I am designing a circuit to switch off the charge current to a battery when the voltage rises above a certain level. Maximum current is 2.3 Amps from a solar panel.

I am using an opto-coupled Omron device (GV3M-41AY, 40V, 2A) as the switch, but it is limited to 2A, so I intend using two in parallel. Can I drive both devices through the one gate-drive resistor?

This thread, Parallel MOSFETs, suggests that separate gate-drive resistors should be used, but does that apply for opto-coupled MOSFETs?

  • 2
    \$\begingroup\$ That relay consists of bidirectional FET switch, activated by LED, thats why it is so low-current. If you do not need these features, may be it is better to replace it by simple MOSFET. It will have better efficiency because of lower on resistance and virtually no control current. \$\endgroup\$
    – Vladimir
    May 15 at 16:20

3 Answers 3


Gate resistors do not apply at all. It does not apply because you do not have direct access to the MOSFET gate in an opto (or an SSR). You're driving an LED, not the MOSFET gate.

That said, LEDs still need current-limiting resistors and you should not not parallel LEDs which is what will happen if you use one current-limiting resistor with two LEDs in parallel. Since their forward voltages won't match identically and one will short out the other such that only one will light up (or not light up evenly). So you need one current limiting resistor per LED unless you series them.

I wonder if there are potentially issues with the SSRs not turning on in sync enough.

  • \$\begingroup\$ Thanks for the feedback. I'll include separate resistors to accommodate the different characteristics of the LEDs. I think I can overcome the possible SSR sync problem by connecting the battery first. The circuit is powered from the battery, and on power-up, the SSRs are automatically turned on, but there will be no current flow. Then I'll connect the solar panel. \$\endgroup\$
    – Guy Burns
    May 15 at 4:08
  • \$\begingroup\$ So, by using one extra resistor (thanks to you) and making sure I connect the battery first, I should have a reliable circuit. \$\endgroup\$
    – Guy Burns
    May 15 at 4:15
  • \$\begingroup\$ @GuyBurns Just wire it up as if you only had one opto. Then exactly repeat the same process with the second one. \$\endgroup\$
    – DKNguyen
    May 15 at 4:22

This works fine.

You put the LEDs in series and the output switches in parallel.

That also means that you control them as if it were a single switch eventually.

  • 1
    \$\begingroup\$ This works most of the time, but now always. Transition times may be profoundly different, on-state resistance can be different as well. Switches of any kind that are good for paralleling are special. \$\endgroup\$
    – fraxinus
    May 15 at 11:24
  • \$\begingroup\$ But in this case, for this application, transition time and on resistance will not matter. Transition time delta will give better on and off transitions, with small pulses well within limits and better response to inductive wires. Load sharing will be mismatched, but well within limits. \$\endgroup\$
    – david
    May 15 at 13:27
  • \$\begingroup\$ @fraxinus almost all MOSFET have a positive TCR, so lend themselves to being parallelled. The same applies to the linked switch. You are right about the switching times, although that would matter only for really fast switching scenarios and not for a relay type application. \$\endgroup\$
    – tobalt
    May 15 at 16:22

Can I drive both devices through the one gate-drive resistor?

If you mean opto driven gates, YES, by putting the LEDS in series and reducing the R value for the remaining R voltage drop and desired current in the datasheet of 10mA @ Vf=1.27V each.


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.