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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):

2-in-1 dimming circuit

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.)".

3-in-1

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 Vext<0.6V. In my test, "external device" is a

schematic

simulate this circuit – Schematic created using CircuitLab

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.

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Thanks to @Rohat for the link to my previous answer. That in turn is a summary of my Adjustable constant-current PSU input articl.

enter image description here

Figure 1. My assessment of the Mean Well dimmable PSU input circuit.

This sounds like they use a constant current source to measure resistance ...

Yes, but I'd word it as, "They use constant current to generate a DC voltage which varies with the applied DC resistance." That way we're all clear that dimming is controlled by DC voltage.

... and I measured 100 µA flowing through any resistance in the range 0-100 kΩ accordingly.

That confirms what I reckoned.

In the above circuit, the current would depend on the connected resistance and so it does not seem to be the circuit Mean Well uses for 3-in-1 dimming.

Your circuit shows a 200 kΩ pull-up. Mine shows a constant current source. In practice it may be a bit of a hack to provide adequately linear response when controlled with a linear pot. That is, it might not be a true constant 100 μA current source.

I'm going to control the dimming input with ... a DAC with amplifier.

This would, in most cases, be overkill as the dimmer can accept PWM directly and most microcontrollers have built-in PWM controls.

As I don't have a negative power supply for the op-amps of the final circuit I would like to add some extra offset to the driver's off-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.

enter image description here

Figure 2. The problem. The PSU's constant current has no return path to the PSU so your DAC acts like an open circuit and (in another clever part of the design) this creates the default mode of 100% current.

Whatever Vext the external device provides in the below circuit, I measure Vdim = 27 V at the driver's dim input.

This confirms my suspicion that they're not using a true constant current source but a resistor fed from a relatively high voltage. Over the range 0 to 10 V this will behave with adequate linearity.

Do you have an idea how the internal circuit look like and how I could apply the desired offset?

Use a rail-to-rail op-amp instead. Better again use PWM. Or ...

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 3. A diode fix.

How it works:

  • With VEXT < 0.5 V R1 pulls the input voltage down to 100μ × 1k = 100 mV. This should turn the lights off as it's below the 0.2 V turn-on voltage.
  • When your VEXT goes above 0.5 to 0.7 V approximately it will start to affect the voltage across R1. For example, at VEXT = 2 V you should see 1.3 V or so across R1.
  • Note that at VEXT = 10 V you will have about 9.3 mA through D1 and R1. You need to ensure that your DAC can supply that much current.
  • To get 10 V out your DAC will need to drive to 10.7 V.
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  • \$\begingroup\$ Thanks for your great answer, I played around with this in ltspice and I believe I fully understand the issue. The output of my device is not dedicated to dimming mean well drivers, but to provide a real analog voltage, that's why I can't simply use PWM. From my understanding, the lm2904 actually is a rail-to-rail opamp, but due to input offset and the following amplification there remains only little margin to the drivers 200mV threshold voltage. \$\endgroup\$ – Sim Son Sep 13 at 10:53
  • \$\begingroup\$ See the update then. \$\endgroup\$ – Transistor Sep 13 at 11:05
  • \$\begingroup\$ Thanks to your explanation I found the exact same solution. You answer was very helpful! \$\endgroup\$ – Sim Son Sep 13 at 12:15
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@Transistor had made a great explanation about Meanwell's clever 3-in-1 dimming technique.

Here's an excerpt from one of my old designs -- inspired by Meanwell, but not the copy of it:

enter image description here

The circuit may be different, but the functioning is the same as Meanwell's.

C305, R314, and R315 form a 1st order LPF /w voltage divider to decrease the 0-10V output down to 0-3.3V for MCU (There's a 3V6 Zener at MCU side, which is not shown here). Removing R315 only gives a 0-10VDC signal.

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  • \$\begingroup\$ Thanks for pointing me to Transistor's explanation! \$\endgroup\$ – Sim Son Sep 13 at 10:54

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