#1
Hey,
i want to add some cosmetic LEDs to my Peavey 5150. Im positive i can just tap into the standby LED and use that to power the extra ones i will be adding on. i ran a multimeter and when the LED is active it has 7.8v running it.

So im not sure how it all pans out but i know if i put a 3.5v LED it will blow it out. so does that mean if i put 2 3.5v LEDs will it be fine?

if thats not the right way to go please let me know. i basically want the lights to turn on with i flip the standby switch.
www.myspace.com/wearedraind
My Rig:
ESP LTD MH401NT
ESP LTD EX-400BD (two of them)
Peavey 5150
Mesa 4x12 Straightback Slant cab loaded with v30's
#2
You have to put a resistor and LED in series in order to keep it from blowing. If the LED is 3.5 volts, then you'll be left with 4.3 volts over the resistor. Now you want to set the current going into the LED with a resistor, say 10-20 mA. So use ohms law to determine what size resistor you are going to need for it (V = IR)
#3
you can put LEDs in series to get away without using a resistor but it'll be a really tacky job. best to use a resistor and 1 LED. But just for future reference, putting LEDs in series will spread the voltage evenly. Say you have three 3-volt LEDs and a 9-volt battery. You can put them all in series and the 9 volts will spread evenly among the LEDs and each one will get 3 volts.

and no need to use math here. most people (including me) are sort of scared away by math.

a 1k resistor will do just fine. lower value=brighter. From what I've seem many LEDs are closer to 50mA but to find that info you'd need a datasheet. Datasheets and Ohm's Law are sort of overkill for just adding an LED to the amp.
#4
Quote by Invader Jim
you can put LEDs in series to get away without using a resistor but it'll be a really tacky job. best to use a resistor and 1 LED. But just for future reference, putting LEDs in series will spread the voltage evenly. Say you have three 3-volt LEDs and a 9-volt battery. You can put them all in series and the 9 volts will spread evenly among the LEDs and each one will get 3 volts.

and no need to use math here. most people (including me) are sort of scared away by math.

a 1k resistor will do just fine. lower value=brighter. From what I've seem many LEDs are closer to 50mA but to find that info you'd need a datasheet. Datasheets and Ohm's Law are sort of overkill for just adding an LED to the amp.


The point of my post was to give TS a way to understand how it works. Just saying "Put a 1k in series with it" is okay... but TS learns nothing other than "1k in series works". If you want to do anything with electronics its a good idea to have a basic understand of ohms law. If simple math scares you, then by god you should fear electricity like the plague.
#5
the kingbright l53 is the typical LED i use, and the top one from smallbear. its what i typically use. all the colors vary slightly, but forward voltage is about 2.3V and the current should be about 25-30 mA. any more current than that and you arent doing anything but shortening the life of the LED, it doesnt get any brighter. i typically aim a bit lower, more like 20 mA, but that again depends on the usage.

for larger LEDs like you might be using (if they are 3.5 V, i assume the current is larger as well) then its going to be a bit different. with 2 of them at 50 mA or so, its not hard to pick a nice resistor.
7.8 - 2(3.5) = 0.8 V over the resistor.
0.8/0.050 = 16 ohms

easier to look at once you pick your LED and stuff, but even something like a 50 ohm resistor should work very well.
#6
Quote by XgamerGt04
The point of my post was to give TS a way to understand how it works. Just saying "Put a 1k in series with it" is okay... but TS learns nothing other than "1k in series works". If you want to do anything with electronics its a good idea to have a basic understand of ohms law. If simple math scares you, then by god you should fear electricity like the plague.

boy do i look and feel like a tard right now. i didnt realize it but i was doing the one thing i wouldnt do. i like to tell people how it works, but here i am giving generic info.