9 to 18v pump charge: Considerations with the led?

[traktop]

I want to build a little box with a pump charge and a switch to swap between 9v/18v for trying with difference pedals.
I know thats ok with tubescreamers and so, and as long as I use 25v caps on my own builds there shouldnt be any problem.
I was wondering if I should have any consideration with the leds on my builds when seeing double the voltage.
Cheers, Gilberto.

[induction]

traktop wrote

I was wondering if I should have any consideration with the leds on my builds when seeing double the voltage.

That depends on a few things, including the specs of the led circuit and your goals.

The led will probably be brighter at 18V. It probably won't burn out, but we can't say for sure unless we know the forward voltage of the led, the size of the current limiting resistor, and the maximum current allowed by the led.

If you just don't want the led to burn out, try it out on the breadboard. Take an identical led and clr, power it with your charge pump, and see what happens.

If you want the brightness to be the same with 18V as with 9V, that can be done in a couple of ways. Let us know if you want instructions.

FYI: If any of your builds use LM386's, or have built-in charge pumps, those could fry at 18V.

[traktop]

Whenever I added a charge pump inside any of my builds, I always isolated it from the led current supply. But having that external little voltage changer box in mind, some instructions about led protection would be more than welcome
Cheers, Gilberto.

[Beaker]

I always understood it as the bigger the limiting resistor, the bigger the voltage it will take.
Therefore if you use a superbright LED and a big limiting resistor (anything up to 10k), it will easily handle 18V, and much more.


If you are in any doubt, you could try these:

http://www.ebay.co.uk/itm/100-LEDs-3mm-Blue-8000mcd-Blue-LED-Blue-PC-Modding-Cars-Auto-Model-/321740721059?var=&hash=item4ae93dcba3:m:mTZ68tjZykXsyRpa0dt8ztw

Ten Euros for 100 pre-wired LEDs from Germany sounds a good deal to me. Just select the largest resistor value they offer (1.2K), and you should be good for 24V.
As they have a extra buyer option, you could ask them for 2.2K or 4.7K instead (saves frying your eyeballs).

I use these high voltage pre-wired Superbright LEDs on all my builds, and I have not had a problem yet with any 18V builds - though admittedly I've only built a couple so far.

[induction]

If you know the forward voltage (Vf) and the maximum allowed current (Imax) from the datasheet, then you can calculate the smallest allowed clr (Rmin) very easily (skip to the bolded part if you don't want the derivation):

- For a given voltage source (Vs), the voltage drop across the resistor is Vs minus Vf. (Because the only things between Vs and ground are the led and the clr, and the voltage drop across the led is Vf, the rest of the voltage must drop across the clr.)

- By Ohm's law (V = IR, where I is current and R is resistance), the current through the resistor will be I = (Vs-Vf)/R. Since the clr is inline with the led, the current through the led must be the same as the current through the clr.

- Given Imax, we can rearrange this formula for the minimum clr: Rmin = (Vs-Vf)/Imax. Insert values for Vs, Vf, and Imax and you're done.



The brightness of the led is determined by the amount of current it consumes. If you want the led to have equal brightness with 9V and 18V, you can use the same formula to calculate the required clr to do so:

- Calculate the current at 9V: I = V/R = (9-Vf)/R1, where R1 is the clr used at 9V.

- Insert this value into the equivalent formula for 18V: R2 = (Vs-Vf)/I = R1*(18-Vf)/(9-Vf).

- When you switch the supply from 9V to 18V, switch the clr from R1 to R2.


Of course, you can just go by eye instead, if you prefer.

[traktop]

You mean a led switch with 2 differents resistors, right¿ That was my first thought.
Anyway, I pretty much always use super bright blue leds with 50k resistors, so if Beaker is right, there shoudln be any problem with those.
Thanks everyone! Gilberto.

[Beaker]

Thanks Induction for the maths, it really helps.

I built a ROG Ginger with an onboard charge pump, and a 9V/18V switch just a couple of weeks ago.

Using one of my pre-wired superbright blue LEDs, and an extra 4.7K resistor on the board, there was no noticeable drop in brightness when switching between the two voltages at all, at least not in daylight anyway.

[induction]

I haven't found a noticeable difference, either, but individual results will depend on the specific leds and clr's used in a particular build. In some cases it may be noticeable, in others it won't. I posted the math mostly to show how to think about the problem, but going by eye is probably just as good, and easier.

Of course, if the charge pump is inside the pedal, then there isn't any need for these considerations. You'll have 9V available to run the charge pump, so you can power the led from there. Then the led sees the same voltage no matter whether the pedal is in 9V mode or 18V mode. And if the charge pump is an external add-on to the adapter, then you can't change the clr with the voltage supply anyway.

So the math isn't at all necessary for a practical solution to this problem.

[Beaker]

You are right of course - in the example I gave, the LED is only powered by 9V.

Basically I was just trying to reassure Traktop that with a suitably large CLR (and he's using huge ones), a superbright LED will handle far more than 18V.

Of course, at 18V a 3V-9V diffuse LED will either go pop, or be so dim that it will be barely visible.