Does a relay act like a capacitor?

As I'm building my electric system up, I found an odd thing (to me). With the positive lead connected from battery to master relay, and the negative/ground cable unconnected, I measure about .7 volts between the (unconnected) negative terminal of the battery and any ground points on the firewall. I measure 0 current and "infinite" resistance at the same places.

I was concerned there might be a leak somewhere, but over the last three days with everything connected, there is no drop in battery volts. I guess this is normal?

Well, there is a tiny tiny tiny amount of capacitance across the contact points. But that should have no effect for D.C. measurements. Since you are trying to measure a potential difference between the battery and a floating (not connected electrically to anything) piece of firewall metal, you're probably seeing the voltage due to the very large but not zero resistance between the battery and airframe. The current is not zero but too small to measure. Try this: repeat the measurement but with a cheap, old-fashioned voltmeter with no electronics (these have a much lower input impedance than modern units). Does the 0.7 volts drop much lower, maybe to zero?

Just for grins, try the same measurement with the meter set to AC volts & report back.

I assume you repeated your DC measurement with the master + terminal disconnected, right?

Thanks for the help.

I re-measured everything with my older voltmeter, though I don't own an analog meter. Might see if I can pick up a cheapo later this week.

I did check everything with both terminals disconnected - no surprises there.

I still see between .7 and .8 volts when I have pos connected to master relay, and neg terminal not connected. No voltage differential present measured from the positive terminal on the battery or the relay to anywhere else but the neg terminal on battery. I tried it with the control wire (grounding) and the diode both disconnected from the relay with same .7 volts showing up.

I tried with the AC meter selected, and see a small voltage - somewhere around .02 V.

I can't find any measurable current still, and neither meter can measure other than max Ohms anywhere that there should not be a path. Measured across the large posts on the relay, from the connected / positive post to the case, ground, or any other grounded part. There is the expected reading across the diode - some resistance but fairly low ohms.

Maybe I did something really dumb. The photo of my firewall is on my facebook link below.

I'm not worried about it - obviously the battery is not draining. When I flip on the master and mess around with the lights (only thing currently connected) I can see a voltage drop to around 12.5 or so, but the battery recovers to 12.76 v after a few minutes with the master switch off.

One other thing I tried - I took a piece of test wire and clipped it to the neg post and the ground post on the firewall to short it. Then I measured the volts from the same two points - 0 volts, and while reading the meter, I pulled the test lead to see if the voltage builds up rapidly, which it does, though it seems to not exactly instantaneously - maybe one second to get from zero to .75 volts.

If you're using a DVM, they have a very high input impedance (typically 10 megohms) and will give you false readings (induction) in situations like that. Use a cheap (or free) analog meter from Harbor Freight.

The voltmeter is not lying. The display just needs to be interpreted correctly. The battery and battery case and meter are in series. A very small current is flowing through this series circuit. Assume the resistance of the meter is 10 meg ohms. The voltage drop across the meter is 0.7 volts. Calculate the current using Ohms law: 0.7 volts / 10,000,000 ohms = 70 nano amps. Assume battery voltage is 12.7 Volts. The voltage drop across the battery case is 12.7 - 0.7 = 12 volts. Now calculate the resistance of the battery case: 12 volts / 70 nano amps = 172,000,000 ohms.
Is anyone still reading this? LOL
Bottom line: connect a load to the circuit to get more meaningful voltmeter readings. An analog meter will load the circuit. Lacking that, a small lamp can be used to load the circuit.

That is not it

Must be strong wifi in the hangar !!!!

Mich48041 said:

The voltmeter is not lying. The display just needs to be interpreted correctly. The battery and battery case and meter are in series. A very small current is flowing through this series circuit. Assume the resistance of the meter is 10 meg ohms. The voltage drop across the meter is 0.7 volts. Calculate the current using Ohms law: 0.7 volts / 10,000,000 ohms = 70 nano amps. Assume battery voltage is 12.7 Volts. The voltage drop across the battery case is 12.7 - 0.7 = 12 volts. Now calculate the resistance of the battery case: 12 volts / 70 nano amps = 172,000,000 ohms.
Is anyone still reading this? LOL
Bottom line: connect a load to the circuit to get more meaningful voltmeter readings. An analog meter will load the circuit. Lacking that, a small lamp can be used to load the circuit.

Click to expand...

Joe, I love your explanations. Takes me back to my college days and DC Theory 101!