...OK thanks, but the 100n was supposed to be in pllel with the 9k, not in seriesFor
C1 = 10n, R2=1k, R1=47K, C3=100n, R3=330, and 100n in series with the 9k divider resistor, unexpected results.
Thank you Klaus, indeed increasing the C3 to 100nF made the plot a little bit better, around 10ms, the 2nd order response is dominant, then the response turns into the first order type as seen in the figure. I tried to connect the C3 directly to the output but it didn't make much of a difference. You are right about the source impedance of 900 ohms, the problem is that R1 is parallel to R3+(1/sC3), so there are no series connected elements otherwise I would have used the source impedance as a part of the transfer function. Here is the resulting plot when I increased C1 to 100nF.Hi,
the feedback voltage divider results in a 900 Ohms source impedance.
I guess R3 with 70 Ohms is rather useless. And C3 can not act as a proper "D" part.
Basically for low frequencies the Error amplifier input impedance is 2k Ohms and for high frequencies about 900 Ohms.
Crossover at about 7200 Hz.
What happens if you just increase C1 to 100nF? (without detailed analysis)
But I think it´s better to connect the right side of C3 to the output.
***
Indeed the ringing is quite expectable because of the integrating part of your error amplifier.
Klaus
Thanks a lot for your help, I forgot to write, the switching frequency is 100kHz, here are the outcomes for the values used. It seems that there are optimal values that are present but I am unable to tune them.Klaus's tip's sound good.
Also, please tell what is Fsw?
Also, try the following to see what it does.....
C1 = 10n
R2=1k
R1=47K
C3=47n
R3=1K
....if still not right, then make C3=100n & R3 = 330R
..if still not right, then add 100n across the 9k upper divider resistor
..if still not right, then make C1=4n7 and R2 = 4k7
I see you are in voltage mode, so you use type 3.
Ill find the voltage mode power stage txfer function in the meantime.
Also, what is your modulator gain?
--- Updated ---
for v mode the modulator gain is some function of sawtooth delta, and control voltage max...ill look it up and get back...then we can make out the bodes.
Hello Kalus, at steady state, the error amplifier output and C1 output voltages are around 2.25 and 1.75 volts respectively. Here is the plot.Hi,
I´d say it doesn´t look bad. 90% within 3ms. No real overshot.
C1 is the integrating capacitor
C3 is the differential capacitor. It has limted function when connelcted like in post#2.
The ringing is typical for an integrating error amplifier (as long as not compensated for this).
Imagine: As long as Vout is not at it´s setpoint the integrator voltage increases. (ignoring the D part) And Vout eventually is at it´s setpoint, then the I part is at it´s maximum --> driving the output even higher. Overshot. And it "needs" the overshot to get the I part do decrease.
I´m interested in seeing
* the error amplifier output plot
* and the C1 voltage plot.
****
You may try to "limit" C1 voltage ... to limit the overshot
Currently your regulator should be in stable state:
* when both error amplifier inputs are 0.5V
* when duty cycle is about 45%
* when error amplifier output is at 2.25V
...and ideally then the C1 voltage is about 1.75V. (Please check this with your simulation)
You could improve the circuit in a way that C1 voltage becomes zero in ideal case.
This could be done by changing the triangle range 0V..1.11V.
Then you are able to limit the C1 voltage with antiparallel diodes.
Klaus
then my mind calculation wasn´t that bad..the error amplifier output and C1 output voltages are around 2.25 and 1.75 volts respectively.
You could improve the circuit in a way that C1 voltage becomes zero in ideal case.Hi,
I´d say it doesn´t look bad. 90% within 3ms. No real overshot.
C1 is the integrating capacitor
C3 is the differential capacitor. It has limted function when connelcted like in post#2.
The ringing is typical for an integrating error amplifier (as long as not compensated for this).
Imagine: As long as Vout is not at it´s setpoint the integrator voltage increases. (ignoring the D part) And Vout eventually is at it´s setpoint, then the I part is at it´s maximum --> driving the output even higher. Overshot. And it "needs" the overshot to get the I part do decrease.
I´m interested in seeing
* the error amplifier output plot
* and the C1 voltage plot.
****
You may try to "limit" C1 voltage ... to limit the overshot
Currently your regulator should be in stable state:
* when both error amplifier inputs are 0.5V
* when duty cycle is about 45%
* when error amplifier output is at 2.25V
...and ideally then the C1 voltage is about 1.75V. (Please check this with your simulation)
You could improve the circuit in a way that C1 voltage becomes zero in ideal case.
This could be done by changing the triangle range 0V..1.11V.
Then you are able to limit the C1 voltage with antiparallel diodes.
Klaus
I didn't account for the ESR of the output capacitor and the inductance model is also non-saturating, this is a first order design of the converter and probably I will have to redesign the compensator, but I finally understood the concept of the effect of the gain on the poles and zeroes of the PID compensator. I will post an update with all the necessary updates, thanks @cupoftea and @KlausST for your helpThanks Zahid, if we start with the almost good values at the bottom of your post #6..
C1 = 4.7n, R2=4.7k, R1=47K, C3=100n, R3=330
...from here, adjust as follows...
Put just 10n across the 9k upper divider resistor....also...
Reduce R3 from 330R to try and damp out the ringings at start-up-rise
...Then increase R1 to 82k and go again from 330R downwards, still with 10n across the upper 9k.
Increasing R1 reduces the gain, and having C3 and R3 across R1 helps to prevent messigng the phase up when the gain is reduced.
Calculate the ringing frequency.....is it equal to the resonance frequency of Lout & Cout?
....If so it means your crossover freq is too close to this resonance, and as such, there is no way you will avoid ringing until you get x'over away from f(res).
[EDIT....sorry i just saw you are aware of this]
....i believe you could also solve the problem by using current mode control.....CMC takes out one of the poles at the output.....and stops it looking like an LC filter.
By the way, what ESR do you have in Cout?....no ESR will make it ring easily.
....
Sure.By putting anti-parallel diodes across the capacitor C1, the capacitor voltage does not go to zero but stays at 0.8 volts.
I don't want to bother you too much (insert sweaty smiley emoji hereHi,
Sure.
Read what I´ve written.
--> You can zero the voltage C1 with adjusting the triangle amplitude.
After that you can limit the C1 voltage with the anti parallel diodes. (limit to about +/- 0.6V)
Klaus
...OK thanks, but the 100n was supposed to be in pllel with the 9k, not in seriesFor
C1 = 10n, R2=1k, R1=47K, C3=100n, R3=330, and 100n in series with the 9k divider resistor, unexpected results.
You are right, 100n was supposed to be in parallel with 9k, i messed up there....OK thanks, but the 100n was supposed to be in pllel with the 9k, not in series
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