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I'm designing a circuit to process a di/dt signal. As the input is the differential of the measured current, I have to integrate the voltage to provide a representation of the current. The duration of the signal pulse is around 50 ns, hence the bandwidth of this signal is of the order of 20 MHz.

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My questions are:

  1. What determines the upper-frequency range of the integrator? If I only consider the amplifier bandwidth GBP≥10 times the maximum frequency (GBP≥10×20MHz), the integrator will function effectively?

  2. What is the role of unity gain cut-off frequency \$f_o\$ here? Does it mean that signals above this frequency will be attenuated? If that’s the case then if the input signal is within \$1/(2\pi R_f C)\$ Hz to \$1/(2\pi R C)\$ Hz range in the hope of getting an integrated output, will I get one?

  3. In the reference, it exploited a single dominant-pole compensated op-amp (AD8045). It says "no feedback capacitor is necessary because the closed-loop response will naturally asymptote to the open-loop response". I don't really understand how does the integration circuit work here. Is it because it has a compensated feedback capacitor inside the op-amp? And how could I calculate the Vout in this case? enter image description here

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    \$\begingroup\$ To be exact - such an integrator can integrate at one single frequency only. That is the frequency where the shift crosses the -90 deg line (real 2- or 3-pole opamp assumed). Hence, the answer to your question depends on the max. error you can tolerate. \$\endgroup\$ Commented Oct 19, 2021 at 8:36
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    \$\begingroup\$ What does it mean "The duration of the signal is around 50ns" ? The signal shall have a frequency in between -3dB and fo from your circuit for correct integration. \$\endgroup\$ Commented Oct 19, 2021 at 8:36
  • \$\begingroup\$ What is the signal source output impedance? Note this signal source must be able to drive the input resistor of your integrator. Does it output voltage or current? If there is a way to make it output current, maybe you do not need an integrator... \$\endgroup\$ Commented Oct 19, 2021 at 9:49
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    \$\begingroup\$ 20MHz is not the maximum frequency of a 50ns pulse ... unless the pulse is a raised cosine pulse, or gaussian. If it is a rectangular pulse and you want to integrate it accurately it needs to integrate over enough bandwidth to accommodate an appropriate number of the odd order harmonics in its spectrum. (the AD8045 looks pretty good if you can tolerate an LF corner about 2MHz ... yes its internal compensation cap provides the integration) \$\endgroup\$ Commented Oct 19, 2021 at 11:29
  • \$\begingroup\$ The bandwidth of a pulse is typically calculated as τr/0.35 where τr is the 10-90% pulse rise/fall time. \$\endgroup\$ Commented Oct 14 at 16:03

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  1. What determines the upper-frequency range? Is GBP>= 10x20 MHz enough? You kinda answered your question. GBP determines you fmax. Anything above will not integrate. You want to hit the range between cutoff frequency and GBP for operation. When your cutoff gets close to GBP it does not behave properly. As a rule of thumb some books say you want 50-100x. Depends on your Bandwidth.

  2. Whats the role of GBP? Technically GBP is the point where the loop gain is 1V/V (0dB). Anything above will not have a usable gain. So yes, it will start attenuating upper frecuencies.

  3. Why does it say "no feedback capacitor needed"? This is tricky but it is related to the open loop gain. An opamp has an open loop gain off $$A_{OL}=GBW/s=GBW/(j2\pi f)$$ in a real life scenario looks more like this: $$\frac{V_{out}(s)}{V_{in}(s)} = -\frac{1}{sRC \left(1 + \frac{s}{GBW}\right) + \frac{s}{GBW}} = -\frac{1}{sRC + \frac{s^2 RC}{GBW} + \frac{s}{GBW}}$$ This acts depending on the dominant pole.

  • At low frecuencies, s<<GBW we can approx to -1/sRC which is an ideal integrator.
  • At high frecuencies, the s squared pole takes over 40dB/dec
  • At the transition, then asyntotes to the open'loop rolloff.

This means the frec response is gonna behave like the combination of integrator and low pass filter you would expect. To calculate Vout just choose RC and C accordingly. e.g: For R=1 kΩ, C=10 pF, $$ f_u = 15.9 MHz$$, $$ f_{transition} \approx \sqrt{\frac{10^9}{2\pi \times 10^3 \times 10^{-11}}} \approx 178 MHz $$ (well above, so ideal up to ~20 MHz). And then apply the formula i just told you before for Vout.

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