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I've got a constant current MeanWell LED driver that has 3-in-1 dimming (100k potentiometer, 0-10V signal or 10V PWM), and would like to control the dimming with a microcontroller.

I found a few suggestions online, the simplest one is feeding a PWM signal to an optocoupler connected to directly to driver's DIM+ and DIM- (without any external power supply).

This works because apparently MeanWell dimming circuit looks like this (courtesy of https://www.eevblog.com/forum/projects/mean-wells-2-in-1-dimming/msg1835957/#msg1835957)

schematic

simulate this circuit – Schematic created using CircuitLab

However I have a couple of additional requirements

1) I need to limit the maximum brightness to about 75%. Just limiting duty cycle is not enough - I want the driver to never output more than 75% of rated current even in case of software bug/controller failure.

2) I need dimming to be 0% when microcontroller is powered off.

If I was using the single optocoupler schematics, I could use 75K resistor in series with opto output, but requirement (2) means I have to invert the output somehow with a BJT/MOSFET.

Could you please suggest a schematic and help with components choice?

PWM frequency doesn't need to be high (100-200 Hz is enough). The driver I'm using is ELG-100-C350B https://www.meanwell.com/Upload/PDF/ELG-100-C/ELG-100-C-SPEC.PDF

PS. Just realised that 75K resistor won't work with opto as it needs to effectively switch from "short" to "75K"

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  • \$\begingroup\$ Check the part number of your LED driver. Some versions of the ELG-100-C have adjustable output current. If yours is adjustable then you could turn it down. That would save you limiting the PWM to 75% - much easier. \$\endgroup\$ – JRE Oct 1 at 11:38
  • \$\begingroup\$ Thanks, unfortunately the one I got doesn't have it - its C350B not AB \$\endgroup\$ – GreyZone Oct 1 at 11:39
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    \$\begingroup\$ Add a 7.5V Zener diode between DIM+ and DIM- with a suitable resistor? \$\endgroup\$ – Finbarr Oct 1 at 11:55
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    \$\begingroup\$ Makes sense. 2 you could solve with a depletion MOSFET like BSS126. \$\endgroup\$ – winny Oct 1 at 12:23
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    \$\begingroup\$ Oh! That complicated things. You can pair an Vishay VOM12xx with said depletion MOSFET and solve it. \$\endgroup\$ – winny Oct 8 at 16:34
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2) I need dimming to be 0% when microcontroller is powered off.

This becomes a bit non-standard immediately. If the driver outputs 0 % even at say 0.7 V, you may be able to cheat around this with a BJT pulling itself up until your MCU can take over and somehow defeat that. A NC relay comes to mind too if you can accept that.

If 0 V is required on the DIM pin when your MCU is not powered, this is an expensive and obscure solution but solid-state.

schematic

simulate this circuit – Schematic created using CircuitLab

You need negative voltage to turn off the depletion MOSFET, but the Vishay VOM1271 can be used "upside down". Again, obscure and expensive solution.

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    \$\begingroup\$ Many thanks for the answer! With regard to "expensive and obscure solution" comment. The straightforward alternative I can see is to go with extra 10V supply and control it from MCU with either voltage or PWM output. I'm not sure it will be cheaper. \$\endgroup\$ – GreyZone Oct 16 at 11:31
  • \$\begingroup\$ BJT approach won't work; according to this graph the output is non-zero at about 0.3V \$\endgroup\$ – GreyZone Oct 16 at 11:50
  • \$\begingroup\$ If it fits your budget and requirements, there you go! 0.3 V will be very difficult. Perhaps some super low Vgsth N-MOSFET. Also, I forgot a pull-up in the schematic. \$\endgroup\$ – winny Oct 16 at 14:27
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    \$\begingroup\$ No, you need negative gate voltage to turn off a depletion MOSFET. \$\endgroup\$ – winny Oct 16 at 16:21
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    \$\begingroup\$ Yes, but a full MOSFET driver is overkill and you still need to generate negative voltage. Feel free to separate the grounds if you wish with the optocoupler. \$\endgroup\$ – winny Oct 17 at 9:10

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