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I own a really bright led lamp that I want to dim. It is powered by a constant current power supply that can supply 37mA between 11V to 17V. If my lamp is connected the voltage settles at 13.58V.

enter image description here Please note, that there are more than four LEDs in unknown wiring in this lamp. I only drew four of them as illustration.

With a bench powersupply I determined, my desired brightness is reached at 1mA where the voltage drops to approximately 12.7V. To limit the current to 1mA while still using the constant current source I've added a parallel resistor like this:

enter image description here

For some reason this wont work. The voltage drops to 2.7V, the current on the branch with the resistor is 7.28mA, and the current on the branch with the led lamp is immeasurably small.

Is there something I'm missing? Or is there another possibility to dim the LEDs besides PWM? I thought abouth changing the sense resistor of the power supply but I don't really want to mess with the power supply.

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    \$\begingroup\$ Your current source looks backwards. and you'll have to know something about the configuration of the LED's (series or parallel) to make any determination of the current. Ideally you'll need a V-I curve \$\endgroup\$ – Voltage Spike Jun 18 at 20:12
  • \$\begingroup\$ The current source is only switched around in the scematic. Do I really need more information about the LEDs? I allready know that they are as bright as I want them at 12.7V 1mA. \$\endgroup\$ – xeetsh Jun 18 at 20:26
  • \$\begingroup\$ You can't use 13.58V for a calculation, the CC source will adjust it's voltage until it reaches 37mA. At 2.7V the total resistance of the loads is 73Ω, which means the LED is a lot lower impedance than you think. Another thing you could do is keep swapping resistors in parallel to find the I-V curve of the LED's \$\endgroup\$ – Voltage Spike Jun 18 at 20:33
  • \$\begingroup\$ Note that when you use the CircuitLab button on the editor toolbar you don't need an account and editable schematics are saved in your post. (This would enable any of us to invert your current source for you.) It also allows us to copy and modify the schematic into our answers. \$\endgroup\$ – Transistor Jun 18 at 21:09
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    \$\begingroup\$ An experiment: try throwing some diodes in series with your resistor. \$\endgroup\$ – Aaron Jun 18 at 21:20
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enter image description here

Figure 1. LED I-V curves.

According to my Figure 1 a typical white LED could be expected to drop about 3.5 V at 37 mA. Four in series would drop 14 V which is in pretty close agreement with your 13.6 V measurement.

Accurate figures for low currents aren't often published but if my graph is any way accurate you should see a voltage drop of about 1.5 V at 1 mA or 6 V across the four in series. You're measuring 12.7 V so your LEDs don't match the graph. They appear to have a much more vertical I/V curve.

Your calculation looks pretty close as far as I can see so I don't know why it doesn't work.

A more efficient alternate solution would be to connect a series resistor to limit the current. The CC PSU will then rise to 17 V and you'll need to just find the series resistor value as you would with any constant voltage supply.

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  • \$\begingroup\$ Sadly, it is impossible to add a series resistor to limit the current, because the power supply want's to stay at 37mA at any cost, even if it means, that the voltage needs to go up to 30 Volts and only pulses because it isn't capable of handling this current at this voltage \$\endgroup\$ – xeetsh Jun 19 at 22:00
  • \$\begingroup\$ I guess the only option is to look at the sense resistor or use a completely different (maybe not current limited) power supply \$\endgroup\$ – xeetsh Jun 19 at 22:00
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PWM may cause current measurement errors and long wires can introduce inductive voltage/current errors.

I suggest you use a TO-220 transistor so it doesn’t get hot, and bias the base current with a Voltage control Rseries or a pot and series R to limit the base drive, in order to dim the LEDs . A voltage divider to 0.4V will turn off the transistor then bias to 1V with 1mA should shunt the collector and turn off the LEDs by choosing the right R ratios and values from Vdc.

This is using it as an active shunt load like your resistor except with short path control between source, control and load in a small loop area to minimize wire inductance. It is the rise time that can cause issues not the PWM rate which might be >>1000x higher bandwidth if running at 2kHz for e.g. If we had some idea of the circuit, I could be more exact.

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This isn't really exactly what I asked for but I solved this by getting rid of the constant current power supply and used an of the shelve power supply with fixed voltage and a current limiting resistor which made it really easy to met my requirements.

I've chosen a 15V Power supply and used a resistor in series with the LEDs:

R = U / I = (15 - 13.58) / 1mA = 1420 Ohm

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