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I am designing a circuit to control a set of 0-10v dimmable lights. The lights have two lines, Dim+ and Dim-. Dim+ will be at 10 volts, and the brightness is controlled based on how much is drained to Dim-. Right now I am controlling the lights with an n-channel mosfet connected to a microcontroller. The microcontroller sends a pwm signal, rapidly turning on and off the mosfet, dimming the lights. Right now, the microcontroller is being powered off of a 5v usb, but I would like to change it so that the microcontroller is also being powered by Dim+, but still have it so that all of Dim+ can be drained to Dim-. Is this possible?

My initial assumption was that it would not be possible, as if all 10v is drained to Dim-, then there is 0v of potential to power the microcontroller. Is it possible to power the microcontroller, so that there is always 3v between 1 and 2, but also 0v between 1 and 3? I know it isn't possible with the configuration I have drawn, but is it with any other config?

enter image description here

Edit: Here is how 0-10v dimming works, and here is the schematic I am currently using. Dim+ can connect to pin 1, 2, or 3, and Dim- connects to pin 6 enter image description here

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  • \$\begingroup\$ There is a schematic editor button when you're writing your question. Please check it out. \$\endgroup\$
    – JYelton
    Commented Dec 11, 2021 at 0:12
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    \$\begingroup\$ how much is drained to Dim-. ... what do you mean by drain? ... what are you draining? \$\endgroup\$
    – jsotola
    Commented Dec 11, 2021 at 0:31
  • \$\begingroup\$ @jsotola its basically a ground, but how bright the lighting is is controlled by how much is drained to it, so if I connect dim+ to dim-, lights are at 0% brightness, if I connect a resistor, causing a 5v drop, then lights would be at 50% brightness, ect \$\endgroup\$ Commented Dec 11, 2021 at 0:35
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    \$\begingroup\$ what are you draining? ... it may be a language translation problem, but it appears that there may be a misconception about electric circuits at play \$\endgroup\$
    – jsotola
    Commented Dec 11, 2021 at 0:42
  • \$\begingroup\$ Can you share a diagram of the circuit you have now that works? That would help us understand which signals are inputs and outputs of the microcontroller and what kind of control the light system expects. \$\endgroup\$
    – The Photon
    Commented Dec 11, 2021 at 0:48

1 Answer 1

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It's going to be tricky, but maybe not impossible. For your existing circuit to work, there must be a series resistor in the output of the light device that delivers the +10 V to your DIM+ input. In order to minimize power consumption, it is probably quite high --- on the order of 1 - 10 kohms.

That means that the amount of power that you could draw from it is pretty low. The maximum power you get will be when DIM+ is pulled down to 5 V. If the pull-up resistor is 1 kohms you'd be getting 25 mW. If the pull-up is 10 kohms you'd only be getting 2.5 mW. But the light circuit would likely interpret this as a command to dim the lights by 50%, which is probably unacceptable.

If you can accept, say, 90% brightness from the lights, then you might get between 0.1 and 1 mA (0.9 to 9 mW). This is probably enough to run a microcontroller if you are careful about it (and don't choose a power pig of a microcontroller).

Something like this has a good chance to work:

schematic

simulate this circuit – Schematic created using CircuitLab

For best efficiency, the REG device might be a buck converter sub-circuit rather than a single linear regulator chip. And again you'll want to make sure your microcontroller is running at under a milliwatt for the best chances of it to work.

Changes in microcontroller load (switching some other GPIOs, for example) might cause the lights to noticeably dim or brighten. If this isn't acceptable, consider using a shunt voltage regulator instead of the series regulator shown at U1. But this will reduce the maximum brightness you can achieve.

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