how will I avoid frying the status LED when using 18V instead of 9V?

Started by KarenColumbo, January 16, 2019, 01:58:13 PM

Previous topic - Next topic

KarenColumbo

... because I fried a few - never thought about the poor buggers when cranking my power supply :( How do those "official" builders do that?
  • SUPPORTER
I see something of myself in everyone / Just at this moment of the world / As snow gathers like bolts of lace / Waltzing on a ballroom girl" - Joni Mitchell - "Hejira"

Marcos - Munky


Mark Hammer

Simple.  Just increase the value of the current-limiting resistor.  Start with 10k-15k, and adjust from there.

Because we rarely store our LEDs in a drawer according to their millicandle rating (and I have plenty of "superbrights" in the same drawer as traditional "garden variety" low-brightness units), one of the things I recommend making and keeping on hand is a little gadget for assessing current-requirements to achieve a given brightness for a given LED.

Get hold of a multi-position rotary switch (something like this is good:  https://www.taydaelectronics.com/electromechanical/switches-key-pad/rotary-switch/rotary-switch-1-pole-12-position-alpha-sr2511-4409.html ) and wire it up to provide roughly equal, or proportional, steps in resistance from 1k to 22k (or let's say 2k2 to 18k, if pressed).  That range should provide enough range to identify a degree of illumination that is neither too dark to see, or allows enough current to instantly fry the LED.  The resistance is inserted between the power source of your choice (9v battery or adapter/power-brick) and a sock you can insert an LED into (one end of which goes o ground, of course).  Start with the max resistance, and switch down in step-wise manner until you hit a brightness that you find acceptable, or close to it.  Select the nearest common resistance value from there that you think will provide a suitable result.  So if 10k is a little brighter than you're aiming for, use a 12k current-limiting resistor for that LED.  If 10k is a little dimmer than you wanted, but the next step down is 6k8, maybe stick a 9k1 or 8k2 in there to get the brightness you want.

This is also a handy way of identifying whether a given LED is superbright, or other.  If it doesn't get bright until you hit 3k3 or thereabouts, it's an oldie but goodie.  If 18k still singes your eyebrows, then it is a superbright.

davent

"If you always do what you always did- you always get what you always got." - Unknown
https://chrome.google.com/webstore/detail/photobucket-hotlink-fix/kegnjbncdcliihbemealioapbifiaedg

ElectricDruid

Since I=V/R, if you liked the brightness you had when you had V =+9V, then you double R when you move to V=+18V, assuming you're using the same LEDs.

HTH,
Tom

Rob Strand

It you don't mind the brightness variation I'm with everyone above, just use a larger resistors.

If you *have* to have a constant brightness LED the way to do it is with a constant current source,

Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

davent

Quote from: Rob Strand on January 16, 2019, 06:24:39 PM
It you don't mind the brightness variation I'm with everyone above, just use a larger resistors.

If you *have* to have a constant brightness LED the way to do it is with a constant current source,


An LM317 plus a resistor? I've used this to set the current for tube heaters in a guitar amp but it's been a few years.


dave
"If you always do what you always did- you always get what you always got." - Unknown
https://chrome.google.com/webstore/detail/photobucket-hotlink-fix/kegnjbncdcliihbemealioapbifiaedg

Rob Strand

QuoteAn LM317 plus a resistor?
Looks fine to me.  There's many ways to do it. 

You might even get a 7805 to work  :icon_mrgreen:  The IQ adds more current but that can be allowed for provided the target current isn't too low.

This is one is essentially the same as the diode one,


You can do a "two wire" version of the diode one was well.
(FWIW I think 1N4148 diodes are a better choice than 1N4003's)

In all cases the active device is the one dissipating the power.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Sooner Boomer

Dan of  ̶9̶  only 5 Toes
I'm not getting older, I'm getting "vintage"

bluebunny



Quote from: Sooner Boomer on January 16, 2019, 08:21:32 PM
LED calculator:

http://led.linear1.org/1led.wiz

Be aware that this ^ will give you the minimum resistance required to avoid zorching your LED, but it will most likely still zorch your retina.  You can comfortably increase the numbers it spits out by an order or magnitude.  Or wear these:  8) *


* Those are welder's goggles, not sunglasses.
  • SUPPORTER
Ohm's Law - much like Coles Law, but with less cabbage...

antonis

Quote from: ElectricDruid on January 16, 2019, 05:22:01 PM
Since I=V/R, if you liked the brightness you had when you had V =+9V, then you double R when you move to V=+18V, assuming you're using the same LEDs.
Not precisely, Tom..  :icon_wink:

Considering LED forward voltage drop constant for a given brightness, we have I = (V - VLED) / R which doesn't equal to
I = (2V - VLED) / 2R ..
(meaning that, for doubling power supply voltage, we need a resistor bigger than double R for the same current as previously..)

