1
\$\begingroup\$

I'm building a circuit that includes a resistive load that I have to switch, which is going to be supplied with ~200 VDC and 0.1 Amps.

I've looked at different mechanical relays, and typically they are specified as "30A@240VAC or 30A@30VDC.".

Ideally, I would like to use one of these. Would they be safe to operate at 200 VDC, since I am using a drastically reduced current?

There is also a range of solid-state relays, but they are only specified at (for example) 250 VAC and 2 A, making no mention of maximal allowable VDC.

As an alternative to the relays (which I guess are generally targeted at AC applications), I suppose a power MOSFET might be usable?

Thank you for any assistance....

\$\endgroup\$
6
  • \$\begingroup\$ How quickly do you need to activate or deactivate the load? \$\endgroup\$
    – Andy aka
    Jul 3, 2015 at 10:14
  • 1
    \$\begingroup\$ Look for relays with an explicit 200V DC (or higher) rating. The difference between switching DC and AC is that if you get an arc going, it'll extinguish (and disconnect the load) at the next zero crossing. With DC, that'll take a while... \$\endgroup\$
    – user16324
    Jul 3, 2015 at 11:20
  • 1
    \$\begingroup\$ There's a good graph on this datasheet showing how a typical relay rated 16A at 25V DC can only break 0.4A at 200V DC. farnell.com/datasheets/1717883.pdf \$\endgroup\$
    – user16324
    Jul 3, 2015 at 11:28
  • \$\begingroup\$ Given @BrianDrummond's link there is a good chance you can use a well-designed European-style relay for 100mA. Be sure to put a flyback diode across the load if there could be any significant inductance. \$\endgroup\$ Jul 3, 2015 at 12:35
  • \$\begingroup\$ 200VDC at 100mA into a resistive load is not too aggressive on full size relay contacts but as mentioned if an arc forms it will possibly not extinguish. There are relays made for this sort of application and they have a wider contact gap when open (sometimes more exotic contact materials). A resistive load such as a wire wound resistor will exhibit some inductance and the flyback diode advice is worth considering. \$\endgroup\$
    – KalleMP
    Jul 3, 2015 at 18:38

1 Answer 1

2
\$\begingroup\$

I would use a MOSFET and the simplest arrangement is an N channel load switcher: -

enter image description here

Ignoring the requirement for switching it on and off at high speed (i.e. kHz upwards), when operating at 200V, certain precautions may be advisable. The first relates to the common earth/0V/ground point. You may want isolation between control signal and the load ground so you may want to go to something like this: -

enter image description here

You'll need to generate a small 12V isolated supply (note B2) and this can be done with a multitude of isolating dc/dc converters such as from Traco and others.

You also need to pick a MOSFET that is rated probably in excess of 300V and there are plenty to choose from here. Given that there may be some back-emf due to load cabling I would still use a reverse protection diode across the load to snub-out any high voltage spikes from damaging the MOSFET.

My 2nd advise is be very careful with the 200V when building something and testing it - make sure it is current limited (fused to prevent a fire) and, if possible find a dc residual current detector to switch off the supply should you get connected to the 200V like this one

\$\endgroup\$
4
  • \$\begingroup\$ Why is it advisable to isolate two circuits from each other when there are high voltages present (I've seen it numeroues times)? \$\endgroup\$
    – Golaž
    Jul 3, 2015 at 13:02
  • 1
    \$\begingroup\$ The basic idea is that a single failure should never cause a hazard such as shock or fire. Isolation is one tool that can help achieve that as part of a safety-focused design. \$\endgroup\$
    – user16324
    Jul 3, 2015 at 13:16
  • 1
    \$\begingroup\$ @Golaž The isolation will reduce the chance that a fault will roast the control circuitry, it may also eliminate the chance of a fault current if the ground potentials want to float with respect to each other (within limits) and prevent a failure. \$\endgroup\$
    – KalleMP
    Jul 3, 2015 at 18:35
  • 1
    \$\begingroup\$ Instead of opto-coupler plus switching circuit plus MOSFET, why not just use an SSR? For example, I see Vishay has some 250-V 150 mA SSR's that sell for less than $2 in onesy twosy. \$\endgroup\$
    – The Photon
    Jul 3, 2015 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.