Yes, but is that a concession or an objective? Avoiding the need for the +/-12V supply would be desirable, no? History tends to dismiss designs that use a lot of power supplies.
According to post #6, https://www.diyaudio.com/community/...ve-this-electronic-filter.395321/post-7256395 , there is a 9 V battery and a 5 kV supply coming from a converter that can only supply a few microamperes. I don't know why there is also a +/- 12 V in some simulation schematics. The op-amp can in principle also work on a single-ended 9 V supply, although the lower the supply voltage, the smaller the input ripple voltage that will drive it into clipping.
It is just for sim convenience: I have not yet decided about power supply arrangements, housekeeping, etc. The opamp supply might be derived from the converter if the headroom is not sufficient, I don't know.
At this stage, I try to get the concepts right. When it's done, I'll care about practical gory details, but for now I want to establish a reliable base before going further
At this stage, I try to get the concepts right. When it's done, I'll care about practical gory details, but for now I want to establish a reliable base before going further
For this circuit
the theoretical stability criterion becomes
C3 < (C1 + C2) R1/(R2 // R3 // R4)
It can be derived very simply from the criterion of post #27 by comparing the transadmittance of R3-R4-C3-R2 of this circuit with the transadmittance of R4-C3-R2 in the circuit of #23 without its R3.
A problem with both circuits is that while they are stable, they ring a lot, especially this one. The same holds for the circuit of post #1 when the ratio of the value of its feedback resistor to R1 is too large. In fact, adding a voltage divider in the feedback path has a similar effect to increasing the feedback resistor's value.
the theoretical stability criterion becomes
C3 < (C1 + C2) R1/(R2 // R3 // R4)
It can be derived very simply from the criterion of post #27 by comparing the transadmittance of R3-R4-C3-R2 of this circuit with the transadmittance of R4-C3-R2 in the circuit of #23 without its R3.
A problem with both circuits is that while they are stable, they ring a lot, especially this one. The same holds for the circuit of post #1 when the ratio of the value of its feedback resistor to R1 is too large. In fact, adding a voltage divider in the feedback path has a similar effect to increasing the feedback resistor's value.
Yes, the two-opamp variant might be preferable: it has practically no stability issue except the opamps internal phaseshift for extreme values, and using a dual opamp is no problem.
It has the filtering capacity of a ~300nF with 2 x 1.5nF physical caps, which should be sufficient in practice
It has the filtering capacity of a ~300nF with 2 x 1.5nF physical caps, which should be sufficient in practice
The Deboo integrator in the loop adds a 90° shift compensating for the differentiating effect of the two 1.5nF. It also adds gain.
The transfer function must contain a square root, becuase the effect the gain is not proportional.
A small feedback from the integrator keeps the circuit DC stable
The transfer function must contain a square root, becuase the effect the gain is not proportional.
A small feedback from the integrator keeps the circuit DC stable
In fact I think you can get approximately the same behaviour out of your original circuit by replacing the 22 Mohm feedback resistor with 2.2 Gohm in parallel with 0.68 pF. The 2.2 Gohm can be replaced with a T network consisting of a 100:1 voltage divider and a 22 Mohm resistor (maybe 68 pF in parallel with the 22 Mohm will then have about the same effect as 0.68 pF across the whole T). Single op-amps usually have the output and negative input on opposite sides of the package, so hopefully the parasitic capacitance is << 0.68 pF.
Yes, I tried it and it worked alright.
I had chosen not to meddle with the Deboo's balance, but the fear was unfounded: it is perfectly equivalent, and saves one resistor. This will probably be my final choice (for the time being: if an even lower impedance is required to manage real world requirements, I may reconsider)
I've calculated the transfer of the circuit of post #1 extended with a capacitor C3 in parallel with the feedback resistor R2. When you replace the op-amp, R2 and C3 with some other circuit that essentially behaves as a transimpedance amplifier with first-order roll-off (for example something with one op-amp and a T-network with a capacitor, or your two-op-amp circuit), substitute the DC transresistance for R2 and the time constant of the roll-off for R2C3.
The result is (sR2C3 + 1)/(s2R1R2(C1C2 + C1C3 + C2C3) + s(R2C3 + R1C1 + R1C2) + 1)
That is, there is a negative real zero that coincides with the pole of the transimpedance amplifier consisting of the op-amp and C3 in parallel with R2. That's logical in retrospect, because for the value of s that kills the local feedback around the op-amp, the loop through C1 and C2 gets infinite loop gain.
Despite the presence of three capacitors, there are only two poles, as was to be expected because the voltage across C3 follows directly from the voltages across C1 and C2 (three capacitors, but only two independent state variables). The natural radian frequency and quality factor of the pole pair are:
ωn = 1/√(R1R2(C1C2 + C1C3 + C2C3)) ≈ 1/√(R1R2C1C2) where the approximation holds when C3 << C1 and C3 << C2
Q = 1/(ωn(R2C3 + R1C1 + R1C2))
Hence, when you choose the values R1, R2, C1 and C2, the natural radian frequency is practically fixed. After choosing a desired quality factor, you can then calculate the required transimpedance amplifier roll-off time constant from
R2C3 = 1/(ωn Q) - R1(C1 + C2)
Example: when
R2 = 2.2 GΩ, like in your two op-amp circuit,
R1 = 1 MΩ
C1 = C1 = 1.5 nF:
ωn ≈ 14.21338 rad/s, fn ≈ 2.26213 Hz, effect of C3 on the natural frequency neglected
For Q < 1: R2C3 > 67.35624 ms
When R2C3 = 1.5 ms, like in your two-op-amp circuit: Q ≈ 15.63472
If my calculations are correct, your two-op-amp circuit should have quite some ringing, somewhere around 2.26 Hz.
