Simple but possibly dumb question

Malsi

Joined Dec 19, 2019
1
So I can't seem to find an obvious answer, but i'm pretty certain the answer is indeed obvious. I searched online for an answer but the keywords tend to supply me with other information that doesn't quite answer my question.

My question is about making a basic circuit with batteries and 2 leds.

The LED Vf is 3 volts, and operates at 20mA
Im using 2 LED (6 Vf)

I have a battery pack of 4 Double A Batteries: 1.5 X 4 = 6V
Obviously when i go to calculate the ohms of resistance i'm going to get 0 because they cancel each other out.

I suspect that IF this were to run, the battery life would run much less efficient but -

Otherwise,do I calculate the resistance needed from the 6V battery alone in which case I use a 300ohb resistor? Or is the voltage portion of the calculation V = 6v-6v

My ultimate quesiton is - Will this burn out the LEDs immediately if I dont use any resistor? OR are there any obvious problems I'm going to run into. I'm a very noobish noob, and I'm not sure if this is even worth wiring up or if It's impossible. If it is impossible, I will 3d print a new battery tray to hold 5 batteries, but this will be larger than my design was intended to be. So if possible, i'd want the 4 batteries to work.

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Papabravo

Joined Feb 24, 2006
21,031
So I can't seem to find an obvious answer, but i'm pretty certain the answer is indeed obvious. I searched online for an answer but the keywords tend to supply me with other information that doesn't quite answer my question.

My question is about making a basic circuit with batteries and 2 leds.

The LED Vf is 3 volts, and operates at 20mA
Im using 2 LED (6 Vf)

I have a battery pack of 4 Double A Batteries: 1.5 X 4 = 6V
Obviously when i go to calculate the ohms of resistance i'm going to get 0 because they cancel each other out.

I suspect that IF this were to run, the battery life would run much less efficient but -

Otherwise,do I calculate the resistance needed from the 6V battery alone in which case I use a 300ohb resistor? Or is the voltage portion of the calculation V = 6v-6v

My ultimate quesiton is - Will this burn out the LEDs immediately if I dont use any resistor? OR are there any obvious problems I'm going to run into. I'm a very noobish noob, and I'm not sure if this is even worth wiring up or if It's impossible. If it is impossible, I will 3d print a new battery tray to hold 5 batteries, but this will be larger than my design was intended to be. So if possible, i'd want the 4 batteries to work.
The Vf of the LEDs will not be precisely 3 Volts. That is an approximate value subject to manufacturing variation and so forth. What you need to do is use an adjustable bench supply to measure the forward drop of both LEDs together with an appropriate resistor. For example let's use a 9 volt battery and try to push 10 mA through the LEDs.
So 9-6 = 3 volts through the resistor.
And 3 volts divided by 0.010 is 300 Ω.
This will get you close and you can measure the actual Vf of the LEDs with a multimeter. So let us now say that we are going to fabricate 5 AA batteries in series for a total of 8 volts and also that our total Vf for 2 LEDs has been measured at 5.6 Volts,
So 8-5.6=2.4
And 2.4/.010 = 240 Ω
The LEDs will wink out when the remaining charge on the batteries no longer exceeds the Vf of the two LEDs

SamR

Joined Mar 19, 2019
5,021
I = V/R = 6V/~0Ω = ∞A Think they can handle that without burnout?

Audioguru again

Joined Oct 21, 2019
6,621
Where will you find a 3V LED? Some are 2.8V and 3V will instantly burn them out. Some are 3.2V and they will be very dim or not light up with only 3V. Buy many LEDs and test them all, then you might find two that are exactly 3.0V.

Doesn't the 1.5V from an AA battery drop as it is used? Then if you find LEDs that are exactly 3.0V you will see them dim until they do not light up anymore but the batteries will still be almost new.

You need more voltage than the LEDs need and you also need a series resistor to limit the current to 20mA.

