Power my loop box LEDs with 18 volt?

[boy howdy]

I built a dual loop box for my pedal board, but the only power option I have left on my power supply is 18 volt. Do I need to modify my resistor value from the usual, what is it, 4.7k? I have one regular LED (red or yellow, haven't decided) and one dual color (red and green). I'm using 3pdt switches. I reckon that about covers it.

Thanks for any assistance.

[bluebunny]

What do you mean the only option left is a 18V supply?  I'm guessing there are others, but these are "in use"?  You can power many things from one power supply, just like you can plug a beer fridge, a lava lamp and a phone charger all into the same mains supply.

[Kevin Mitchell]

If it only needs power to light some LEDs then perhaps you should consider an internal battery and perhaps a power switch just so it doesn't drain when you're not using it.

Otherwise - if the issue on hand is regulating/cutting down power, simply get a regulator that fits the bill. A 5volt one should do.

[midwayfair]

http://ledcalc.com/

Calculator. Follow the instructions and be sure to read the question marks.

[davent]

For a simple passive loop pedal i used a 2 cell, AAA battery holder inside the 1590B enclosure to power the two LED's, power switched in the usual manner with the input jack.

In the amps i've made i used the high voltage supply, plus 300v and a resistor to power the indicator LED.

dave

[GibsonGM]

If it only needs power to light some LEDs then perhaps you should consider an internal battery and perhaps a power switch just so it doesn't drain when you're not using it.

Otherwise - if the issue on hand is regulating/cutting down power, simply get a regulator that fits the bill. A 5volt one should do.

Any of the other methods would be fine ("external" battery, just using the (lower voltage) power supply that's nearby....

You would THINK that 18V would be just GREAT!  Use a regulator, off you go.   NO.....to drop 13V, that regulator will get VERY VERY VERY HOT, and waste all that voltage just to do that tiny scut work...too much difference between regulated and input voltage.    Even a simple DIVIDER would be better than that.

Luckily, as Jon posted the link to the calculator...it's not hard to run an LED off most ANY voltage.   Ohm's Law teaches us about voltage drops....

[bluebunny]

Whether you drop the voltage with a regulator or a resistor, you have to waste (dissipate) the same amount of power.  Both are wasteful (which is why I mentioned other - presumably 9V - outputs) and both items will get hot.

[GibsonGM]

Whether you drop the voltage with a regulator or a resistor, you have to waste (dissipate) the same amount of power.  Both are wasteful (which is why I mentioned other - presumably 9V - outputs) and both items will get hot.

Yes, you do have to dissipate the same amount of power.  My comment had more to do with spending $4 US for a regulator vs. 10 cents for a dropping resistor.   There is zero need for Regulated power to LEDs (big overkill);  close is close enough. 

[bluebunny]

Yep, good point Mike.  I love it when we agree!   

[GibsonGM]

Yep, good point Mike.  I love it when we agree!   

Don't most all of us?!       Mr. Ohm leaves little room for bending the rules, ha ha!

[ElectricDruid]

Putting some values into that calculator is quite enlightening:

If you're using 4.7K as your "typical" value on a 9V supply (is that right?) then you've got 1.5mA through your LED. If that gives reasonable brightness, it's a low current LED. Which is a good choice for a pedal, I'd say.

If you put 18V supply and 10mA current into the calculator, you get 1.8K as the series resistor. That'd be enough for a more "old school" LED. If you drop the current to the same 1.5mA you had with a 4.7K on 9V, you come out with 12K. In all cases, a 1/4W resistor is sufficient. Keeping the LED current low helps reduce the amount of heat generated/power wasted by the resistor, so the low current LED is a good idea.

All these calculations assume a 2V voltage drop (Vf) for the LED.

Definitely don't use a regulator. Like the others have said, massive overkill.

HTH,
Tom

[EBK]

[OT]Continuing with the idea of massive overkill, isn't there a way to derive a low-power constant current source to efficiently drive the LED from the 18-volt source?  [/OT] 

[GibsonGM]

Yes.  A 5 cent resistor.

Eqn:   Vsupply - Vf LED / I LED = Resistance to use.   
 
Vf LED = LED forward voltage.  I will assume 3V here for grins, check your data sheet or determine experimentally   
I LED = desired current for LED; I am assuming 5mA for the heck of it, YMMV


18V - 3V = 15V/.005 = 3K

This is what the calculator does...

[EBK]

Yes.  A 5 cent resistor.

Eqn:   Vsupply - Vf LED / I LED = Resistance to use.   
 
Vf LED = LED forward voltage.  I will assume 3V here for grins, check your data sheet or determine experimentally   
I LED = desired current for LED; I am assuming 5mA for the heck of it, YMMV


18V - 3V = 15V/.005 = 3K

This is what the calculator does...

Yes, I understand how resistors and LEDs work (but I do at least appreciate the patient thoughtfulness of your explanation).

