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I want to dim some hundred LEDs with analog dimming (current control, not PWM). There are various LED drivers that are able to analog dim. Two examples: A6211 and AL8860. Can I control about 20 of them in parallel with only one direct analog signal, no op-amps or additional circuits?

A6211:

enter image description here

AL8860:

enter image description here

enter image description here

Some additional information:

  1. It will be a copper-core PCB (only one layer). That's why the circuit should be as simple as possible.
  2. PWM is really not an option. (Request by the client.)
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    \$\begingroup\$ But you and your customer are aware, that the LED-Driver basically don't do anything else than PWM at the output? \$\endgroup\$
    – kruemi
    Commented Nov 20 at 14:57
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    \$\begingroup\$ That's not true. Analog dimming is really adjusting the output current without turning the LEDs off and on. See for example here from page 3: ti.com/lit/an/snva605/… \$\endgroup\$
    – BruderTom
    Commented Nov 20 at 15:27
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    \$\begingroup\$ @kruemi Both of those circuits adjust the set point of a constant current buck converter. They do do not PWM the output. \$\endgroup\$
    – user1850479
    Commented Nov 20 at 15:32
  • \$\begingroup\$ @BruderTom loock up "Buck Converter". Sure, they don't switch on and off the LED but they are still switching converters. The output of both converters is caled "SW" for "Switch". Just because they rely on an inductor to smooth out the switching, they still produce EMI. \$\endgroup\$
    – kruemi
    Commented Nov 21 at 5:51
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    \$\begingroup\$ @BruderTom ok. That's a fair point. From your statement, that PWM is not an option for the control I falsely assumed that it was EMI reasons. On "constantly on" I'm with you. In my opinion PWMing LEDs is a crime (ecause you only need an inductor and a diode to make it right). (Yes, addressable LEDs don't have the space for that, I see that point). \$\endgroup\$
    – kruemi
    Commented Nov 21 at 7:23

1 Answer 1

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Short answer: Yes

Slightly longer answer: Your driver needs to be able to deliver 5.2V with a current of 5.2V / (25kOhm + 1kOhm) = 0.2mA per circuit you want to control (with the A2611) which should not be an issue.

For the input load of the ctrl pin of the AL8860 you can assume that it doesn't ake more than 1mA per chip.

So make sure your driver has enough driving capability to feed all the loads.

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  • \$\begingroup\$ Thank you for the quick answer! Do you think it's also no problem, when the analog signal has to travel through some meters of wire first? Will adding a capacitor at every input help with the accuracy? \$\endgroup\$
    – BruderTom
    Commented Nov 20 at 15:21
  • \$\begingroup\$ @BruderTom That chip will take a PWM input signal to adjust the output current as well as analog, so if you're concerned about noise/accuracy I would use that. \$\endgroup\$
    – user1850479
    Commented Nov 20 at 15:31
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    \$\begingroup\$ Actually no I'm wrong about that it would PWM the output. Drive the analog signal with a low enough impedance then. Can filter it if you want. \$\endgroup\$
    – user1850479
    Commented Nov 20 at 15:47
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    \$\begingroup\$ @BruderTom a capacitor might help with noise but it won't do anything for wire losses at DC, if there's no AC component it's like the capacitor isn't there. Let's say you have 20 units each drawing 0.2mA, that's 4mA total. If you use 28ga wire it's around 65Ω/1000ft so 1000ft of cable (2 conductors) would lose 2*65Ω*4mA=0.52V. Only way to reduce that loss is a shorter cable or heavier wire, going up to 16ga would give you 0.032V loss. If you only need a couple of meters of cable it's probably not going to be a problem. \$\endgroup\$
    – GodJihyo
    Commented Nov 20 at 17:09
  • \$\begingroup\$ As @GodJihyo said: a capacitor would might only help with reducing noise. What will help if you put in a buffer (for example an opamp with unity gain) to provide a high impedance input for the analog signal and a low impedance output towards the LED drivers close to the LED drivers. But at the currents involved, you're not likely to see a big drop in voltage. \$\endgroup\$
    – kruemi
    Commented Nov 21 at 5:54

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