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I need to equalize the ground voltage between the input and output of an isolated DC-DC power supply while still maintaining galvanic isolation. Is there a way to do this?
The reason I need this is so that I don't exceed the maximum isolation voltage rating on the DC-DC power supply.
I need to equalize the ground voltage between the input and output of an isolated dc-dc power supply while still maintaining galvanic isolation
If I had a black box with two wires going into it and was told that the two wires ARE NOT galvanically connected but, V1 always equals V2, then the only reasonable conclusion I could draw is that they ARE galvanically connected and they are indeed shorted.
Imagine putting your multimeter across the two wires - it would read zero ohms. It couldn't possibly read anything else because V1 = V2.
Andy's right. Your specification contradicts itself: "Galvanic isolation is a principle of isolating functional sections of electrical systems to prevent current flow; no direct conduction path is permitted." - Wikipedia
Without a conduction path, charge built up on one circuit causes the voltage to trend away from the voltage of the other circuit. Without the ability to balance the charge, the voltage difference will always remain.
If you don't actually need galvanic isolation, you may try putting a high value resistor between the two grounds to allow excess charge to transfer between the two circuits at very low rates. This would tend to keep the grounds near the same point over time. Rapid movement of charge in either of the circuits would cause an imbalance though which would self correct given enough time through the resistor.
EDIT
Another option would be to add a capacitor rated to a high voltage along with a high value resistor. From the standard capacitor equation it becomes obvious that it would help mitigate the voltage difference you seek to squash.
$$ C= \frac{Q}{V}$$ or $$V= \frac{Q}{C}$$
If you have a constant charge between the plates, as you increase your capacitance between your two circuits, the result is a smaller voltage difference. Adding in that resistor would still allow you to drain the voltage difference slowly. The capacitor would allow you to absorb large transients of additions of charge differences between the two circuits.
Andy could be wrong, since:
BLACK BOX
/
+-------------------------+
| |
| +-------+------+ |
| | | | |
| | [R1] [R3] |
| | | | |
| | | +----------->E1
| [SUPPLY] | | |
| | +------|----------->E2
| | | | |
| | [R2] [R4] |
| | | | |
| +-------+------+ |
| |
+-------------------------+
as long as R1/R2 = R3/R4, then E1 will always equal E2 even though the two wires aren't galvanically isolated.
However, since an ohmmeter is a voltage - or current - source, once connected across E1 and E2 it would unbalance the bridge, causing charge to flow, and it would indicate some resistance.
In the attached images, the fixed resistors in the bridge - the ones in the gold anodized housings - are 120 ohm 3% wirewounds.
The ones on the right are in series, with about 25V of excitation across them from the little HP6216 on top of the decade resistor box, and the one on the left is in series with the decade resistor box and a 50 ohm pot used to balance the bridge with the same excitation supply across that string.
The best I could do was to balance the bridge to 3.1 millivolts on its output, as shown on the Fluke, and the Wavetek was set to 200 ohms, and wasn't connected to anything.
In the second image, the ohmmeter was connected across the output of the bridge and, as you can see, the bridge was unbalanced by about 106 millivolts and the ohmmeter read 116.1 ohms
