Driving 12 - 15 LEDs using 4 CR2450 batteries.

[k7elp60]

This is the schematic and a picture of the project I built that uses a number of LED's from a CR2450 battery. I connect a number of LED's in parallel. To do this I match thieir Vf with a constant current source. The constant current source is about 3 to 4 Ma.
Each letter all the LED's are in parallel, with one current limiting resistor per letter.
I use a LT1054 to boost the 3V battery to about 5V because the blue LED's are over 3V.

 

[darkside501st]

I have some name-brand Fairchild red LEDs. At 20mA their forward voltage is spec'd at anywhere from 1.5V to 2.4V. Any LED resistor calculator will be completely wrong.
I can't find a part number for a Name-Brand white LED to see its datasheet.

LEDs are DIMMED by pulsing because then their average current is reduced by pulse-width-modulation.

Here is a couple of datasheets for a white flat top 5mm LED with similar ratings:
https://pdf1.alldatasheet.com/datasheet-pdf/view/136620/MARKTECH/MT4730NF-WT.html
https://www.electro-tech-online.com/custompdfs/2012/05/551LW7C.pdf

Here is the info from the ebay auction I am looking at:

5mm Flat top white Super Bright LED Wide Angle
Forward Voltage(V) If=20mA - Min = 3.2 Typ= 3.4
Dominant wavelength(mm)K If=20mA - Min = 5000 Typ = 6500
MCD If=20mA - Min = 16000 Typ = 20000
Reverse current(uA) Vr=5V Max - 10
Angle (deg) - 120-140

Would you mind explaining how to calculate the correct resistor in more detail?

In your OP you said you wanted to drive 10 to 15 LEDS.

Yes but I was going to run the LEDs in parallel so each LED would get the 3.7v and a resistor to control the current.

Also, I am not going to do the pulsing thing so we don't have to debate that.

 

[audioguru]

Your ebay white LEDs do not have the maximum forward voltage listed like all other LEDs. Let us GUESS that it is 3.6V.
To calculate the resistor for 20mA maximum current:
1) The maximum battery voltage from a rechargeable lithium cell is 4.2V.
2) The minimum forward voltage at 20mA is 3.2V.
3) Then the resistor value is (4.2V - 3.2V)/20mA= 50 ohms. The nearest common value is 51 ohms.

But what happens when the battery is discharged a little and the LED has the maximum forward voltage:
1) The battery is 3.5V.
2) The LED is 3.6V at 20mA and is 3.3V at 4mA or 5mA.
3) The current is (3.5V - 3.3V)/51 ohms= 4mA so this LED will be very dim but other LEDs will be fairly bright.
4) You might see this LED getting dimmer and dimmer as the battery runs down.

 

[darkside501st]

Ok I think I am following you somewhat. This is a whole other level of technical detail on LEDs than what I have gotten into before. So you are saying that you have to calculate the resistor using the max figures so that you don't burn the LEDs out but you have to consider the current once the voltage drops on the power supply because it will effect the brightness of the LEDs.

So these LEDs I am looking at will be pretty dim at 4mA once the voltage on the battery drops to 3.5v. You also said that other LEDs would still be bright at 4mA but wouldn't you need a different resistor value for other LEDs? If you had an LED that runs well on 4mA then the max would be like 10mA (just a hypothetical guess) so that LED would need a 100 ohm resistor (4.2 - 3.2)/10mA=100 ohms. Also, should we be using the min voltage on the battery for this calculation?

So if the battery drops to 3.5v then we would see that these LEDs actually have 2mA (3.5v - 3.3v)/100ohm=2mA.

I think I did that right based on what you showed me. Anyways, my point is that according to what you are telling me I don't think any LEDs would still be that bright once the voltage drops to 3.5v. So the real question is how long till the battery drops to a level where the LEDs will not be bright anymore.

