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To measure a voltage in a (potentially dangerous) external system, people typically use a differential amplifier or even an isolation amplifier, something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

This way, the voltage going into the safe system isn't directly connected to high voltages, and we can happily put our ADC there and supply it a safe input.

I want to do this the other way around. I am generating a voltage \$V_{out}\$ under 5V, and I would like to apply this voltage across an element of the other circuit.

  1. Is there a typical setup for doing this?
  2. I believe that the low side of my output will be connected to the ground of the external circuit. That means it would be possible to do this without a differential output - I could just put my regular ground-referenced output into the external circuit. Am I correct to think that it's unsafe to simply connect the two grounds together, or is this the much simpler solution?
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  • \$\begingroup\$ The reason to use an amplifier in your example is purely functional. You want to separate the measurement instrument from the measured circuit, by inserting infinite input impedance buffer between, since your instrument might have a finite input resistance and will affect the measured value. \$\endgroup\$
    – Eugene Sh.
    Jan 21, 2015 at 16:59
  • \$\begingroup\$ Why did you show the diff amp? If your big scary voltage has a ground reference I think it will fry your opamp. How big a voltage are you trying to add the 5V to? \$\endgroup\$ Jan 21, 2015 at 17:19
  • \$\begingroup\$ @GeorgeHerold - I'm connecting to a battery that's plugged into a very Chinese charger. I don't know what this charger does with the 120V from the wall, and I don't really want to find out the hard way. \$\endgroup\$
    – Greg d'Eon
    Jan 21, 2015 at 17:30
  • \$\begingroup\$ There is no problem to connect two grounds together if only they are at the same voltage in regards to a third common ground (the earth). Otherwise, a big current will flow between them. Regarding the Cninese charger, it should be not connected to (isolated from) the neutral point (the ground) of the net. \$\endgroup\$ Jan 21, 2015 at 17:30
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    \$\begingroup\$ Do you have a DMM? You can measure to see if there is any AC between plug ground and the battery terminals. \$\endgroup\$ Jan 21, 2015 at 17:39

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The most common solution for your problem is to use an isolation amplifier such as Analog Devices AD210. Note that there are many different models available from many suppliers.

Isolation amplifiers are generally broken into two categories: 2-port & 3-port. The number of ports specifies how many isolated sections are in the amplifier. A 3-port isolation amplifier has the input separated from the output with both sections separated from the power supply input. A 2-port isolation amplifier has the power supply in the same section as either the input or output, depending on that particular model. Obviously, 2-port isloation amplifiers cost less than 3-port units.

Also be aware that the isolation voltage is different between ports. The AD210 that I mentioned earlier has higher isolation voltage rating between the input port and both other ports whereas the output port has a lower isolation voltage between itself and the power supply port.

A quick Google search for the search terms "isolation amplifier" will show you the plethora of options available.

Note also that I as speaking of Analog isolation amplifiers. If you have the ability to convert a digital bitstream into your desired result at the destination, there exists another plethora of digital isolation devices.

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  • \$\begingroup\$ Aha! When I looked up isolation amplifiers, I found some chips that work as inputs (as per my circuit above), but not as outputs, and I assumed that was the case for all isolation amps. I'll see what I can find if I dig a little deeper. \$\endgroup\$
    – Greg d'Eon
    Jan 21, 2015 at 17:34

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