I was just having fun, suggesting (too subtly -- my fault) that there are some unnecessarily complex ways to drive the LED and waste less power (more along the lines of a constant current switching mode power supply), which is why I tagged my comment as off-topic. 

[ElectricDruid]

In this vein, you could try using persistence-of-vision to reduce the required current. If an LED is flashed brightly fast enough, persistence of vision makes it look as if it is on all the time, at almost the same brightness as if it was on all the time. Since we've got bags of voltage to use, a bright flash would be easy. Perhaps we could have a 555 astable running the LED in this mode, giving (say) a 10% duty cycle.

Of course, this is impractical because it's likely to cause all kinds of hideous noise as the 555 switches a honking great LED current, but I kind of got the idea that was the type of suggestion you were after...

T.

[EBK]

In this vein, you could try using persistence-of-vision to reduce the required current. If an LED is flashed brightly fast enough, persistence of vision makes it look as if it is on all the time, at almost the same brightness as if it was on all the time. Since we've got bags of voltage to use, a bright flash would be easy. Perhaps we could have a 555 astable running the LED in this mode, giving (say) a 10% duty cycle.

Of course, this is impractical because it's likely to cause all kinds of hideous noise as the 555 switches a honking great LED current, but I kind of got the idea that was the type of suggestion you were after...

T.

A quick search turns up lots of power-efficient LED driver ICs that employ both a constant-current SMPS and PWM control.  Looks like we both failed at being impractical!  . At least we can agree it's not the appropriate solution to boy howdy's original problem. 

[GibsonGM]

Yes.  A 5 cent resistor.

Eqn:   Vsupply - Vf LED / I LED = Resistance to use.   
 
Vf LED = LED forward voltage.  I will assume 3V here for grins, check your data sheet or determine experimentally   
I LED = desired current for LED; I am assuming 5mA for the heck of it, YMMV


18V - 3V = 15V/.005 = 3K

This is what the calculator does...

Yes, I understand how resistors and LEDs work (but I do at least appreciate the patient thoughtfulness of your explanation).

I was just having fun, suggesting (too subtly -- my fault) that there are some unnecessarily complex ways to drive the LED and waste less power (more along the lines of a constant current switching mode power supply), which is why I tagged my comment as off-topic. 

Yes, there are, and some that take up too much real estate, but can be done (and ARE done, depending on what the designer needs to do). 

In my thickness, I didn't 'get' your sarcasm, LOL....but maybe my explanation will help the OP to play with LEDs or something?

[Transmogrifox]

Something cool I saw done was a design that operated the LED's on their maximum efficiency curve for 30us every 10ms (100Hz pulse train of 30us).  These were high efficiency LED's to start with but the average current was about 60 uA at 3.3V, or 198 uW.

You can run a very small battery for a long time on 198 uW. 

Also there is the Joule Thief which operates on a similar principle, possibly better since series resistive losses are much lower than you get in a current limiting resistor.  Also the Joule Thief would be a good candidate for 18V operation since this is only a matter of transformer turns ratio.

RF noise  a possible issue if it has audio frequency components that can be rectified into the signal path.

Other ways to save power at 18V is to stack all LEDs in series with a single current limiting resistor or a transistor current source.  You short across each LED you want to turn off, and this is where the constant current source helps maintain equal brightness regardless of who you short.

Joule Thief could be run with all LED's in series and a switch only shorts the unused LED.

Just some brain-diorrhea in case you can find a kernel.

[EBK]

In my thickness, I didn't 'get' your sarcasm, LOL....but maybe my explanation will help the OP to play with LEDs or something?

I didn't intend it as sarcasm, per se.  More of a realization that the original question was answered in a way that poked at my engineering brain.  Two of the best solutions each had the "problem" of throwing energy away as heat ("problem" being an overstatement).  My thought was: Now that we've got the "right" answers, could we address the power dissipation issue, regardless of practicality of the solution?  More of an academic exercise than anything serious. 

[EBK]

Something cool I saw done was a design that operated the LED's on their maximum efficiency curve for 30us every 10ms (100Hz pulse train of 30us).  These were high efficiency LED's to start with but the average current was about 60 uA at 3.3V, or 198 uW.

You can run a very small battery for a long time on 198 uW. 

Also there is the Joule Thief which operates on a similar principle, possibly better since series resistive losses are much lower than you get in a current limiting resistor.  Also the Joule Thief would be a good candidate for 18V operation since this is only a matter of transformer turns ratio.

RF noise  a possible issue if it has audio frequency components that can be rectified into the signal path.

Other ways to save power at 18V is to stack all LEDs in series with a single current limiting resistor or a transistor current source.  You short across each LED you want to turn off, and this is where the constant current source helps maintain equal brightness regardless of who you short.

Joule Thief could be run with all LED's in series and a switch only shorts the unused LED.

Just some brain-diorrhea in case you can find a kernel.

My brain thrives on such diarrhea, it seems (eew.... ). I always appreciate hearing about the cool things others have seen done.  I've heard of the Joule Thief a long time ago, but at the wrong time to fully appreciate it.