It has been a while since I have done any algebra calculations (a long while) so I am not too familiar with how to swap the formula around but I will give it a shot:

(x - 3.3v)/15mA=50ohm

x=4.05v

So if I am following what you said correctly and my math is right then once the voltage drops to like 4v then the LED will be down to 15mA. I am guessing that at 15mA the LEDs will still be decently bright but much less than that and we would probably start to see them dimming. Just a guess.

So am I understanding you correctly?

This doesn't seem right, I mean I have used all different kinds of LEDs powered on batteries with many different resistor setups... I have used them in parallel and in series and the LEDs are usually bright till nearly the very end of the batteries usable capacity. Take a regular 1.5v battery... I have used them till there is less than 1v left in the battery and then I have tried to swap them into something else that uses less power and there is nothing left in them at that point. so on a 3v setup that is like a 1v drop whereas we are looking at a .7v drop and saying that the LEDs will have already been dim after like a .4 volt drop (another guess). Why is that?

Another thing I want to add is that it is possible that these LEDs will be too bright at their full potential. I have some 5050 SMD LEDs that are around 10000mcd and they are really bright. So it could be that I could cut down the current so that the LEDs would be closer to 10000mcd instead of 20000mcd.

Ok, I just bought these LEDs:
**broken link removed**

 

[audioguru]

So you are saying that you have to calculate the resistor using the max figures so that you don't burn the LEDs out but you have to consider the current once the voltage drops on the power supply because it will effect the brightness of the LEDs.

Yes.

So these LEDs I am looking at will be pretty dim at 4mA once the voltage on the battery drops to 3.5v.

LEDs have a RANGE of forward voltage. The ones with a low voltage will be bright and the ones with a high voltage will be dim. You don't know which LED is high or low unless you measure them.

You also said that other LEDs would still be bright at 4mA but wouldn't you need a different resistor value for other LEDs? If you had an LED that runs well on 4mA then the max would be like 10mA (just a hypothetical guess) so that LED would need a 100 ohm resistor (4.2 - 3.2)/10mA=100 ohms. Also, should we be using the min voltage on the battery for this calculation?

To avoid burning out the LEDs you must calculate the resistor value using the highest battery voltage and the lowest LED voltage.
Your problem is that the battery voltage is too close to the LED voltage so a small change in voltage causes a large change in current.

So the real question is how long till the battery drops to a level where the LEDs will not be bright anymore.

Try it to see.

(x - 3.3v)/15mA=50ohm. Then x=4.05v

So if I am following what you said correctly and my math is right then once the voltage drops to like 4v then the LED will be down to 15mA. I am guessing that at 15mA the LEDs will still be decently bright but much less than that and we would probably start to see them dimming. Just a guess.

Maybe your LEDs are very very bright. Then very very bright is the same as very bright.

This doesn't seem right, I mean I have used all different kinds of LEDs powered on batteries with many different resistor setups... I have used them in parallel and in series and the LEDs are usually bright till nearly the very end of the batteries usable capacity. Take a regular 1.5v battery... I have used them till there is less than 1v left in the battery and then I have tried to swap them into something else that uses less power and there is nothing left in them at that point. so on a 3v setup that is like a 1v drop whereas we are looking at a .7v drop and saying that the LEDs will have already been dim after like a .4 volt drop (another guess). Why is that?

Look at the datasheet for one of your LEDs. There is a graph of forward voltage change with current change. There is another graph of brightness change with current change.
For example:
1) At 30mA the forward voltage is 3.3V.
2) At 5mA the forward voltage is 2.9V. Then a change of only 0.4V causes the current to change 6 times.
3) At 30mA the intensity is 1.4.
4) At 5mA the intensity is 0.25. So a change of 0.4V causes the intensity to change 5.6 times.
Note that the LED forward voltage is different from LED to LED (unless you are lucky) and also changes when the current changes.
The battery voltage also changes.

The brightness of an LED is determined by its current.
If you want no brightness changes then power the LEDs from a constant current source where the current stays the same even when the voltages change.