The result is (sR2C3 + 1)/(s2R1R2(C1C2 + C1C3 + C2C3) + s(R2C3 + R1C1 + R1C2) + 1)
That is, there is a negative real zero that coincides with the pole of the transimpedance amplifier consisting of the op-amp and C3 in parallel with R2. That's logical in retrospect, because for the value of s that kills the local feedback around the op-amp, the loop through C1 and C2 gets infinite loop gain.
Despite the presence of three capacitors, there are only two poles, as was to be expected because the voltage across C3 follows directly from the voltages across C1 and C2 (three capacitors, but only two independent state variables). The natural radian frequency and quality factor of the pole pair are:
ωn = 1/√(R1R2(C1C2 + C1C3 + C2C3)) ≈ 1/√(R1R2C1C2) where the approximation holds when C3 << C1 and C3 << C2
Q = 1/(ωn(R2C3 + R1C1 + R1C2))
Hence, when you choose the values R1, R2, C1 and C2, the natural radian frequency is practically fixed. After choosing a desired quality factor, you can then calculate the required transimpedance amplifier roll-off time constant from
R2C3 = 1/(ωn Q) - R1(C1 + C2)
Example: when
R2 = 2.2 GΩ, like in your two op-amp circuit,
R1 = 1 MΩ
C1 = C1 = 1.5 nF:
ωn ≈ 14.21338 rad/s, fn ≈ 2.26213 Hz, effect of C3 on the natural frequency neglected
For Q < 1: R2C3 > 67.35624 ms
When R2C3 = 1.5 ms, like in your two-op-amp circuit: Q ≈ 15.63472
If my calculations are correct, your two-op-amp circuit should have quite some ringing, somewhere around 2.26 Hz.
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The two-opamp circuit doesn't have a 2.2G feedback resistor, be it physical or virtual. The integrator and its DC stabilization link might influence the apparent value of the 22M, but I do not see clearly how or how much.
However, it does exhibit a strong peaking at around 2Hz, which is problematic since it falls right into the hand-effects range. This makes it unsuitable for my application.
The plain-vanilla single 22M resistor circuit only has a modest peaking, at a higher frequency, ~15Hz, but unfortunately it has a significantly lower performance.
Back to the drawing board
However, it does exhibit a strong peaking at around 2Hz, which is problematic since it falls right into the hand-effects range. This makes it unsuitable for my application.
The plain-vanilla single 22M resistor circuit only has a modest peaking, at a higher frequency, ~15Hz, but unfortunately it has a significantly lower performance.
Back to the drawing board
When you have the circuit of post #38, but with R5 connected in parallel with C4 and with R4 replaced with a short, you have a transresistance amplifier with 22 Mohm feedback resistor followed by an integrator with a DC gain of 101. Together that's a DC transresistance of 2.222 Gohm from the input of the first op-amp to the output of the second op-amp.
I haven't calculated the precise value for the original circuit of post #38, but it must also be close to 2.2 Gohm: about 22 Mohm in the first stage and a gain of 100 in the second stage.
Damping the resonance is easy, just reduce the bandwidth of the amplifying part (increase the integrator time constant for the two-op-amp version), but damping it without making the filtering far less effective is another matter.
By the way, what kind of leakage do your capacitors have at 5 kV?
I haven't calculated the precise value for the original circuit of post #38, but it must also be close to 2.2 Gohm: about 22 Mohm in the first stage and a gain of 100 in the second stage.
Damping the resonance is easy, just reduce the bandwidth of the amplifying part (increase the integrator time constant for the two-op-amp version), but damping it without making the filtering far less effective is another matter.
By the way, what kind of leakage do your capacitors have at 5 kV?
At 5kv usual ceramic caps can leak anywhere between 1pA to 1uA...which makes the usual cap multiplier with bleeders an obvious option for simple minded people like me who wouldn't have the skills to calculate such a circuit in a thousand years from now.
I need to get born again to have a chance at understanding this thread...
I need to get born again to have a chance at understanding this thread...
I do not see this composite as 2.2G transresistance: the integrator has its gain of 101 at DC only.
Absolutely negligible: they can remain charged for an hour or more if left open
I have tried a different approach (in sim only for the time being): this time, I multiply the value of the 1.5nF, which causes less issues as it becomes a first order system.
There is an ugly quirk at ~0.6Hz, when the bootstrap kicks in, but it appears harmless.
The lead network C6 R8 is not strictly required, but it keeps the synthetic capacitor more perfect at higher frequencies by compensating the phase lag introduced by the two opamps
There is an ugly quirk at ~0.6Hz, when the bootstrap kicks in, but it appears harmless.
The lead network C6 R8 is not strictly required, but it keeps the synthetic capacitor more perfect at higher frequencies by compensating the phase lag introduced by the two opamps
When you have a transimpedance amplifier consisting of an op-amp with a 2.2 Gohm feedback resistor in parallel with a 0.68 pF capacitor, you ideally have a transresistance of -2.2 Gohm at DC and a roll-off with a 1.496 ms time constant.
With the two op-amp solution, you have a transresistance of -2.222 Gohm at DC and a roll-off with a 1.5 ms time constant.
Seems pretty similar to me.
Great, because with a 9 V supply and 2.2 Gohm transresistance, a few nA of leakage would drive the circuit into clipping.
Clipping is one of the issues I am worried about: with 15V supplies, a 50Hz, 6~7µA peak current will be sufficient to drive the opamp into overload. I may need to use higher voltage supplies derived from the converter, and a high-voltage, low-current post-amplifier stage