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I have to measure a DC voltage and an AC voltage around 100V. I found the circuit with an AD629 and one OP177.

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I simulated everything in Spice and it works fine, at least in the simulation. In reality the power supply (V_IN) is isolated. It doesn't share a ground with the rest of the measurement circuit. V_OUT should be converted by an ADC which doesn't share a ground with V_IN.

Is it still possible to use this circuit because of the high input impedance? If so should pin 1 (Ref-) of the AD629 connect to the V_IN ground or to the shared analog ground with the rest of the circuit?

And do you think there are problems when measuring DC voltages instead of AC voltages? In the simulation it doesn't matter.

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  • \$\begingroup\$ "In reality the power supply (V_IN) is isolated. It doesn't share a ground with the rest of the measurement circuit. " , but it has to. \$\endgroup\$
    – Marko Buršič
    Jul 21, 2019 at 20:22
  • \$\begingroup\$ @marko Won't Vin self-bias through the AD629 internal resistors? The currents from Vin should find their way around from one terminal to the other. \$\endgroup\$
    – DKNguyen
    Jul 21, 2019 at 20:43
  • \$\begingroup\$ You desperately need to rethink this. A power supply with the ground not connected (but the supply lines are) is not "quasi isolated" - it's a disaster waiting to happen. And the only way to have a circuit with the output ground not connected to the input ground is by using magnetic or optical means to transfer information. \$\endgroup\$
    – WhatRoughBeast
    Jul 22, 2019 at 4:05
  • \$\begingroup\$ @WhatRoughBeast If I connect pin 1 to ground of the circuit (analog ground). So to the same ground as pin 3 of OP177 and the same ground as the ADC. Then the "isolated" ground isn't galvanic isolated anymore. But there is an high impedance (resistor network) between the isolated and analog ground. A differential probe works similar or not? It's also not galvanic isolated, but you can probe between a +500V and a +200V point without an isolated oscilloscope, because the probe has a high impedance input. \$\endgroup\$
    – Momchilo
    Jul 22, 2019 at 14:22

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