 

[darkside501st]

Ok so maybe I should do a simple constant current circuit like this:
http://www.instructables.com/id/Circuits-for-using-High-Power-LED-s/ - skip to step 6

That would provide a constant current to the LEDs, right? Or one thing he said is that this circuit has a low drop out and the power source could be as low as .6v higher than the output voltage. So does that mean if we set it up to put out 3.4v to the LEDs then once the battery drops down to 4 volts then the circuit won't work anymore?

He talks about hi power LEDs but also mentions that this circuit can be applied to regular LEDs in series or parallel.

Also, here is a stupid question for you. If you connect LEDs in parallel then they each require 20mA of current so if I use 12 LEDs then that is 240mA total. Say this is on a 4.2 v power source. Now compare that to connecting 3 of them in series so there will be 4 rows in parallel which means that the LED array will be using 80mA total. This would have to be connected to a power source greater than 10.2v. So I am correct in thinking that the LEDs connected in series will be feeding off of the same current therefore this LED array will be using less power than the LEDs connected in parallel? So if that is the case then the same battery would last 3 times longer. Is that right?

Ok so what about if I use a voltage booster to bring the voltage on the 3.7v battery up to 11 or 12 volts. How does that work? Will the battery still last 3 times longer or does one action cancel out the other and would result in the battery still being drained as fast as the LEDs connected in parallel?

It kind of sounds like regardless of whether or not I connect the LEDs in series or parallel I will need to boost the voltage on the battery (if for a parallel array boost to 6v? and for a series array boost to 12v?) Then it wouldn't matter as much as the voltage drops, right?

So how about these options:
http://www.reuk.co.uk/DC-Voltage-Multiplier-Circuit-Plans.htm
http://chemelec.com/Projects/Voltage-Boost/LT1372.htm

At first glance I don't really understand how these work so I will have to do some more research but on the first link it shows the MAX756 chip that boosts the voltage but why do you have to have the capacitors and power inductors? What functions do they serve... I am only asking in the intrest of saving space. I think this design is getting to be too complicated, not in the sense that I can't understand it, but rather I am concerned that there will be too many components and that it will not all fit in the limited amount of space. I can't add another layer of depth so we are looking at just the space that is left on either side of the battery which is only about 5 or 6mm tall. So this LED array may not have enough space to make it as efficient as possible. I would like to make it as efficient as possible in the amount of space that is available.

So I have been asking some specific questions and I am curious about the answers but I may not be able to implement them. To get kind of a final solution I would ask you, if this were your project how would you set it up using 10 - 12 LEDs, the mobile phone battery (unless you have a better option), and the ability to connect it to a usb port for charging/continuous run time? To help you I am going to make a little schematic drawing of the arc so that you understand the space limits a little better. I will post it soon

 

[darkside501st]

Ok so here is the schematic I drew up:
View attachment 64538

Here is a picture of the screen used prop:
View attachment 64539

And here is the video of it in action:
https://www.youtube.com/watch?v=o5LHIR8O3G8

Another video with close ups and the prop master talking about it a little bit:
https://www.youtube.com/watch?v=JuuR58l9xl0 - skip to 58 sec

I would love to know how they did their unit. You can see that all their circuits are behind the LEDs and yet their arc reactor is only like 1/2 inch thick. I have no idea how they got such great light diffusion in such a small space. And their circuit is way more complicated with an RF receiver and everything too. It can't be an el sheet... it is way too bright for that. Also, then you would need a transformer to power it (I think).

 

[audioguru]

One 3.3V LED at 20mA uses a power of 3.3V x 20mA= 66mW from the battery.
Two in parallel use 3.3V x 40mA= 132mW from the battery.
Two LEDs in series use 6.6V x 20mA= 132mW from the battery. When battery is twice as big as the battery for one LED then it lasts for the same amount of time.

If you use a circuit to boost the battery voltage then the battery current increases because the output power is higher (as shown above). When the battery current increases then its duration is less unless it is bigger.