P.S.
Overkill, of course, for low order scaling of high voltages but risky enough for the reverse proportion.. :icon_redface:
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

merlinb


ElectricDruid

Quote from: antonis on January 17, 2019, 05:12:50 AM
Quote from: ElectricDruid on January 16, 2019, 05:22:01 PM
Since I=V/R, if you liked the brightness you had when you had V =+9V, then you double R when you move to V=+18V, assuming you're using the same LEDs.
Not precisely, Tom..  :icon_wink:

Considering LED forward voltage drop constant for a given brightness, we have I = (V - VLED) / R which doesn't equal to
I = (2V - VLED) / 2R ..
(meaning that, for doubling power supply voltage, we need a resistor bigger than double R for the same current as previously..)

Yep, of course. Antonis is right. I was forgetting the LED volt drop, which is constant despite changing supply voltage. It must have been late here...

T.

deadastronaut

https://www.youtube.com/user/100roberthenry
https://deadastronaut.wixsite.com/effects

chasm reverb/tremshifter/faze filter/abductor II delay/timestream reverb/dreamtime delay/skinwalker hi gain dist/black triangle OD/ nano drums/space patrol fuzz//

duck_arse

here's a way to use all those j201's you have laying about .....




[.... I'm not really a welder.]
You hold the small basket while I strain the gnat.

amz-fx

My LED calculator also tells you the wattage of the resistor needed:

http://www.muzique.com/schem/led.htm

If you are powering with 18v, a resistor of 2.2k or larger in 1/4w will be okay. That allows over 7ma for the LED, which is a lot with more efficient modern indicators. A 4.7k or 5.1k might be a good compromise if you are switching between 9v and 18v.

regards, Jack

Mark Hammer

Quote from: bluebunny on January 17, 2019, 02:37:08 AM
Be aware that this ^ will give you the minimum resistance required to avoid zorching your LED, but it will most likely still zorch your retina.  You can comfortably increase the numbers it spits out by an order or magnitude.  Or wear these:  8) *
That's why I prefer the how-much-do-I-need-to-get-what-want method of the resistor substitution box.  I've run into plenty of LEDs that I was testing out on a near-death 9V that posed not even a trickle to my tongue, and the LED damn near blinded me.

Like I say, if we stored them in drawers or bags that listed their specs, that would be one thing.  But if you're like me, maybe you have a little bag of all your green ones, another little bag of your red ones, another for blue, white, yellow, RGB, etc., with no distinction made between the various LEDs contained.  Some of thse little 3mm green ones might need a 12V lantern battery to be visible against anything other than a pitch black pedal top, and others will leave you with afterimages for an hour, but they look identical.

amz-fx

Quote from: Mark Hammer on January 17, 2019, 11:49:01 AM
Some of thse little 3mm green ones might need a 12V lantern battery to be visible against anything other than a pitch black pedal top

You and I both need to clean out our component boxes and get rid of those 30-year-old LEDs that need 20 ma to be seen in daylight!!!  :icon_mrgreen:

Most modern LEDs that I have bought off Ebay recently are plenty bright with a milliamp or two!

Best regards, Jack

Derringer

Are you running an 18 volt wall-wart? Or are you using a charge pump to get 9volts up to 18?

If the latter is the case, just tap your LED power from the incoming 9 and leave the 18 for the rest of your circuit.

KarenColumbo

Woo! Took me quite some time to like ALL those posts. Very informative reading here, thx!

And yeah, I mean those pedals you can power from either a 9V or a 18V wall wart (or anything between). I never had a closer look at LED brightness when doing this - but I guess I should go the way of a good compromise for both voltage extremes - 4k7 min which will neither delete my cone cells nor make the status LED invisible even when all the light fails me.

  • SUPPORTER
I see something of myself in everyone / Just at this moment of the world / As snow gathers like bolts of lace / Waltzing on a ballroom girl" - Joni Mitchell - "Hejira"