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ADG794 Synchronously Rectified Charge Injection

Hello.

I'm trying to synchronously rectify and integrate a high-impedance AC signal using an ADG794 switch in the circuit shown below.

It works for lower values of R1, but once the current drops below about 100nA RMS, the integrator goes the wrong way (i.e., the rectified current is in the opposite direction of what it should be).

The signal frequency (V1) is much higher than the R1 C1 time constant, so the phase shift should be negligible. I've tried offsetting the switch and signal frequency clocks in time. The system is effectively dual supply and nothing exceeds the ADG794 supply rails. The issue does not seem to be DC leakage, because the integrator holds a constant charge for a long time if the switch remains off. Using a low input-offset op amp does not seem to improve anything. The only thing I've been able to come up with is that the switch's charge injection is also being synchronously rectified.

Does this make sense? Is there a better explanation I could test? Is there a model for this switch or a similar one that would show the problem? Is there a solution besides adding a current amplifier before the switch?

Thanks,

Dan

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  • Hi Sean,

    The purpose of the circuit is to determine the magnitude of a high impedance AC source (V1 in the first schematic I posted). The signal can have a large, variable DC offset, which I do not want to affect my measurement (hence C1).

    The idea is that the current is integrated for only half of the AC the waveform. When V1 is low, current flows out of the integrator, causing the output voltage to go up.

    Then, during the other half of the waveform, the current is shunted to ground and the output of the integrator remains constant. Without the switch the integrator would just reverse direction and end up back where it started (ignoring leakage).

    The circuit works as intended for lower impedance sources (e.g. R1 = 100kOhm), but for larger R1 (or if R1 is disconnected, as in the diagram I posted on 5-9), the switch seems to inject current into the integrator that overwhelms the desired signal.

    While I expected current to be injected during switching, I did not anticipate a net charge over the course of a whole cycle (whatever is injected during the positive switching transition should be canceled out by the opposite charge being injected during the negative transition, I thought).

    Unfortunately, I do not have access to a good scope, so I can't tell if the problem is on the positive or negative switch transition (or both).

    You asked about rise time, but I'm not sure which rise time you're referring to. If you mean the total rise time of the integrator, it is slow (a few volts / second, over many switching cycles).

    Let me know if anything is still unclear.

    Thanks,

    Dan

Reply
  • Hi Sean,

    The purpose of the circuit is to determine the magnitude of a high impedance AC source (V1 in the first schematic I posted). The signal can have a large, variable DC offset, which I do not want to affect my measurement (hence C1).

    The idea is that the current is integrated for only half of the AC the waveform. When V1 is low, current flows out of the integrator, causing the output voltage to go up.

    Then, during the other half of the waveform, the current is shunted to ground and the output of the integrator remains constant. Without the switch the integrator would just reverse direction and end up back where it started (ignoring leakage).

    The circuit works as intended for lower impedance sources (e.g. R1 = 100kOhm), but for larger R1 (or if R1 is disconnected, as in the diagram I posted on 5-9), the switch seems to inject current into the integrator that overwhelms the desired signal.

    While I expected current to be injected during switching, I did not anticipate a net charge over the course of a whole cycle (whatever is injected during the positive switching transition should be canceled out by the opposite charge being injected during the negative transition, I thought).

    Unfortunately, I do not have access to a good scope, so I can't tell if the problem is on the positive or negative switch transition (or both).

    You asked about rise time, but I'm not sure which rise time you're referring to. If you mean the total rise time of the integrator, it is slow (a few volts / second, over many switching cycles).

    Let me know if anything is still unclear.

    Thanks,

    Dan

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