What resistance do I need when powering a 3v LED with a 3v Battery?

Thread Starter

AstraObscura

Joined Oct 25, 2021
5
I have 3x 3v White LEDs that are to be powered with a 3v battery (3x CR2032 in series).
The forward voltage of the LED is 3.0v-3.2v. From my understanding, you always want to use a resistor in the circuit. SO with this setup the LED would be slightly dimmer.. but what ohm resistor do I need to use?
 

Alec_t

Joined Sep 17, 2013
12,318
Welcome to AAC!
1) What is the rated maximum current for the LED? The running current needs to be less than that, unless the LED will be driven only occasionally, with extremely short pulses.
2) What current do you want through the LED? A cheapo LED will probably be happy with about 20mA. The current determines how much series resistance is required.
3) If your LED specimen happens to have a Vf of 3.2V it might not even light, or will light dimly. with a 3V supply.
4) A CR2032 can provide only a small current and has a short life-time. If you have space, use bigger battery cells.
5) 3 x CR2032 in series give 9V initially but that will droop. They will drive 2 LEDs in series, at about 20mA, using a series resistor of (9-2 x 3)V/20mA = 150 Ohms.
 
Last edited:

dl324

Joined Mar 30, 2015
13,535
Welcome to AAC!
From my understanding, you always want to use a resistor in the circuit.
CR2032 is an exception to the rule. It can't provide much current for very long. People have been taping LEDs across CR2032 batteries for a long time. I went to a Maker Faire and that was one of the activities for children (and the young at heart).

Cheap LED flashlights put LEDs in parallel, with no ballast or current limiting resistors and they seem to have a decent lifetime.
SO with this setup the LED would be slightly dimmer.. but what ohm resistor do I need to use?
It depends on the LEDs. There are standard brightness to ultra bright.
 

BobTPH

Joined Jun 5, 2013
4,051
Your battery is 9V, not 3V, but then, so are the three LEDs in series.

There is no point in adding a resistor, it is already there inside the battery. If I recall correctly, there is about 20Ω in each cell, so 60 total. If you measure it, I think you will find that there is something like 2.5 to 2.8V across each LED.

Bob
 
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MisterBill2

Joined Jan 23, 2018
9,849
The purpose of he resistor is to drop the voltage so that the current is not excessive. With that particular battery there is no need because the voltage and internal resistance of the battery are well suited to driving a common LED. So with a CR2032 battery an external resistor is not required, using 3 in series for one LED is a waste.
 

Thread Starter

AstraObscura

Joined Oct 25, 2021
5
Welcome to AAC!
1) What is the rated maximum current for the LED? The running current needs to be less than that, unless the LED will be driven only occasionally, with extremely short pulses.
2) What current do you want through the LED? A cheapo LED will probably be happy with about 20mA. The current determines how much series resistance is required.
3) If your LED specimen happens to have a Vf of 3.2V it might not even light, or will light dimly. with a 3V supply.
4) A CR2032 can provide only a small current and has a short life-time. If you have space, use bigger battery cells.
5) 3 x CR2032 in series give 9V initially but that will droop. They will drive 2 LEDs in series, at about 20mA, using a series resistor of (9-2 x 3)V/20mA = 150 Ohms.
I am thinking of using the https://lighthouseleds.com/1-8mm-2mm-led-white-ultra-bright.html It says to use a 1ohm resistor for a 3v supply. Would that be 1 resistor per LED or for the circuit?
I am thinking something like this?

1635195282094.png
 

dcbingaman

Joined Jun 30, 2021
498
Because there is no 'head room' with the voltage being slightly less than the LED quoted forward drop of 3.2V, they may or may not light up and you will also find variations from one LED to another even with the same part number. You really need a higher voltage.
 

MisterBill2

Joined Jan 23, 2018
9,849
LEDs will certainly illuminate at a bit less than the rated voltage! Not full brightness, but certainly some. And certainly that relationship is far from linear.
 

