#### eli-ott

Joined Sep 22, 2022
7
Hi, I have a problem where a simple bush button makes a 3.2V into a 2.8V.
I wonder if there is any way to fix this or if every button does this.

#### AlbertHall

Joined Jun 4, 2014
12,328
Welcome to AAC, @eli-ott
Where does the 3.2V come from?
What is the button connected to?
How much current flows through the button?

#### eli-ott

Joined Sep 22, 2022
7
Welcome to AAC, @eli-ott
Where does the 3.2V come from?
What is the button connected to?
How much current flows through the button?
The 3.2 come from 2 AA batteries, and the button is connected to a led. Here's the datasheet of the led: datasheet.
I tried to just link my button to the batteries and a multimeter and the voltage output is the same as the input.
And the push button used is an On/Off push button, I don't have any datasheet for this one.

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#### Marley

Joined Apr 4, 2016
501
Have you got a current-limiting resistor in series with the LED?
If not, the LED will pull the battery voltage down when you press the button - and draw a lot of current.

#### eli-ott

Joined Sep 22, 2022
7
Have you got a current-limiting resistor in series with the LED?
If not, the LED will pull the battery voltage down when you press the button - and draw a lot of current.
I don't have a current limiting resistor in series, is it necessary ? If yes may I have a little bit more information on how to do it ? Because the led is rated at 3.1V, so a 100 ohm would be a little too much I think. Or a 0 ohm can maybe do the same work (it's probably dumb)?

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#### Marley

Joined Apr 4, 2016
501
The current needs to be controlled somehow.
Looking at your LED datasheet I notice that the LED draws 3.0A at 3.1V (or 3.3V depending on the type). This is more current than a pair of AA batteries can provide. That explains why the voltage is dropping.
So, without a series resistor the current is being limited by the internal resistance of the battery.
Now, this may be OK - but the battery might get warm and it will not last very long.
These high-power LEDs are not really designed for battery operation.
With a much larger battery (a car battery for example) or some kind of mains supply power supply you will need some sort of driver circuit that will regulate the current at 3.0A. This will be "switchmode" based so as to be efficient.

#### eli-ott

Joined Sep 22, 2022
7
These high-power LEDs are not really designed for battery operation.
Pretty bad news not going to lie. Is there any option to still make it decently work with batteries. Like maybe with a 100/200ohm resistor?

#### eli-ott

Joined Sep 22, 2022
7
So, without a series resistor the current is being limited by the internal resistance of the battery.
The thing also is that I still want to have a good light, and I tried with a 100 ohm resistor but the light decrese by a lot. I was also wondering if a 0 ohm resistor can regulate the current or not ? Or I have 10/20/47 ohm resistor.

#### Marley

Joined Apr 4, 2016
501
I would use larger batteries, say a pair of C cells or D cells in series.
With this high-power LED, there is probably no need to have a series resistor - just rely on the internal resistance of the battery.
A 0 ohm resistor is just a piece of wire!

Problem is, when the battery voltage drops even by a small amount, the LED will dim quite quickly.
This is because the LED needs a minimum voltage to function (conduct electricity).
White LEDs are in fact a blue or UV LED covered with phosphor. These LEDs have a forward voltage of 3 or more volts. So only a new battery will have enough voltage to make it work.

It would be possible to make a circuit to drive the LED correctly from a wider range of battery voltage but this is a complex circuit beyond your expertise. Sorry to say that - but I can tell you do not know much about electronics!

#### eli-ott

Joined Sep 22, 2022
7
It would be possible to make a circuit to drive the LED correctly from a wider range of battery voltage but this is a complex circuit beyond your expertise. Sorry to say that - but I can tell you do not know much about electronics!
You're absolutely right. And also the problem with C cell or D cell battery is their sizes, like the size of an AA battery is the maximum to make something decent. I'm also curious with your circuit do you have any links so that I can take a look by curiosity please ?

#### Marley

Joined Apr 4, 2016
501
If you want to stay with AA cells then I would suggest a much smaller LED. Say a half-watt size instead of 8watts.

#### eli-ott

Joined Sep 22, 2022
7
If you want to stay with AA cells then I would suggest a much smaller LED. Say a half-watt size instead of 8watts.
Yeah I think I'm ready to sacrifice the AA size for the C cell batteries. But if I use C cell batteries do I need a current-limiting resistor or the batteries are good on their own?

#### Marley

Joined Apr 4, 2016
501
Here is a link to a document discussing driving LEDs with Buck-Boost converters.

#### djsfantasi

Joined Apr 11, 2010
9,151
Yeah I think I'm ready to sacrifice the AA size for the C cell batteries. But if I use C cell batteries do I need a current-limiting resistor or the batteries are good on their own?
You will need a current limiting resistor regardless. Otherwise when you supply more current, the LED will burn out!!!

There is a way to calculate What resistor you need.

But you’ll need to know a few things first.
• The forward voltage of the LED. This is a specification provided for each LED. I’ll put a variable name in parentheses after each item. Like this (Vf)
• The current required by the LED. The datasheet for your LED lists 3A maximum (which is the same as 3000mA) (it’s rated for 750mA minimum). You don’t want to operate an LED at its maximum current, so let’s choose an operating current of 2A (since you want a bright light source. In other cases, I’d choose 1A) (I)
• The supply voltage or your battery supply. Note it should be larger than Vf. In your case, I’d choose 4.5 volts. Don’t skimp here or your circuit may not work. If you’ve seen LEDs without resistors, the battery itself acts like a resistor. This only works in simple cases and I don’t recommend it (Vs)
• The current limiting resistor value. This is what we’re calcuiating
The formula is simple
R = (Vs - Vf) / I​

Choose the closest standard resistor. Note you also have to consider the resistors power rating. More on that later.

Using your case (with my recommendations) as an example…

R = (4.5 - 3.1) / 2​
1.4 / 2​
0.7 Ω​

Or a 1Ω resistor. You may need a little more headroom in your power supply - maybe 6V - but there’s another problem.

I mentioned power rating. The resistor will see 1.4V across it with a current of 2A. So what should it’s power rating be? This is measured in Watts which is calculated by multiplying volts by amps. In this example, 1.4x2 is 2.8W.

A 2.8W resistor really needs to be a higher wattage to prevent overheating. And their physical size is a consideration, too. A 3W (almost too small) resistor will take close to 0.75” length on you board. Maybe 1/3 (I’m guessing) of a single battery.

Something has to give. I would look at different LEDs, running at the minimum current or building a larger device, to fit all those batteries and high wattage resistors into.

#### Audioguru again

Joined Oct 21, 2019
6,606
When you talk about batteries, there are some good powerful Alkaline batteries made by Name-Brand companies and there are cheap and old style Carbon-zinc batteries made by no-Name-Brand places. Some cheap weak batteries are even wrongly labelled "Super Heavy Duty" but they are junk.

The datasheet for a name-Brand Alkaline battery shows how much current it can supply.