With a 6V battery and one LED then the resistor for 20mA in a 3V LED is 6V-3V)/20mA= 150 ohms.
If the LED is actually 2.8V then its current with new batteries is (6V-2.8V)/150 ohms= 21.3mA which is fine.
If the LED is actually 3.2V then its current with new batteries is (6V-3.2V)/150 ohms= 18.7mA which is fine.
When the battery drops to 4.5V then the current in a 3V LED with the 150 ohm resistor is (4.5V- 3V)/150 ohms= 10mA which is a little dimmed but is fine. You calculate the current when the battery has dropped to 4V.
Use one 150 ohm resistor for each LED.

MrChips

Joined Oct 2, 2009
30,523
It is not a simple or dumb question. It is simply a dump title for a thread.

Please use a title that is more descriptive of the topic. For example "How to drive two 3V LEDs with four 1.5V batteries" would be much better than "Simple but possibly dumb question".

MrChips

Joined Oct 2, 2009
30,523
And that reminds me. This is the third such question in two months. I really need to finish my blog on the need for a series resistor.

cmartinez

Joined Jan 17, 2007
8,182
Buy many LEDs and test them all, then you might find two that are exactly 3.0V.
And that would be only for a while. LED's forward voltage changes with time too... so a 3.0V LED won't stay that way for long.

Bernard

Joined Aug 7, 2008
5,784
You might drop one battery to give 4.5V & use 2 Rs with 2 LEDs. Use measured data to figure Rs. if Vf are almost same , say +- 50 mV, you could try 1 R & parallel LEDs.

Audioguru again

Joined Oct 21, 2019
6,621
I have never seen a battery holder for only 3 AA battery cells. If a battery holder is found for 3 cells then the LEDs will be seen dimming as the battery runs down.

Ylli

Joined Nov 13, 2015
1,085
You need to have a 'reasonable' series R to convert the fixed voltage source into a pseudo-current source. Stay with the 6 volt battery pack, but connect the LED's in parallel, each with it's own series resistor. If you want to push 20 mA through the LEDs, then you would need the series resistor to be (6 - 3)/0.02, or 150 ohms. If you want 10 mA, then go with 300 ohm resistors.

Bernard

Joined Aug 7, 2008
5,784
Battery container, 3 AA, Keystone 2465, 2464, Digi-Key. 3-AAA, 2479.

Audioguru again

Joined Oct 21, 2019
6,621
The 3-cells holder would be good if the cells are the very expensive Energizer "Ultimate Lithium" throw-aways that hold up their voltage to the very end.

Tonyr1084

Joined Sep 24, 2015
7,788
Is there a reason why you want two LED's in series? Can you have them in parallel, each with its own resistor? Then you can stick with the 6 volts. Am I wrong?

Bernard

Joined Aug 7, 2008
5,784
Think of the millions of LED flash lights operating on 3 AA or AAA & not a resistor in sight.

Audioguru again

Joined Oct 21, 2019
6,621
My cheap Chinese flashlight has 24 matched white LEDs in parallel, no resistor and was powered by three "super heavy duty" cheap Chinese AAA cells. The battery cells wore out soon and I replaced them with modern and powerful Energizer alkaline cells then the light was so bright that I am lucky the LEDs did not burn out. I added a resistor.

jjw

Tonyr1084

Joined Sep 24, 2015
7,788
Here's what I mean:  wait a minute, I think I miscalculated. 20 mA through each LED but the overall circuit is 40 mA. Sorry for the mis-step.

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DickCappels

Joined Aug 21, 2008
10,128
Do as Audioguru again suggested and measure the voltage across your LEDs when running at the current at which you want them to operate in the target application.That will tell you whether you need a resistor or not and if needed, what the value needs to be.

If you have the LED datasheet, many manufacturers not only publish the voltage drop range, but also bin the LEDs into different ranges of voltage drops for a given current. The hitch might be that the manufacturer and part number not known. Using Audioguru's method will get you the needed information.

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