New question about voltage

[RC_Racer]

I have a question about voltage in my led lights. I have 2-3.6v led's from radio shack. I originally had them hooked up parallel to 2- AA batteries. Last night while I was bored I decided to see what difference a third battery would make. It definitely gave me the brightness I need for my project but I am worried about blowing the bulbs if it is too much voltage for them. They are used as headlights for a remote control car and I really need them to be very bright. Is the third battery going to be too much for the bulbs?
Thanks in advance for any help.

 

[wmmullaney]

I don't think so, but I'm almost sure the leds are going to blow almost immediately if they are given to much power.

 

[Krumlink]

Well if we use ohms law:

3.6V LED Light
20mA each

So 40mA total

3.6V / 0.04 = 90 ohm's

It would be safe if you put a 90 Ohm resistor on there for safety.

Oh and don't worry about too much Voltage: Try running a LED across 25kV and 20ma. The LED Lives to work another day.

LED's Run off Current, but as voltage increases they start to consume more current, giving the common misconception that too much Voltage makes them blow. Just be sure to use ohm's law and you will never go wrong.

**broken link removed**

 

[audioguru]

Well if we use ohms law:
3.6V LED Light, 20mA each
So 40mA total
3.6V / 0.04 = 90 ohm's

It would be safe if you put a 90 Ohm resistor on there for safety.

No. The resistor limits the current with the difference in the voltage across it of the supply voltage and the LED voltage. You forgot the supply voltage in your calculation.

Three 1.5V cells are 4.5V when new. If the LEDs are 3.6V and you have two in parallel with 20mA each, then the calculation is (4.5V - 3.6V)/40mA= 22.5 ohms.

The LEDs are not exactly the same. The one with the lowest voltage will hog all the current and will burn out at 40mA. Then the other one will burn out.
Each LED needs to have its own current-limiting resistor of 45 ohms.

 

[OutToLunch]

you've got things a little messed up there. you don't apply ohms law to the forward drop of the LEDs, you apply it to the difference between the forward drop and the battery supply. 3 AA batteries yields 4.5V, so you can't put the LEDs in series, so an LED and resistor need to be put in series - there will be 2 of these series circuits in parallel across the 4.5V battery.

so, if you want 20mA through the LED, you also want 20mA through the resistor:
(4.5 - 3.6)/.02 = 45 ohms

90 ohms would only yield 10mA.

edit - was replying to krumlink, not guru...

 

[Krumlink]

What if the AA batteries were rechargeable? Then no voltage drop.

Also OutToLunch, you forgot that he wanted 2 LED's in parallel, yielding 40mA.

 

[OutToLunch]

What if the AA batteries were rechargeable? Then no voltage drop.

what are you talking about? i t matters not if the batteries are rechargeable. the voltage across the battery minus the voltage across the LED is the voltage drop across the resistor. divide that voltage drop by the desired current and you get the required resistance.

Also OutToLunch, you forgot that he wanted 2 LED's in parallel, yielding 40mA.

you don't put LEDs in parallel. you can string them in series and have a single current limiting resistor but you do not put LEDs directly in parallel with one another. since there is not enough headroom for putting the LEDs in series, then each LED needs its own current limiting resistor. with 45 ohms, you 20mA per LED.

 

[Krumlink]

what are you talking about? i t matters not if the batteries are rechargeable. the voltage across the battery minus the voltage across the LED is the voltage drop across the resistor. divide that voltage drop by the desired current and you get the required resistance.

you don't put LEDs in parallel. you can string them in series and have a single current limiting resistor but you do not put LEDs directly in parallel with one another. since there is not enough headroom for putting the LEDs in series, then each LED needs its own current limiting resistor. with 45 ohms, you 20mA per LED.

Doh I forgot about voltage drop

 

[audioguru]

Three Ni-MH rechargeable cells in series are 4.5V fresh from the charger which quickly drops to 1.2V each (3.6V total) for most of the charge.
They don't provide enough voltage to reliably light "3.6V" LEDs.

A "3.6V" LED might be 3.4V, 3.8V or any voltage in between. If it is only 3.4V then it will burn out. If it is 3.8V then it won't light.

 

[RC_Racer]

Please forgive my ignorance but I'm actually a little more confused now....Do I wire the lights in a series or parallel? If I do them in a series am I correct that I only need one 45 ohm resistor and if I go parallel then I need one for each light?
One other thing, I may not be able to get a three battery holder so I may have to go to four. I assume that if I do this I would need to take the resistors up to 120 ohms? Not sure if I am doing the math right...one post shows (4.5-3.6)/.02 and another shows dividing by .04...again, forgive the ignorance, very new to this but I want to learn it. Thanks again

 

[Pommie]

You should calculate each LEDs current limiting resistor using the 0.02 value. (6-3.6)/0.02 = 120Ω. So, connect a 120Ω resistor in series with 1 LED and it will work fine on your 6V battery. Build as many of these resistor/LED circuits as you desire and connect them all to the battery in parallel.

Mike.

 

[RC_Racer]

mike thanks....one more question....as far as wiring the series...positive wire out of the battery-to the resistor- then to the positive side of led then negative out of led back to battery....is this correct?

 

[Torben]

mike thanks....one more question....as far as wiring the series...positive wire out of the battery-to the resistor- then to the positive side of led then negative out of led back to battery....is this correct?

Yes, that should do it.


Torben