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I have a MeanWell IDLV-65A-12 LED driver that I would like to control brightness via the 0-10 VDC dimming circuit. Spec sheet here.

I have a basic understanding of electrical circuits, and successfully dimmed similar drivers using an external 10 VDC power source. But this time, I need to do so without an external source. Note this "A-Type" model has an unregulated auxiliary DC output (last page of spec sheet) that may be useful for this.

I tried connecting a 100K logarithmic potentiometer as a series resistor (only two terminals) between DIM+ and DIM-. That works fairly well for dimming, going down to minimal brightness. The problem is this does not achieve 100% brightness; 100K ohm causes dimming. I tested a variety of fixed resistors between DIM+ and DIM- in series, and found even 10 M ohm still causes dimming.

First Question: Am I correct to conclude that simply introducing resistance to the dimming circuit is not how it should be used? Rather that a voltage must be introduced between DIM+ and DIM-?

Second Question: Is it safe to use the auxiliary VDC output to drive the dimming circuit? Specifically I would connect the potentiometer as a voltage divider; potentiometer outer pins to AUX+ and AUX- (0.12 mA current in AUX circuit), potentiometer center pin to DIM+, AUX- to DIM-. I should also be able to swap the CW/CCW control by alternatively center pin to DIM-, and AUX+ to DIM+.

This seems correct to me, but I'm asking for confirmation and advice before I ruin an expensive driver. Thanks for the help :).

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  • \$\begingroup\$ The spec sheet says to use a voltage source. You are using a resistor. Why do you think this will work? \$\endgroup\$ – WhatRoughBeast Mar 25 at 12:07
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You need a real voltage source for dimming, so just a potentiometer is not enough. A floating input might see an unpredictable voltage and that's why it seemed to work. But as you have the A-type it's fairly easy: as you suggested you can use the auxiliary output as voltafe source for your potentiometer. To be save you can add a high-side series resistor to divide the Vaux (~12V) to the required 10V, but I doubt that leaving this step would actually harm the device.

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  • \$\begingroup\$ I did exactly as you suggested with the 100 K log pot, and it worked perfectly. Decent full-range dimming control, and full brightness. \$\endgroup\$ – Tyson Hilmer Apr 2 at 12:06

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