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I designed a relay network board. The idea is to select one differential measurement points from a high voltage device. In order to do that, I used these solid state relays.

You will see the board below:

PCBA

The board has two differential probes actually. I got the schematics from this address. And I've seen that this is pretty basic in differential probes but I will give the block below for your reviews. The block is working fine doing its job.

diff probe

What I did in this board is to connect the probes inputs to 8 relays for positive measurement point selection and 8 relays for negative measurement point selection.

diff probe with relays

In the datasheets of the relays, it states the input to output cap is 1 pF.

So, when one channel is being measured, and there is a sine wave on as input, the other with square wave is not measured and its relay is open. This square wave interferes the measurement via this cap and distorts the output. I made the simulation:

simualtion

And measured the same case and verified:

real measurement

My question is, what can I do about it? Is there a solution to prevent this capacitive effect on the relays?

Before you suggest to say use different amplifiers for each, I have to say this will become a product and needs to be cost effective and reliable. Before you say 'Why not convert it to single end?' I have to use differential output because the receiving end is a fixed differential input digitizer.

winny
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Alper91
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  • What is the frequency of the signal and the design bandwidth of this circuit? – bobflux Oct 07 '21 at 16:11
  • What solid state relays? There is no such thing as a solid state relay - there are only solid-state devices which try to act like relays, but aren't relays. – user253751 Oct 07 '21 at 16:16
  • and the capacitor is just there to simulate the real SSR's worst-case input-to-output capacitance, yes? it's not something you actually put in your circuit? – user253751 Oct 07 '21 at 16:17
  • @bobflux 10mhz bandwidth but the signal here is 1Mhz – Alper91 Oct 07 '21 at 17:19
  • @user253751 yes it is to simulate the caps effect. not in design. – Alper91 Oct 07 '21 at 17:20

1 Answers1

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In the datasheets of the relays, it states the input to Output cap is 1 pf.

That's a minor consideration.

My question is, what can I do about it? Is there a solution to prevent this capacitive effect on the relays?

It's worse than you think: -

enter image description here

Not only do you have \$\color{red}{\text{6 pF}}\$ across the "contact" when open but, you might have a current path of a few μA too. But, it's even worse than that if the blocking voltage of the solid-state relay is low: -

enter image description here

With a disabled relay and 1 volt across the "contacts", the capacitance is typically over 30 pF.

My advice is the use reed relays for this type of job (as I have done so in the past to overcome exactly this problem).

I have to say this will become a product and needs to be cost effective and reliable

Then you have a serious product design problem because reed relays probably cost as much as the chips you have used (if not more). Good luck.

Andy aka
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  • What about these signal relays: https://www.digikey.com/en/products/filter/signal-relays-up-to-2-amps/189?s=N4IgTCBcDaIE4FMA2BDAniAugXyA – Alper91 Oct 07 '21 at 17:53
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    That link is 273 pages of relays with 25 relays per page so, I'm going to take a guess that at least half a dozen would be good for what you need. Look for sealed contacts if the operating current is very low (as I expect it is) hence why I suggested reed relays. – Andy aka Oct 07 '21 at 18:37
  • @Andyaka - The input to output capacitance is 1pF as the OP states. The 6pF is the output capacitance when the LED is energized, not across the contact. Presumably it is mainly to the control circuitry. – Kevin White Oct 08 '21 at 00:41
  • @KevinWhite I don't think so. Sure they state an If of 2 mA but they also use an If of 2 mA for the off-state leakage current. Hence, I believe that the data sheet might be incorrect here because it makes no sense to specify an off-state leakage current when the LED is activated. Neither does it make any sense to specify some value of capacitance to some undisclosed control circuitry. – Andy aka Oct 08 '21 at 07:51
  • @Alper91 you can research some highend opto-mosfet relays which are offered by e.g..Toshiba or Panasonic. there are special RF types with ultra-low off capacitance. But these cost and in the end it can be more economical to indeed use Reeds+driver despite all the trouble that comes with actual relays – tobalt Oct 08 '21 at 17:28
  • @Andyaka - I agree. They do seem to have some confusion in their datasheet. I think by input to output capacitance they mean from the LED to the MOSFET switches, rather than between the two terminals of the relay. – Kevin White Oct 08 '21 at 17:52
  • Yes, that is the 1 pF mentioned in the question. – Andy aka Oct 08 '21 at 18:01
  • @KevinWhite I think it is standard nomenclature across all manufacturers to call the LED 'input' and the two MOSFET drains 'output' in such MOSFET optoswitches. – tobalt Oct 10 '21 at 06:10