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In the thread "Using a variable resistor to dim an LED", Phil Frost provides almost the answer I need, and even provides a great diagram, but he's working from a 5V supply.

My question concerns what the resistor values would be for a similar circuit running a red LED from a 3V power supply. I'm pretty sure the fixed resistance would be less, and I know how to calculate the 'drop' resistance for a fixed-brightness LED, but I'm not an EE, and I'm not sure I understand the relationship between the fixed and variable resistance values in the circuit.

My project is a simple dimmer for a "unit-power finder" to be mounted on a astronomical telescope. The LED does not need to reach its full brightness as it's being used to back-illuminate a film reticle which is being projected against the night sky. Fine control at the dim end of the range would be a plus.

I'd use an Arduino based dimmer/pulser circuit, except all the good toys are still packed from a recent move, and I need something quicker and dirtier for the time being now that we're approaching prime observing weather.

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  • \$\begingroup\$ YOu would get much better control with a current adjuster \$\endgroup\$
    – Trevor_G
    Commented Mar 15, 2018 at 15:54
  • \$\begingroup\$ @Trevor_G The problem is that the OP wants a non-linear behavior (perhaps like an audio or anti-audio pot?), doesn't want full operational brightness, and probably would like to be able to have something to allow a way to set the shape of the behavior so that the final result is close. So the details will matter here, I think. The referenced answer from Phil is not a bad idea. See: Telrad, I think. Something like that? I'm not sure why the OP doesn't just buy one. \$\endgroup\$
    – jonk
    Commented Mar 15, 2018 at 16:07
  • \$\begingroup\$ Why aren't you just buying one of the cheap solutions like Telrad or Rigel's Quik-Finder? And don't you also want the ability to pulse? Or do you only want continuous? \$\endgroup\$
    – jonk
    Commented Mar 15, 2018 at 16:17
  • \$\begingroup\$ Good on you for asking a new question instead of piggybacking on the previous questions or answers! Seen too many of those.. \$\endgroup\$
    – pipe
    Commented Mar 15, 2018 at 16:38
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    \$\begingroup\$ I'd suggest just starting with the same values and experimenting. Given variables of your particular LED's characteristics, the level of brightness you want, and the "feel" of the controls, you'll spend more time doing math than you would sticking three components into a breadboard. \$\endgroup\$
    – Phil Frost
    Commented Mar 15, 2018 at 16:43

2 Answers 2

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It's been a long time since I did any amateur astronomy. Other things impinged. (I built my own telescopes in the late 1960's, early 1970's. But also spend thousands of hours of work and testing and refiguring. Two reflectors and one Maksutov with a secondary spot on the meniscus.)

Too bad it seems so difficult today to get hobbyist quantities of various kinds of glass. Used to be dozens of suppliers for hobbyists making telescope mirrors and eyepiece lenses. No more. Most are gone and the few I've called wanted to know "How many tons of that did you want?" Old Willmann-Bell now pretty much offers nothing at all. (I actually bought my Maksutov meniscus from them, way back when.) If you know of anyone offering various types of glasses in hobbyist quantities, I'd love to know about them.


I think Phil's approach is great. It's very simple and you can adjust the curve by changing out resistor values. I could, but I'm not going to do it, prepare a complete set of partial differential equations so that you could optimize it perfectly for some intention. But Phil's recommendation that you just swap values of resistors just makes so much sense. So that's my recommendation.

The one thing I didn't like seeing in Phil's schematic was that there was no provision to manage the case where the potentiometer reaches one end of its sweep. So, I'd add a small resistor there to avoid the risk of directly applying \$3\:\text{V}\$ to your LED. Something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Start out with \$R_S=100\:\Omega\$ and \$R_P=1\:\text{k}\Omega\$. See what that does for your red LED. You can adjust \$R_S\$ to be lower if you need to increase the peak brightness upward. Then adjust \$R_P\$ to get the curve you want. (But you may need to re-adjust \$R_S\$ again, if you make a lot of change to \$R_P\$.)

Start there. There's nothing particularly complicated and it is very easy to set this up on a protoboard and look at it in a dark room (after 10 minutes of adaptation on your part.) It should not take you too long to work it out.

This really isn't something we can calculate using some flawless equation. There is a great deal of variability in LEDs, human responses to light, and limitations in the batteries you use, and more. Also, because you are running a voltage supply that is very near a red LED voltage (near \$2\:\text{V}\$), variations in different red LEDs will have more impact than otherwise. So this really is an experimental thing for you.


It would be possible to arrange things so that there more of a precision circuit that delivers very close to the same red LED current regardless of the vagaries of the red LEDs you use and which would reproduce the same pot-driven behavior curve every time. But it would involve active parts and/or ICs. And unless you were making a commercial product and willing to go through testing with amateur astronomer customers to get their various opinions about the "best feel" for the controls, Phil's circuit with the added \$R_S\$ is probably good enough for many uses.

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The battery voltage (Vbat) minus the LED forward conduction voltage (Vf) equals the voltage across the resistor, or combination of resistors in series. When the variable resistor is 0 ohms, the fixed resistor sets the maximum LED current. Calculate that based on the LED datasheet. If you have a guess for the minimum LED current you want, subtract 33% (for margin), use Ohm'Law to calculate the total resistance for that current, and then subtract the fixed resistor value. That gives you the variable resistor max. value. There almost certainly is not a pot available with that exact value, so get the next larger one.

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