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I need to perform a ground offset test. Ground offset = 1V Please tell me whether my approach is correct.

Typ Input power supply to the circuit is +12V Typ Load current for the circuit is 0.5A

So, I connect the power terminal of the power supply to positive input of the circuit. I, then, connect a 2ohm resistor between the ground of my power supply and negative(return) terminal of the circuit.

2 ohm because, V=I * R. To have, 1V drop between negative and ground, 0.5A*2Ohms = 1V.

So, If i place a 2 ohm resistor between the ground of power supply and the negative terminal of the circuit, I will have achieved a 1V ground offset between the input and the return terminal of the circuit.

So, Positive terminal of the circuit will have +12V and the negative (return) terminal of the circuit (top of the resistor connected to negative terminal and bottom of the resistor connected to negative terminal of battery)will have +11V

Am I right?

Please provide a connection diagram to help

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  • \$\begingroup\$ If the load current is stable. Better to use a zener diode to create the 1V drop \$\endgroup\$
    – Swedgin
    Oct 31, 2019 at 8:50
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    \$\begingroup\$ @Swedgin For testing purposes, better use something that is a stabilized voltage source instead of a simple zener. \$\endgroup\$ Oct 31, 2019 at 9:11
  • \$\begingroup\$ Load current is not stable. Is my approach current? \$\endgroup\$
    – user220456
    Oct 31, 2019 at 11:15
  • \$\begingroup\$ Please tell me the procedure if I want to use zener diode? And let me know whether my connection is right \$\endgroup\$
    – user220456
    Oct 31, 2019 at 13:07
  • \$\begingroup\$ The proper way to do this would be 2 power supplies in series, 1 set to 1V, the other set to 11V, this way the offset is consistent across most loads \$\endgroup\$
    – Reroute
    May 23, 2020 at 10:19

1 Answer 1

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The proper way to have a consistent offset over most load conditions would be to have 2 power supplies in series, This way no matter how the load changed, it will be a consistent offset

As your device normally runs off 12V, you would want atleast 1 floating supply, in a pinch this could be a battery that only your circuit is connected to

To the negative of that supply, you would have another outputting 1V to offset you circuit against any external signals to the normal ground it would normally connect to, this makes any external signals look 1V less than the circuits negative.

The main caution with this setup is to make sure the lower supply is protected by a fuse or circuit breaker, if you accidentally short the "ground" and circuit positive, that lower supply will not enjoy the encounter. this should only be carrying the I/O currents you have to ground, so I expect this to be fairly low for most circuits.

enter image description here

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  • \$\begingroup\$ Thank you for the answer. Do i need to connect a simple wire from the second 1V battery negative to ground? \$\endgroup\$
    – user220456
    May 23, 2020 at 11:13
  • \$\begingroup\$ It would be a power supply not a battery, but for a short period a single battery cell might work for you, the negative supply would have its positive to the "circuit -" and its negative to where that negative would normally connect with the rest of the devices it talks to. \$\endgroup\$
    – Reroute
    May 23, 2020 at 11:16
  • \$\begingroup\$ Like , I have a dual power supply. I put the power supplies in series mode. And I connect the first power supply as per the diagram you have shown. The second one, I take second wire from the negative of the power supply and connect to the ground? \$\endgroup\$
    – user220456
    May 23, 2020 at 11:33
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    \$\begingroup\$ Yes, that would be correct, that connects to what your offsetting from, \$\endgroup\$
    – Reroute
    May 23, 2020 at 11:42
  • \$\begingroup\$ Thank you for your answer \$\endgroup\$
    – user220456
    May 23, 2020 at 11:43

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