i made a little gem amp. small 1 w ss beginner amp. im looking to add a on/off switch and an LED indicator to the circuit. i understand that the switch can be placed bassically anywhere in the circuit but im going to put it in between the battery and the circuit.

the LED however, some people told me that i need to run a certain resistor in series with it. the circuit is powered by a 9v battery. the LED MIGHT be 3.6v. if it isnt 3.6v, i still want to know how to find out the value of the resistor.

so how do i calculate the value of the resistor?
for a 9v battery.. you need a 330ohm resistor in series with the battery.

if you connect it directly.. the led will burn out.
Opus Pocus =]
you will probably need a 100ohm or higher resistor looking up google will help
ok, if the LED is 3.6V and you have 9V, then that leaves 5.4V that would be going over the resistor. we use ohms law to find out your resistor from that. we know that R = V/I, so we take our 5.4 and divide it by the current that we want. i dont know what LED you are using and what current it is rated for, so i am just going to make one up. lets say we want 100mA, so we divide 5.4 by .1 and get a 54 ohm resistor. now 100mA is probably way too much, its going to be closer to 10mA or 20mA most likely. so if we use 20mA, that would give us a 270 ohm resistor and 10mA would give us 540 ohms. something in between there would probably work, depending on the LED of course.
Quote by jof1029
ok, if the LED is 3.6V and you have 9V, then that leaves 5.4V that would be going over the resistor. we use ohms law to find out your resistor from that. we know that R = V/I, so we take our 5.4 and divide it by the current that we want. i dont know what LED you are using and what current it is rated for, so i am just going to make one up. lets say we want 100mA, so we divide 5.4 by .1 and get a 54 ohm resistor. now 100mA is probably way too much, its going to be closer to 10mA or 20mA most likely. so if we use 20mA, that would give us a 270 ohm resistor and 10mA would give us 540 ohms. something in between there would probably work, depending on the LED of course.

^ This guy here is exactly right.

90% of what you neeed to know, math-wise, can be summed up in two equations. Everything else derives from these.

First, Ohm's Law:

E = I * R

E: voltage in volts
I: current in amps
R: resistance in ohms

And the Power Equation

P = E * I .

P: power in watts
E: voltage in volts
I: current in amps
i already done the ecuation.. you need 330ohms xD

can anyone disprove me??
Opus Pocus =]
thanks to all for posting

so the LED packaging will tell me what current it needs to run on?

do i put it =

9v > on/off Switch > resistor for LED > LED > circuit ?

so if invader jim is right, then it means that i can use a resistor with a "close enough" value?
Quote by Luxifer

do i put it =

9v > on/off Switch > resistor for LED > LED > circuit ?

No. If you run it that way then your circuit will only get the volts that the LED is getting. You need to keep the LED seperate. Get a DPDT switch and wire your circuit to one side, and wire the resistor and led on the second side of the switch.

That way the full 9v gets to your circuit, and the resistor is only affecting the LED's voltage.

And yes Jim is right about the close enough. If its too dim you could decrease the value of the resistor a bit to let more voltage through. Vice-versa if its too bright.
Last edited by Matt420740 at Jul 26, 2009,
Quote by josuetijuana
i already done the ecuation.. you need 330ohms xD

can anyone disprove me??

I am more than happy to disprove you

do you have the exact specifications of the LED in question? if you dont, then you cannot possibly know for a fact that you need exactly 330ohms.

this LED that i just pulled up with a search requires 2V and 20mA. if he is using a 9V battery it would require a 350 ohm resistor. this LED typically takes 3.2V at 20mA, which results in a 290 ohm resistor.

so without knowing what LED is being used, you cannot say for certain that 330 ohms is the ideal resistor.
^^^ More or less, you need to find a common ground - somewhere between the resistor being too small so the LED burns out, or the resistor too big that the LED isn't bright enough. 1k should be fine though. Remember that each LED is different though, so try a few out.
Last edited by BGSM at Jul 26, 2009,
can i somehow work around an SPST switch?
what if i put the resistor and the LED at the end of the circuit? i dunno if that will make a difference... just putting it out there.

by pole you mean switch???
Quote by Luxifer
can i somehow work around an SPST switch?

Yes. The lug you get you power from is at 9v. Leave your circuit hooked there, and soldier your resistor to that lug as well. Run the resistor to your LED. The circuit will still be getting 9V from the lug, and the LED's voltage will be cut down from the resistor.
wait so how many lugs do SPST switches have?

so im guessing it has 2 lugs. so 1 lug is connected to the negative terminal of the 9v. the 2nd lug is connected to the circuit and the led circuit SEPEREATLY?
Quote by Luxifer
wait so how many lugs do SPST switches have?

so im guessing it has 2 lugs. so 1 lug is connected to the negative terminal of the 9v. the 2nd lug is connected to the circuit and the led circuit SEPEREATLY?

2 Lugs. Negative is connected to ground, not a lug. The POSITIVE goes to one lug. The resistor and circuit should go to other lug. I'll upload a pic for you in a minute and show you how it should be wired.
Quote by Matt420740
2 Lugs. Negative is connected to ground, not a lug. The POSITIVE goes to one lug. The resistor and circuit should go to other lug. I'll upload a pic for you in a minute and show you how it should be wired.

thanks so much not many people have patience for a noob like me

Battery + should go to one lug, and the other lug gets a wire and resistor. You would run the wire to your circuit. Its 9V. Run the end of that resistor to your LED. The wire will still have full 9V for your circuit, and the resistor will cut down the voltage for your LED. The battery - should be tied to the same ground point as everything else.

I started building Amps a while back, and if it wasn't for people helping me out at first I would have never gotten anywhere. So don't feel bad.
Last edited by Matt420740 at Jul 26, 2009,
so i connect the wire from the switch AS the positive? i should still ground the LED and connect the negative terminal to the circuit's negative needs?
Quote by Luxifer
so i connect the wire from the switch AS the positive? i should still ground the LED and connect the negative terminal to the circuit's negative needs?

Yes. You could make a second ground point for the battery, but its easier to just use the one you already have. If your using a metal chassis then its all grounded to the same thing even if you do use multiple grounding spots. If your not using a metal chassis and you use multiple grounds then you should tie them together.
right now, my current understanding of "ground" is connecting to the negative terminal of battery.

do i just connect multiple wires from the negative terminal of the battery to create different ground spots?

so to rephrase my previous question: one side of the LED to the resistor, the other side to the negative terminal of the battery? and all the components in the circuit that need to be "grounded", should be also connected to the negative terminal of the battery?
ok so i put a 1k resistor with the LED. works fine. the LED is 3.6 volts 20mah.

when i play my guitar, the LED flashes according to when i strum. which is cool sometimes but is there a way to stabilize it?