1
\$\begingroup\$

The context of the problem is that I am trying to switch a ~200 VDC Meanwell Constant Current Supply @ 1.4A, to selectively power on a number of COB LEDs connected in series. My initial approach was to switch the power using a 30 VDC 10A optocoupler relay module.

I understand that this will certainly cause arcing during switching if the line is hot. However, I can cut off the mains input to the power supply right before the switch, and prevent any arcing since the relay would switch on a cold line. I'm sure that would work for mechanical switches, but is my understanding correct when we are talking about optoisolators?

I know this would be a risky way of implementing this, and I'd certainly go with a properly rated relay module for 200+ VDC, but being a newbie in this field I'm not well equipped to research for the most cost effective option here. If there's anyone here who can provide a recommendation for a SPDT relay at these ratings I'd really appreciate it. Thanks in advance!

\$\endgroup\$
3
  • \$\begingroup\$ Hi @anon.chou, Welcome to Electrical Engineering Stack Exchange! Why not switch only the AC line? \$\endgroup\$
    – vu2nan
    Sep 12 '20 at 10:53
  • \$\begingroup\$ Hey @vu2nan, good to be here. Thing is, switching the AC line would toggle the entire LED array at once, whereas I'm looking for selective toggle. To provide a little more detail, I'm stringing a series of 5 COB LEDs with Vf @ 36V. At a given time, anywhere between 1 and 5 of the LEDs are going to be active. As per the relay input, the series is going to either run through, or bypass a particular COB. \$\endgroup\$
    – anon.chou
    Sep 12 '20 at 11:21
  • \$\begingroup\$ Got it! Thanks. \$\endgroup\$
    – vu2nan
    Sep 12 '20 at 16:50
0
\$\begingroup\$

Switching off the AC input won't switch off the output until the power supply capacitors have discharged. This may take 0.2 to 2 s or so depending on the load so it is unlikely to provide you with a solution to your problem.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Equivalent circuit.

Arcing is particularly destructive with un-snubbed inductive loads as they generate a high voltage across the opening contacts in an effort to keep the current going through the inductor. In this case you don't have any inductance and you're not changing the current - you are only diverting it from the LED to the switch.

Your switch, whether mechanical or electrical, only has to be rated for the LED current (DC) and the maximum voltage to be switched.

\$\endgroup\$
0
\$\begingroup\$

With a relay de-energised, the voltage across its open contacts would be 36 V DC. With a relay energised, the current through its closed contact would be 1.4 A.

enter image description here

Since no inductive switching is involved, general purpose 5 V DC relays with 2 A contacts would suffice.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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