I've calculated the characteristic polynomial of this circuit
where the circle is a nullator (ideal op-amp input port) and the two circles a norator (ideal op-amp output port).
If I did it properly, the result is:
a3s3 + a2s2 + a1s + 1
with
a3 = R1(R2R3 + R2R4 + R3R4)C1C2C3
a2 = R1R3C3(C1 + C2) + R1(R2 + R4)C1C2
a1 = R1C1 + R1C2 + R3C3
I've tried getting nicer results by applying a star-triangle transformation to R2, R3 and R4 and replacing two of the transformed resistors and C3 with an inductance with series resistance, but it didn't make the equations any clearer.
where the circle is a nullator (ideal op-amp input port) and the two circles a norator (ideal op-amp output port).
If I did it properly, the result is:
a3s3 + a2s2 + a1s + 1
with
a3 = R1(R2R3 + R2R4 + R3R4)C1C2C3
a2 = R1R3C3(C1 + C2) + R1(R2 + R4)C1C2
a1 = R1C1 + R1C2 + R3C3
I've tried getting nicer results by applying a star-triangle transformation to R2, R3 and R4 and replacing two of the transformed resistors and C3 with an inductance with series resistance, but it didn't make the equations any clearer.
Normally you will want a1, a2 and a3 to have some desired values, for example
a3 = 1/(2 pi fc)3
a2 = 2/(2 pi fc)2
a1 = 2/(2 pi fc)
if you want a third-order Butterworth response with cut-off frequency fc. Assuming you also want to choose R1, C1, C2 and C3, you can calculate the rest, but the results may be negative or complex. In that case, you have to change the chosen values and try again, or prove it isn't possible to get anything sensible out.
R3 = (a1 - R1C1 - R1C2)/C3
p = R2 + R4 = (a2 - R1(C1 + C2)R3C3)/(R1C1C2)
q = R2R4 = (a3 - R1(R2 + R4)C1C2R3C3)/(R1C1C2C3)
R2,4 = (p +/- sqrt(p2 - 4q))/2
a3 = 1/(2 pi fc)3
a2 = 2/(2 pi fc)2
a1 = 2/(2 pi fc)
if you want a third-order Butterworth response with cut-off frequency fc. Assuming you also want to choose R1, C1, C2 and C3, you can calculate the rest, but the results may be negative or complex. In that case, you have to change the chosen values and try again, or prove it isn't possible to get anything sensible out.
R3 = (a1 - R1C1 - R1C2)/C3
p = R2 + R4 = (a2 - R1(C1 + C2)R3C3)/(R1C1C2)
q = R2R4 = (a3 - R1(R2 + R4)C1C2R3C3)/(R1C1C2C3)
R2,4 = (p +/- sqrt(p2 - 4q))/2
When I try it using a spreadsheet, I can get positive real values out, but ones that won't be of any practical use...
I'll give it some more thought. Maybe a capacitor straight between the op-amp input and output can stabilize it without throwing away all the benefits, maybe not.
I'll give it some more thought. Maybe a capacitor straight between the op-amp input and output can stabilize it without throwing away all the benefits, maybe not.
Thanks for your admirable efforts. I haven't written down that kind of equation in more than 20 years, and I would have been totally incapable of deriving something useful.
I have made some tests, and the stability appears to depend heavily on the value of R3 relative to the // value of the two others: at first sight, it cannot be smaller than half that parallel value. Other factors probably play a role
I have made some tests, and the stability appears to depend heavily on the value of R3 relative to the // value of the two others: at first sight, it cannot be smaller than half that parallel value. Other factors probably play a role
If the op-amp were ideal, this combination should be stable:
R3 = 0
C3 < (C1 + C2) R1/(R2 // R4), where // means in parallel with.
It should also work with a real op-amp when its gain-bandwidth product is much greater than the filter bandwidth and you don't use a C3 value that's too close to the edge.
R3 = 0
C3 < (C1 + C2) R1/(R2 // R4), where // means in parallel with.
It should also work with a real op-amp when its gain-bandwidth product is much greater than the filter bandwidth and you don't use a C3 value that's too close to the edge.
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Damn...I have a feeling this page is going to be another important page where I won't be able to understand a single phrase so I'm writing here one that I can understand. This way I'm gonna be able to claim I was also part of history 🙂 just like Forrest Gump caught in the middle of Watergate scandal ...
Is there any way that mere mortals can understand this calculus?
Is there any way that mere mortals can understand this calculus?
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Basically we are looking for an improved version of the very clever second-order ripple filter for low-current, high-voltage supplies that Elvee posted in post #1. I had an idea to make it third order, but it is easy to dimension it such that it either becomes an oscillator, or a third-order filter that barely filters any better than Elvee's original circuit.
Post #27 is my latest attempt to come up with a way to dimension it such that it works better than the original. I haven't found any major issues with this proposal yet, but it could very well be that there is something wrong with it anyway.
Post #27 is my latest attempt to come up with a way to dimension it such that it works better than the original. I haven't found any major issues with this proposal yet, but it could very well be that there is something wrong with it anyway.
A long time ago, I played with a 3-part totem pole circuit where the upper output drove the lower rail capacitor and the lower output drove the upper rail capacitor. But I'd have to spend some time to recreate it. Maybe you could use two op-amps, but it was originally discrete. This assumes the voltage being filtered is also the amplifier supply.
I have tested the R3=0 version, based on the initial build. It works for C3~<650pF, and the improvement in filtering is about the same as for the RC version: ~10dB. Useful, but not huge
They were the ones shown in the first post, with the 22 meg resistor split in two. The frequency was 50Hz
Interesting. It should start oscillating from 545.4545... pF onward, but you had it working up to about 650 pF.
The values aren't hugely accurate, and it is built on a breadboard, which probably provides some capacitance between the output an the (-) input
I have tested this 2-opamp variant:
It allows an improved performance without stability constraints, thanks to the integrator that compensates other poles in the circuit; it is more complex, obviously
It allows an improved performance without stability constraints, thanks to the integrator that compensates other poles in the circuit; it is more complex, obviously