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I am using this instrumentation amplifier to measure the output voltage from a solar cell. This IC has the input protection by using a resistor Rlim, and the datasheet includes the equation for calculating the value of that resistor, shown in below screenshot. My Vs is 3.3 V, and this Vs shares the same GND with the amplifier. My V+ and V- connect to the solar cell positive and negative terminals. The maximum output voltage from the cell is 0.5 V. That means I don't know what value V+ and V- referenced to the amplifier's GND is. If I'm using 0.5 V for Vover and 3.3V for Vs, I will get a negative number. So, I wonder if I am doing anything wrong here.

input protection.

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    \$\begingroup\$ You have to define the potential of the solar cell with respect to the amplifier one way or another. The easiest way is to connect one terminal of the solar cell to the amplifier's ground. Otherwise it will be floating, as you've already determined, and that will mess up your measurements. \$\endgroup\$ Commented Mar 20, 2023 at 18:45
  • \$\begingroup\$ Is there another solution for this if I can't connect one terminal of the solar cell to amplifier's GND? I am using an isolated Dc/DC to step up 0.5V to 3.3V and use this 3.3V to power the amplifier. Since the DC/DC is isolated, I can't connect one terminal of the solar cell to amplifier GND... \$\endgroup\$
    – Alex
    Commented Mar 20, 2023 at 19:03
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    \$\begingroup\$ Why wouldn't you be able to connect the grounds? The fact that the DC/DC converter is isolated makes this even easier - just connect GND to V- of the solar cell and it should all work. Since the solar cell and amp are connected anyway, there's no proper isolation in the first place, only a badly floating voltage that skews your measurements. \$\endgroup\$ Commented Mar 20, 2023 at 20:26
  • \$\begingroup\$ that makes sense. Thanks ! \$\endgroup\$
    – Alex
    Commented Mar 21, 2023 at 0:53

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In the text you have provided, from the data sheet, it mentions that the inputs are clamped to the supplies, and if the input voltage(s) exceeds VS you should use the RLIM as shown. Clamping in this case means that the "clamping diodes" shown connected between the input and the power supply pins conduct when the input voltage magnitude is greater than the power supply pin magnitude. You need to determine what the largest overvoltage can be, and then use the equation from Ohms law I=E/R or, in this case R=E/I. The data sheet says limit the current to 10 mA so the I is 0.01. The E is the largest voltage you expect (negative or positive) and the E is the difference from the appropriate supply. (The clamping diode has to be forward biased) I am ignoring the forward voltage of the diodes.

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  • \$\begingroup\$ Thank you ! That helps. Do you think the equation above is wrong? because right before the diode, we should have Vs+0.7. So the equation should be (Vover-(Vs+0.7))/10mA= Rlimit? \$\endgroup\$
    – Alex
    Commented Mar 20, 2023 at 19:39

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