Audioguru again

Joined Oct 21, 2019
3,853
CR2032 batteries are small and weak. In the circuit with 1 ohm resistors, if the LEDs need 3.4V then they will be dim and if they need 3.0V then they will be bright for a few minutes then gradually dim for another few minutes until they do not produce any light.
 

Thread Starter

AstraObscura

Joined Oct 25, 2021
5
Because there is no 'head room' with the voltage being slightly less than the LED quoted forward drop of 3.2V, they may or may not light up and you will also find variations from one LED to another even with the same part number. You really need a higher voltage.
hm. do they make LEDs that are less then 3v? I am limited on space as far as battery size goes (really can't go much larger than a 2032. My only other option would be to find lower voltage LEDs than.
 

bertus

Joined Apr 5, 2008
21,681
Hello,

The forward voltage of a LED depends on the color and material it is made of.
The following chart will give you an indication of forward voltages:
led color guide.png
As the voltage is dependend on the current, I have found this graph:
IV-curves-all-colours.png
Bertus
 

ElectricSpidey

Joined Dec 2, 2017
1,971
I just tested (using the pinch method) a standard 20mA white LED rated 3 to 3.6 forward voltage with a 2032 battery (fresh) and it was very bright.

I have no idea how long it would stay that way, because I need to keep the battery for my use.

That was a TopBright brand LED.

PS: I wasn't trying to prove anything...just curious to see if a random white LED would light.
 
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dcbingaman

Joined Jun 30, 2021
498
hm. do they make LEDs that are less then 3v? I am limited on space as far as battery size goes (really can't go much larger than a 2032. My only other option would be to find lower voltage LEDs than.
The voltage that creates a given color (frequency) is directly related to energy of a photon:
e=hf. Where h is planks constant. The energy is in joules and the frequency is in hz.
Another way of stating it:
e=hc/wavelength.
Red light has much longer wavelengths to the number of joules required to create a photon of that light is not as high.

Voltage is just joules per coulomb of charge. You need less joules of energy to create a lower frequency light thus a lower voltage.

In short red is the longest and has the least energy. Thus if you use red LED's you should need less voltage.
 

MisterBill2

Joined Jan 23, 2018
9,849
I have a number of solar powered lights in the yard and some of them use what appears to be a single NiCad AA cell. They produce a white light and I have not analyzed the circuit that powers them, other than to observe that it is small and simple and covered witha small blob of black epoxy. I should investigate and see what voltage it takes to light those LEDs.
 

neonstrobe

Joined May 15, 2009
180
White LEDS are nearly all made from blue LEDS with phosphors to convert some of the blue light to green and red, usually as a yellow mix which is why they appear yellow.
These all need about 3V to run. That is the quantum potential of blue light, effectively. Red is about half that, so red LEDs run at 1.5V
I think that the whole idea of running an LED using a battery's internal resistance is "not cool". Some cells may have lower impedances than another. Also the LED light will dim as the cell drains.
Every LED torch I have made (some more than a decade ago) have used simple flyback converters which operate from 1.5V cells (for one to a few low power LEDs) or 6V (for higher power LEDS). The input power is controlled, and therefore the output power is also regulated.
If you want your LEDs to last a long time it is better to ensure that the current they operate on is in their design range.
Next best is to use a resistive dropper but that would need to have at least a volt if not 3 to set the current properly, which essentially throws away the advantages of LED lighting!
But many people throw LED torches and lights together with batteries with a nominal internal impedance because it is cheap.
I don't recommend it, but it obviously has been done.
But to answer your question, the resistance needed is (Vcc -VLED)/ILED where Vcc is your supply voltage, vLED the LED voltage and ILED the LED current. Pretty simple, really. BUt this shows if Vcc and VLED are close, small variations will give large variations in current as the resistance is small. And the LED might not even light if VLED is greater than Vcc.
 
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