Small current through relay when open - Page 1
Forum section: electronics / beginners

author: Guille
author_rank: Regular Contributor
timestamp: September 25, 2023, 10:45:21 am
content: I'm using a certain relay board:https://docs.rs-online.com/db6c/A700000009611047.pdfto disconnect a pulsed current source from a load like so.The problem is I found out with the osci that I'm getting some current at the load after disconnecting, the following 2 screenshots show the voltage at the load when connected and disconnected to the source:Source connected to load: 111V peak to peakSource disconnected from Load: 1.94V peak to peakAll cabling was done with shielded BNCs. Disconnecting the cable manualy removes the problem. I was thinking this could be capacitve coupling at the relay but a senior engineer claimed it's impossible with the level of isolation common relays provide. How would you solve this? The current is significant for this use case.

author: wraper
author_rank: Unknown
timestamp: September 25, 2023, 10:51:00 am
content: Quote from: Guille on September 25, 2023, 10:45:21 amSeniority does not necessarily mean knowledge or skill. Also as I understand it's not a sine signal, so it contains components with way much higher frequency than 5kHz.Also you should not use relays like these for very low currents. You need signal relays with gold plated contacts for that.

author: Guille
author_rank: Regular Contributor
timestamp: September 25, 2023, 11:34:18 am
content: Yep the signal is a 5KHz square wave, 1 to 0.1mA amplitude limited to 48V. Do signal relays guarantee better isolation of higher frequencies at lower currents? If so could you point me to some of them?Thank you for your feedback

author: Chalcogenide
author_rank: Regular Contributor
timestamp: September 25, 2023, 01:19:27 pm
content: If it really is a current source, why not short it to ground via the other relay pin when not connected to the load? This should get rid of most of the capacitive coupling across the relay.

author: Guille
author_rank: Regular Contributor
timestamp: September 25, 2023, 01:22:33 pm
content: That's good advice, however this is only a test for a further implementation where the current source will be switched to different loads, shorting it to ground would bypass the connected load

author: mikerj
author_rank: Super Contributor
timestamp: September 25, 2023, 01:36:02 pm
content: Quote from: Guille on September 25, 2023, 11:34:18 amPower relays require a minimum "wetting" current through contacts to ensure a low resistance connection. Without this the connection can become unreliable over time. Signal relays use different contact materials to enable reliable operation at very low currents.To get better isolation you could use a DPDT relay and use the other set of contacts to short the unusedloadto ground (rather than the source). This will shunt any capacitive coupling to ground.

author: Alex Nikitin
author_rank: Super Contributor
timestamp: September 25, 2023, 01:48:26 pm
content: Quote from: mikerj on September 25, 2023, 01:36:02 pmOr just connect the relay differently - pin 11 to output, pin 12 to common and pin 14 to the source.CheersAlex

author: Guille
author_rank: Regular Contributor
timestamp: September 25, 2023, 02:00:37 pm
content: Quote from: mikerj on September 25, 2023, 01:36:02 pmThat's a great idea, I tried comparing the waveforms for an open relay configuration compared to an open relay with a cable shorting both terminals at the load and the peak to peak voltage at the load dropped from 2v to 780mV. Now I wonder where this voltage might be coming from, specialy since the osci leads are effectively shorted out. If I unplug them from the load and short them I get no waveform so it can't be some pick up at the probe.Open RelayOpen Relay. Shorted Load

author: Guille
author_rank: Regular Contributor
timestamp: September 25, 2023, 03:05:00 pm
content: Hi everyone, I solved it by grounding the sample whith the same relay like soThanks everyone for the help.

author: mikerj
author_rank: Super Contributor
timestamp: September 25, 2023, 04:14:28 pm
content: Quote from: Alex Nikitin on September 25, 2023, 01:48:26 pmThat works fine, but OP did imply he would be using the relay to switch the source to different loads.

author: ejeffrey
author_rank: Super Contributor
timestamp: September 25, 2023, 04:19:38 pm
content: Quote from: Guille on September 25, 2023, 10:45:21 amThe first thing to do is to calculate the size of the parasitic capacitance (or generally conductance) that could cause this. If I did my math correctly on my phone it's about 0.5 pF. That's totally in the range of parasitic capacitance. You could get that in the relay body, but also on the PCB traces around it.The inferred conductance is something like 55 megahm which is also in the range of a poorly cleaned board with flux residue or other leakage paths. So I wouldn't count out resistive leakage either.In either case, shorting the load when it's isolated should eliminate the current. If that is allowable for the load, I would go that route. If it's not you will have to track down the leakage more carefully or use multi stage isolation.

author: Kleinstein
author_rank: Super Contributor
timestamp: September 25, 2023, 07:07:53 pm
content: The waveform comming through looks very much like capacitive coupling. There is current when the input voltage changes and very little in the times in between.

author: Guille
author_rank: Regular Contributor
timestamp: September 25, 2023, 09:51:32 pm
content: Hi, thanks for your reply, I should marked this post as solved somehow since one user recomended a similar method and it works (as I wrote in the comments). Your response caught my eye though, if you don't mind how did you perform those calculations?