getting analog switch common mode down to -0.10V


I need the ADG709 input to go 1/10V below ground, so ourplan is to use a diode as a shunt regulator on V- (-15V at 10KOhm bias). The V+ supply pin is at 5V. Is this a viable approach? What are the pitfalls? See attached.


-Tim Starr on behalf of AK.G

  • 0
    •  Analog Employees 
    on Jun 30, 2011 12:50 PM

    Hi Tim,

    We haven't characterized the device with VDD=5V and VSS=-0.5V as you suggest in your circuit, so we cannot guarantee its performance and reliability. However, the device will be functional and I expect similar performance to the single 5V supply dataheet specifications.

    The only problem I can see with your approach is the variation of the forward bias voltage of the device(part to part variation and variation over temperature). To be able ho handle the -0.1V signal, the forward bias voltage should never go below 0.1V and above 0.5V. Because of this I recommend using a regulator instead of a simple diode. If a 0.5V regulator is not available then you should regulate the 5V rail to 2.5V and the -15V rail to -2.5V. This way the switch is able to handle the -0.1 signal and its performance is available in the datasheet.

    If your signal exceeds 2.5V on the positive side, you  may want to use a different switch such as the ADG1609(+/-5V supply) or ADG1409(+/-15V supply).

    As a separate comment to your circuit, I was wondering if you have considered the variation of the ON resistance of the switch when calculating the gain of the amplifier. Since you have large resistances in the feedback loop, the gain variation cause by the switch is small, but not zero.



  • 0
    •  Analog Employees 
    on Aug 2, 2018 4:23 PM
    This question has been assumed as answered either offline via email or with a multi-part answer. This question has now been closed out. If you have an inquiry related to this topic please post a new question in the applicable product forum.

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    EZ Admin