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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.

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    \$\begingroup\$ What are you doing that makes you think you will exceed the maximum isolation voltage? \$\endgroup\$ – Phil Frost Jun 12 '14 at 17:43
  • \$\begingroup\$ I'm stacking a few DC-DC bricks in series, so the total output voltage will be fairly high. \$\endgroup\$ – TaroKako Jun 12 '14 at 17:45
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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

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.

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    \$\begingroup\$ For having voltages to remain precisely equal when any current flows between systems would indicate that they are not galvanically isolated. On the other hand, since it would be rare for there to be absolutely no leakage path between two devices, things are often said to be isolated even when a slight leakage path exists. Even a slight leakage path will suffice to keep things within a few volts of each other in cases where there's no significant current trying to push them further apart. \$\endgroup\$ – supercat Jun 12 '14 at 19:05
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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.

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  • \$\begingroup\$ Thanks for the feedback. Unfortunately, the inherent leakage in the components could potentially be too slow to equalize the potentials. I currently have a 10 MOhm resistor across, to limit current flow. I appreciate the feedback! \$\endgroup\$ – TaroKako Jun 12 '14 at 20:38
  • \$\begingroup\$ @TaroKako From your comment about marine galvanic isolators a thought came to mind. Why don't you add a capacitor between your circuits as well as a high value resistor. The capacitor would be able to absorb any transient excess charge difference while the high value resistor could slowly drain the excess charge on one side of the circuit or the other. The capacitor requires significant charge difference for a change in voltage. (Make sure you use a very high voltage rated capacitor.) \$\endgroup\$ – horta Jun 12 '14 at 21:19
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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.

enter image description here

enter image description here

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

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    \$\begingroup\$ I'm not wrong, the crucial part to the question is "need to equalize the ground voltage between the input and output" - this means keep constant and un-moving. Your answer is related only to quizzing my answer and does not provide an answer to the OP other than to imply "ignore Andy, your question is reasonable"!!! Equalized, constant and un-moving requires a short-circuit as my answer has pointed out. \$\endgroup\$ – Andy aka Jun 12 '14 at 21:01
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    \$\begingroup\$ Andy's answer has some hyperbole in it, but the general idea is correct. It doesn't have to be 0 ohms, it can be 1000 ohms and would still be classified as not galvanically isolating. Circuits with galvanic isolation have non-infinite resistance to them, but it's really about the value and quality of that resistance that matters. \$\endgroup\$ – horta Jun 12 '14 at 21:14
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    \$\begingroup\$ @EMFields Sorry dude but you are wrong - if the two voltages are equalized then they are equal. This means there is a mechanism that makes them equal so, let's say that one probe on the ohm meter sets a voltage on input A then "the mechanism" (whatever that is) sets the same voltage on input B (because it is equalized) and, this means the other probe will "see" the voltage on input A (but on input B) and this means the meter will register an exact short circuit, zero ohms. \$\endgroup\$ – Andy aka Jun 12 '14 at 22:20
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    \$\begingroup\$ @horta no hyperbole in my answer as far as I'm aware - it cannot be any ohmic resistance other than zero if the two voltages are equal under all circumstances. The question is flawed in principle. \$\endgroup\$ – Andy aka Jun 12 '14 at 22:23
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    \$\begingroup\$ @EMFields The point is that you cannot have galvanic isolation if the voltages are equal. The op's question is flawed if he expects equalized potential on both sides and the "equaling" mechanism can be no other than a short circuit. \$\endgroup\$ – Andy aka Jun 12 '14 at 23:33

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