In Meanwell led drivers like the ones from the XLM, HLG or LCM families, how does the internal circuitry for the 3-in-1 dimming feature (0-10V, 10V PWM, 0-100kOhm) look like? I use a LCM-25 in my specific case.
There already is a question on that topic here, but the circuit the OP provides there appears to belong to 2-in-1 dimming (0-10V, 10V PWM):
From my understanding, dimming this circuit with a variable resistance between DIM+ and DIM- would in deed work, but there is this sentence in the datasheet: "dimming source current from powersupply: 100µA (typ.)".
This sounds like they use a constant current source to measure resistance and I measured 100µA flowing through any resistance in the range 0-100kOhm, accordingly. In the above circuit, the current would depend on the connected resistance and so it does not seem to be the circuit Meanwell uses for 3-in-1 dimming.
The reason why I'd like to understand how the circuit works:
I'm going to control the dimming input with an external device. For my tests I use a bench power supply but finally it will be a DAC with amplifier.
As I don't have a negative power supply for the opamps of the final circuit I would like to add some extra offset to the driver's off-voltage (normally, the driver is on when the dimming voltage is above 0.2V, but I'd like it to switch on at a higher voltage). I expected to achieve this by simply adding a diode between DIM- and the external device's ground. For my surprise, now the driver does not switch off at all. Whatever
Vext the external device provides in the below circuit, I measure
Vdim = 27V at the driver's dim input. Instead I expected
Vdim to be 0V for
In my test, "external device" is a
Do you have an idea how the internal circuit look like and how I could apply the desired offset?
As an additional info: the DIM input has an input resistance of 300kOhm.