3v led and resistor value question

[01downer]

I want to splice into my amps power lamp jewel and install 4 leds inside the amp for the "glowing" effect
My question which watt and value resistors should i use? I believe the current being supplied to the power lamp is 120v ac (standard 3v pedal style leds will work with ac current as well correct)?
So im thinking just string along some leds in series tapped into the power lamps hot and ground terminals. Which value and what resistors would i use? And will this even work?
The amp is a v1 carvin v3

Thanks

[italianguy63]

When I did mine, I used a "cold cathode" LED kit.  (about $10)  It is a neon light tube, with a voltage converter.  The converter switches 120VAC to 12VDC for the light.  I added a 2PDT toggle on the case back to turn it on or off.  You tap right into the 120V jack on the input from the wall outlet..  MC

[01downer]

So my method i was thinking of going wont work?

Btw you were a huge help in suggesting how to install the voltage starve mod on my gcb95 circuit. Thanks! Ill try to get a pic up of the results

[italianguy63]

Oh it could!  Look up LEDs in series/parallel calculator in Google.

You'll need to know the voltage requirement of the particular LEDs, how many, and the supply voltage.

I'm not sure, but I think it would work on AC.  Maybe somebody else will chime in...

MC

[01downer]

I read from a 4 different places the 3mm standard leds will work on both ac and dc currents. Im thinking 4 leds but would be fine with 2.
Splicing them into the amps power indicator lamp with the proper resistors paired with the leds seemed like it would work to me. But alot of projects in that past that seemed like they would work for me havent worked so well....haha

[italianguy63]

Probably the easiest will just be hooking 4 in a chain, and figuring out the resistor value.  It's not rocket science.

I'm just GUESSING here, but say using 4 super-bright LEDs, some clear/color variety, with a 300R to a 2K2 resistor will work well.  Of course, test it before you affix everything!!

MC

[01downer]

Thanks. I tried the conversion calculator and it was saying 2 watt 5.5ohm resistors.
I will try both methods though. Thanks for sharing some knowledge

[italianguy63]

That's probably right!!  I forgot it is 110V not 9V!

(That's probably why I used the cold-cathode kit when I did mine... but it was a while back).

[GibsonGM]

1) totally do-able and not difficult
2)  Just do a little research:  https://www.youtube.com/watch?v=Obomwdk3fZk


The only issue I see here for you is how you run 4 of them, which changes the resistor value.

Vsource - Vled/I led

For 120VAC use, I've always *assumed* something like a 1.5V drop for ONE LED.    That comes out to:  120-1.5/.020 = 118.5/.020 ~~ 6K. 
Looks like 01Downer is on the right track!   There is some room for forgiveness. 

If I were running 4 in ||, I'd think along the lines of current * 4....use an online calculator.

[01downer]

For 2 of them i went supply voltage 120
Forward voltage 3 since each led is rated at 3v
And forward current 2.14

And it give me .25 1/4 watt 54k resistors to use.

Would this be right guys?
Ive got tons of resistors but they are all quarter watt and some half watts here and there

[01downer]

http://www.digikey.com/us/en/mkt/calculators/led-series-resistor.html

Thats the calculator im using

[R.G.]

Guys, this is serious stuff. Any time you even think about touching the AC power wiring, you are opening the possibility of killing yourself or others by electrocution, or starting fires. We'll just start with that. In general, it's not practical to learn to do AC mains wiring from questions on an internet forum. If you don't already know how to do this safely, don't do it.

With that as background, on to the LEDs.

LEDs are diodes. They happen to convert some of their forward current into light. They have a forward voltage that is characteristic of their color. Red ones are about 1.2-1.5V, getting bigger through orange, yellow, green and blue. Blue LEDs are about 3+ volts to 4V forward. They also have a reverse breakdown voltage, which is often only 5V or so. They often *die* if you break them backwards.

120Vac is a 120VRMS sine wave, it peaks at 120*1.414 = 170V. LEDs see the instantaneous voltage. Your LED will see, forwards and backwards, *every* voltage between +170 and -170V across them as the AC wave changes, 60 times per second. To keep from killing the LEDs (quite aside from wiring safely to keep from killing your amp or YOU) you have to limit their forward current to less than the LED's absolute maximum specified current from the datasheet and limit their reverse voltage to the max value for that from the datasheet.

Standard LEDs in the 5mm diameter T 1 3/4 package have an average forward current max of about 20ma, and a peak current that is dependent on the LED of about 80-100 ma as long as the peaks don't average out to more than the 20ma power limit. So your calculations for a limiting resistor need to limit the forward current with 170V peaks to less than the peak current and the average to less than about 20ma. If you have four "average" LEDs at about 2.2V each, that's 8.8V, leaving 170-8.8 = 161.2V across the limiting resistor.

The average value of a half wave rectified sine is Vpeak/pi volts or Ipeak/pi amps for current. The RMS value is the peak divided by two. So the average current needing to be less than 20ma means that the peak current must be less than Ipeak = Iavg * pi = 0.02 * 3.14 = 0.068A, and the dropping resistor must be bigger than 161.2/0.068 = 2371 ohms or the LEDs will overheat. This is below the usual peak current for a 5mm LED, so the peak current has taken care of itself.

