Looking at driving two LED strings with a DC/DC constant current LED driver module. The constant current side can't be referenced to ground of the control circuit, so I need to have an isolated switching to choose which of the two strings I would like to drive.

I'm "only" driving 300mA current, so I was thinking of using two optocouplers to select the string I want. I am having trouble finding a optocoupler that can take 300mA continuous current. How can I change my circuit so the optocouplers can drive 300mA? Can I add another transistor?

Schematic of what I had in mind:


(note - the driver I'm using cannot handle no-load situations, so I can't just use a SPDT relay, as I will have no-load conditions during switching causing voltage spikes and god knows what other issues)

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    \$\begingroup\$ Add a schematic. Like this, it's not clear how your optocoupler would relate in any way to the LED strings and your supply. \$\endgroup\$ – Marcus Müller Apr 11 '17 at 10:27
  • \$\begingroup\$ I will do so in a few minutes. I thought the question would be enought to make the topology clear. \$\endgroup\$ – Joren Vaes Apr 11 '17 at 10:30
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    \$\begingroup\$ I'd use the optocouplers to drive a $0.10 N-MOSFET or NPN-BJT switch on the ground-end of the LED strings. $0.20 (max) price tag, and problem solved ;) \$\endgroup\$ – Robherc KV5ROB Apr 11 '17 at 10:32
  • \$\begingroup\$ Schematic has been added. @RobhercKV5ROB: Do you mean pull down the gate of a NMOS with a resistor, and drive a current through the resistor with the optocoupler? \$\endgroup\$ – Joren Vaes Apr 11 '17 at 11:03
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    \$\begingroup\$ @JorenVaes Shopping questions are off topic, I changed it to a design question so the other mods won't close it. \$\endgroup\$ – Voltage Spike Apr 11 '17 at 15:14

The opto-coupler outputs don't need to switch the 300 mA current directly. Use a transistor or two to amplify the few mA thru the opto to the 300 mA thru the LED string.

In this case, I'd probably use the opto to switch the gate of a FET. Something like this:

You didn't say what the LED supply voltage is, so it's up to you to make sure that the opto output can withstand it when off. You also have to adjust R2 allow enough current thru it to raise the FET gate to 12 V when the opto is on, but not waste too much extra current. I'd aim for the zener drawing around ½ mA when the FET is on.

  • \$\begingroup\$ Thanks! I'm looking at building something like this. I'm going to play around on the breadboard with the few optocouplers I have. The thing that is throwing me off is the fact that my led circuitry is all driven by the constant-current source, and I am unsure how it will influence my design. I guess that's what LTSpice and breadboards are for right? \$\endgroup\$ – Joren Vaes Apr 11 '17 at 11:20
  • \$\begingroup\$ I breadboarded this circuit and got it to work quite easily. Now I can do some propper math and select the components for the final design. Thanks a lot! \$\endgroup\$ – Joren Vaes Apr 11 '17 at 11:34

Optocouplers are not built for large currents (e.g., 4N32: 150 mA, 6N138: 60 mA). The purpose of Darlington optocoupler is not necessarily to allow larger output currents but smaller input currents.

To drive larger loads, use an optocoupler to provide the base current to a larger transistor:

optocoupler with transistor amplifier

(The base current is limited by the optocoupler's current transfer ratio.)

  • \$\begingroup\$ The thing I'm unsure about here is that the driver is constant-current. Wont using the topology you show cause all current to flow through the optoisolator and then into Q1? \$\endgroup\$ – Joren Vaes Apr 11 '17 at 11:15
  • \$\begingroup\$ You wouldn't connect pin 4 to the driver. Updated the schematic. \$\endgroup\$ – CL. Apr 11 '17 at 11:15
  • \$\begingroup\$ But, as stated in my original question, the driver has to be isolated from the control side - hence, all my components on the LED side have to be referenced to the driver only. \$\endgroup\$ – Joren Vaes Apr 11 '17 at 11:17
  • \$\begingroup\$ VCC = VBAT. In any case, the optocoupler does not allow a large current if the LED current is small enough (it has a maximum CTR). \$\endgroup\$ – CL. Apr 11 '17 at 11:19
  • \$\begingroup\$ +1 for pointing out the following: "The purpose of Darlington optocoupler is not necessarily to allow larger output currents but smaller input currents." \$\endgroup\$ – Enric Blanco Apr 11 '17 at 15:21

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