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I'm working with an AL5810Q, which is a linear LED driver where the current can be set with only one external resistor following $$I_{LED}=\frac{750}{R_{SET}}.$$ With that, since I need a current of 10mA I calculated an external resistor of 75kΩ. (Vin = 12V)

I did a simple test following the schematics provided in the datasheet, which are the following enter image description here

So far, so good. 10mA Sharp.

However, the application I have in mind considers a very high line impedance (≈100Ω/mm) so I wanted to model that impedance by means of a resistor of 330Ω, which would leave the previous schemes like this:

enter image description here

by implementing these resistors the current is no longer constant and becomes a ratio to the voltage.

To be honest, I was not expecting this behavior, but it doesn't surprise me. The block diagram is as follows (with resistors)

enter image description here

can anyone shed some light on what might be happening and what I can do to fix this.

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    \$\begingroup\$ What's V_in and R? It can only be compliant up to a certain voltage. Edit the details into your question. \$\endgroup\$
    – Transistor
    May 17, 2021 at 13:21

2 Answers 2

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The minimum voltage drop across the device is 2.5 V. Add in a couple of green LEDs and you've got a minimum voltage drop of 2.5 + 2 × 2.2 = 6.9 V.

If your supply is, say, 12 V then the maximum cable voltage drop is 12 - 6.9 = 5.1 V giving a maximum resistance of V/I = 5.1/10m = 510 Ω. At 100 Ω/mm that would give you a max of 5 mm of conductor before you're in trouble.

You can recalculate for your supply voltage.

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  • \$\begingroup\$ makes sense to me. However, I have done two independent tests, one with a red led and the other with a white one (both using the same resistor I mentioned above 330 ohms.) which have a voltage drop of 1.8 and 3 respectively, so there should be enough space to not have this behavior? \$\endgroup\$ May 17, 2021 at 14:42
  • \$\begingroup\$ Also, what I see from playing with the source is the current makes a curve, peaking at the desired current (10mA) when Vin = 7 to 9 V. Before and after it decreases. \$\endgroup\$ May 17, 2021 at 14:51
  • \$\begingroup\$ You haven't supplied the detail requested under your question and the information in the comments above is a bit vague. Tabulate it properly in your question along with details of the series resistance. \$\endgroup\$
    – Transistor
    May 17, 2021 at 15:13
  • \$\begingroup\$ I did but in the question itself. Anyways. R= 330Ohms and Vin = 12V. Now I'm making some test with a white LED (OSRAM LW Q38E-Q1OO-3K6L-1) and with an R-value of 200Ohm. Vin<6.8V, I<10mA. when 6.8<Vin<8.9V, I=10mA and when Vin>8.9, I<10mA. \$\endgroup\$ May 17, 2021 at 15:40
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Any constant current source must have a minimum voltage drop to regulate the feedback and internal drop. Adding excessive resistance will defeat this requirement.

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