You have to somehow keep that reverse 170V from breaking down the LEDs. You can do this with a series or shunt diode. A series diode allows current to flow only when the LEDs are on, a reverse shunt diode lets approximately a sine wave of current to flow in the indicators, but the diode takes the reverse current off the LEDs because its forward voltage is smaller than their reverse breakdown. You get to pick. Series cuts down on the total power spent running the LEDs and heating the resistor, shunt means that 160V half-sine-wave signals are not running around all over your amplifier and inducing hum, but maximizes the power spent in the dropping resistor.

The series diode setup means a current of the same 0.020 average, 0.034 rms (roughly) in the resistor, or a power of 0.034*0.034*2371 = 2.74W. You need at least a 3W resistor, and that will get to about 100C surface temp with 2.74W in it. Make it at least a 5W resistor to avoid a burn hazard. That's for the series diode.

For the shunt diode setup, you have (roughly) 162Vpk/1.414 = 115Vrms across the resistor, and the resistor power will be 115*115/2371 = 5.6W, and you'll need a 10W resistor to keep it cool enough not to be a burn hazard.

Also note that EVERY SINGLE WIRE ON THE LEDS AND RESISTOR(S) IS LIVE TO THE AC POWER LINE AND IS AN ELECTROCUTION HAZARD.

I've been trying to outline some of the problems with running LEDs from the AC power line. There is the whole electrical safety issue, the amp noise issue, and the wasted power issue. LEDs are impractical to run from such a high voltage. It can be done, but is very, very wasteful and hazardous.

It is much safer and more practical to run them from any other voltage in the amp that's a secondary voltage. I ...STRONGLY... advise you to think about doing that.

[01downer]

Thank you for sharing rg
Ive now decided to just leave it alone. The technicals you laid out went way above my u.derstanding. dont want to hurt myself or anyone else trying to have fun with my amp

[R.G.]

You don't need to leave it alone - the point of all that is that 120Vac is an impractical and dangerous supply voltage for LEDs. Look at the schematic of your amp and find some other voltage on the secondary side of things and use that. You can probably easily tap off the first DC power filter cap and run them. That power is now isolated from the AC power line and is not an electrocution hazard, and the resistors are no longer as high a power. Neither is the current to the LEDs pulsing and a noise source.

Try a little thinking about that. If you can find and post the schemo to the amp, I'll help figure out values and such.

[PRR]

Why do you want _four_ LEDs inside a jewel?

One is *plenty* of light to know the amp is "on". (OTOH if you need a puddle of work-light in front of the amp, you need dozen(s) of LEDs.)

You won't easily be able to see that there are "Wow! four LEDs!".

(Yes, by moving your head and interpolating the various patterns between LEDs and jewel-facets, you "can" count the sources; but this isn't an electron microscope.)

Yes, fifty _K_ is a reasonable trial number for 120V (NOT "5.5ohm"). That won't be super-bright, but you are already up to Watt-size resistors, so you shouldn't get greedy.

Remember that resistor-dropping 120v to 1.7V is wasting-off 98.5% of power as heat. Efficiency wise, this is very bad.

> tapped into the power lamps hot and ground terminals

Neither wire on a 120V lamp is "ground". You have two wires hot to each other, with an uncertain relation to dirt (other metal), and it is bad to leak current from either wire to dirt (or metal or chassis or amplifier).

It is VERY difficult to get FOR-SURE insulation between a flock of LEDs and the chassis bezel. Remember that a fault on either 120V wire "may" be fatal, or cause fire.

Worth repeating: EVERY SINGLE WIRE ON THE LEDS AND RESISTOR(S) {as proposed} IS LIVE TO THE AC POWER LINE AND IS AN ELECTROCUTION HAZARD.

LEDs only work one way. They "do" light on AC, but only half the time. Flicker. The other half-wave, if over 5V, they break-down. This does no great harm at low currents. However if you were pushing the maximum forward current, the breakdown heat on the reverse half-cycles can burn the LED.

If you want steadier light and no reverse-cook on AC, and accept multiple LEDs, you can wire two reverse-parallel on one resistor. One lights one way, one lights the other way, and neither one goes into reverse breakdown.

I've done this, didn't like it. The back-and-forth flicker isn't obvious if you stare at the LEDs, but if you move your head rapidly you get funny dotted streaks. If you also have old chemical brain-strain you get flashbacks.

Without seeing the amp.....

It would be much gooder to find a LOW-voltage source associated with the part of the amplifier you touch.

I dunno every Carvin. Many recent amps have +12V DC or +/-15V DC supplies. These can usually spare a few mA without strain.

Go from a 12V-15V point, through 2.7K, wire, LED, wire, 2.7K, to power common. Good light, and the two resistors make damage un-likely even if the LED leads short in the bezel.

[01downer]

Thank you rg and prr
Your knowledge well exceeds mine and would feel safer just leaving it alone.  I only wanted 2 leds inside the amp to give it a glowing effect.  Looking at the schematic would only cause more confusion for me as im still an advanced rookie at understanding them

When time permits it will probably just install i 9v dc power adapter on the side of the amp out of plain view and just use a battery or wallwart to power them. That way i dont dont to tap into the amplifiers power source