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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

  • Hello Dan,

    Could you show me the waveforms created by the above circuit and give me an indication of the values you are using for the components?

    You mention that the system is dual supply, while the ADG794 is single supply only. Can you confirm the rail voltages you are using for the switch and the input voltages?

    Find following a link to an article about charge injection:

    http://www.analog.com/library/analogDialogue/archives/31-3/ask.html

    Kind regards,

    Sean

  • Thanks for the prompt response.

    I am using 3.3 V rails for the switch, but the 'ground' shown in the diagram is at 500mV to be within the switch and op amp input range. The important thing is that nothing exceeds the switch rails.

    C1 is 100nF, C2 is 10nF, and V1 is a 80mV square wave at 100kHz. The circuit works fine when R1 is 100kOhms, but if I increase it to 2M (or disconnect it) it does not.

    I'm using an old analog scope, so I can't give you a good waveform, but here's what it's supposed to look like (e.g., what it looks like when R1 is 100kOhms).

    The switch is in sync with the excitation (V1 in the schematic). The issue is that the integrator output reverses direction (integrates down instead of up) when R1 is 2MOhms.

    Both your article and my previous understanding of charge injection tell me that the charge injection should not result in a DC current like the one I'm seeing, but I don't have any other explanation for what could be happening.

    Thanks,

    Dan

  • Hello Dan,

    I am unable to simulate the issue you are seeing here. Please can you confirm the following for me:

    • ADG794 supply voltages: Vdd = 3.3V, GND = 0V
    • Signal Voltage: 0.5V - 0.42V Square wave 100Khz

    What do you see when you probe the output of the switch, when it is not connected to the op-amp? Do you see a glitch that could account for the reverse in integration?

    Kind regards,

    Sean

  • That is correct. However, the problem is still there even without any excitation signal. The switch output with the integrator unconnected shows the expected charge injection square wave.

    However, I think I figured it out. The circuit below uses 3.3V rails, and the switch switches at 100KHz. Cpar is the parasitic trace on my board.

    Any difference in charge injection between the two switches will be dumped into the integrator each cycle. Additionally, any op amp input offset will have the same effect, but with the op amp I'm using that effect is much smaller.

    If this makes sense to you, are you aware of a solution other than amplifying my signal to overwhelm the problem?

    Thanks for all your time.

    --Dan

  • Hello Dan,

    I would like to understand the application of this circuit better.

    Could you explain to me the role of the switch in this circuit? Would the integrator not work as you depicted above without the switch?

    What rise time are you trying to achieve? When the integrator begins to integrate in the opposite direction, is it during the on pulse or the off pulse?

    I am also unsure about C1.I've not seen this in the "traditional" integrator circuit.

    Regards,

    Sean

  • 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

  • Hello Dan,

    Thank you for the clarification. We think that the issue could be either charge injection or leakage, or a combination. Though it is difficult to identify which is the actual cause. In this case it is probably best to do as you suggested, using a current amplifier before the switch.

    If you would like to try different parts:

    • The ADG774A in the QSOP package is a pin for pin replacement and has the leakage values production tested. It also has slightly lower charge injection.
    • The ADG12xx family of parts have very low charge injection, but they require a supply range greater than you've mentioned previously.

    I'm sorry that I couldn't be of more help.

    Regards,

    Sean