# Powering 2.4v bulb with 3.7V lithium battery

Simple question to those who know I think:

I’m trying to fit some electronics inside the handle of a torch. The space is very tight. In order to save space I powered the 2.4v 600 mA xenon bulb directly from the battery, thinking it is normally powered from two 1.5V C batteries… The bulb blew.

I can’t find a small 2.4/2.5V step down that is rated to 600 mA.

My question is, given The bulb wattage is constant, and the lithium battery voltage is fairly constant, can I just use a single resistor as a voltage divider?

Kenneth

First question are the C batteries alkaline or rechargeable. From what you say they are alkaline. If you are looking for a battery you are able to charge the simplest way would be 2 C size rechargeable cells @ 1.2V each. = 2.4V
3.7 - 2.4 = 1.3
1.3 / .06 = 2.166Ω
Nearest preferred value is 2.2Ω
Power dissipated in resistor 1.3 * 0.6 = .78W
Would need to be at least 1W resistor as in enclosed space.
Could get warm or even hot and be somewhat larger than usual .25W resistors.
I think your safest would be 2 C 1.2V rechargeable.
Cheers Bob

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Thank you both for your replies. The torch (it’s actually a hospital laryngoscope) takes either alkaline or rechargeable batteries. The torch would only be used for 60 seconds or so at a time but the handle is enclosed aluminium.

With my math I get blub resistance R = V/I = 2.4V/600mA = 4Ω

Vout = Vin (r2 / (r1 + r2))
r1 = (Vin.r2 - Vout.r2)/Vout
r1 = (3.7x4 - 2.4x4)/2.4
r1 = 2.17Ω

Thanks for your help. Just to confirm do you think a resistor is a good solution here in place of a step-down regulator? If so I’ll get a 1W 2.2Ω resistor. The bulbs are \$30 each, hope I don’t blow another :-/

thanks Kenneth

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well i went by your topic heading which confused me such .i came up with many different figures and scenarios . and there are many different ways of calculating these type of outcomes the ohms law triangle the full- circle version …after all it gets confusing in the end… kirchoffs triangle the ohms law triangle an the one that has four corners all in one big circle…then there is just plain old transformation of formula…it goes on for ever the theory…of ohms law …and algebraic as taught these days,trying to save face … i`m over it but will with draw my incorrect post…

Different method, same result,
Me, being a simple person, took what voltage I needed to drop, the current to drop this voltage and ohms law did the rest.
2.2 Ω is the nearest preferred value (the value you can buy off the shelf).
It may be a little bit large.
It may get a little bit hot.
You are trying to replace 2 off C cells @ 1.5V each is that correct? If recharging is what you are after why not 2 off Ni-Mh cells @ 1.2V each as I suggested above. That to me is the simple way to get the battery rechargeable.
EXACTLY which type of 3.7V lithium battery are you trying to use. They come in lots of flavours and sizes. If it is only the extra longevity you are after 1.5V NON REGHARGEABLE batteries are available in AA size but I don’t know about C size. Adaptors to adapt AA to C (and D) sizes are also available. I have got some.
Cheers Bob

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The torch normally takes 2 C batteries (alkaline or rechargeable). I’m exchanging them for a 3.7 lithium battery because I need to free up the space to put my electronics (Arduino nano). The entire setup needs to fit inside a 3d printed custom sleeve that is the size of the original two C batteries

kenneth

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Hi Kenneth,

As Robert has said you’ll need to calculate the particular resistance that is required based on the input voltage. There’s actually a few online tools that you can use to do this automatically depending on the forward voltage of the bulbs (note that this will change if the arrangement is not in series with a supply relative to each light)

https://www.digikey.com.au/en/resources/conversion-calculators/conversion-calculator-led-series-resistor

Do you have the product number or any more information on those batteries? It would be very helpful in helping us determine what the optimal setup would be to compact the project down to that scale.

Hi Kenneth
That puts a different light (no pun intended) on things. A resistor will consume battery power (heat) so may be too wasteful so if possible a converter may be the best way to go if you can get one small enough. If you have only got the space of 2 C cells there is not much to play with so I am assuming the lithium is AA as I don’t think you would have a hope with 18650. Not sure of Arduino Nano size but it can’t be that small. Maybe they have a Pico??.
Anyway you could check and see how hot a 2.2Ω resistor gets but I have a feeling it will be too hot in that confined space.
Good luck
Cheers Bob

Hey Bob,

I agree you should be able to determine the maximum theoretical loss from P = I^2R where P represents the power loss to heat. As for sizes, the Nano comes in at approximately 43x18mm and about 1.3mm thick, and the Raspberry Pi Pico at 51x21mm 1mm thick, hopefully that helps

https://www.rapidtables.com/calc/electric/watt-volt-amp-calculator.html
dissipation…which was not stated originally …so there were many faults in the op`s question exactly what available current was there to play with all that was really stated was the fact that the bulb was indeed rated at 2.4 volt and required 600 Ma .. this left me to think that he was using an 600Mah 3.7 lipo battery..if he hooked the bulb to theoretically 3 volts unless it was actually a semiconductor and he hooked it up in reverse..the bulb did not have the ability to protect its self and now i am wondering if he meters the bulb is it open or is it in fact shorted like most semi`s fail in general only 2 ways under normal conditions…

Already done above. 0.6(A)*1.3(V)=0.78W or in round numbers 1W.
43 X 18 X 1.3, allowing for a bit of wall thickness of this printed enclosure that would just about fit into a C cell profile. Remembering that 3.7V AA lithium batteries are a bit longer than a standard C or AA due to battery protection circuitry there would not appear to be much space for anything else. If the NANO has a USB connector this may just fit. One must assume there will be some connections to the outside world also. A small voltage converter might just fit alongside a AA size lithium in the other space but I have no idea what is available. I think Kenneth has to do a search and experiment here. I think in this confined space and in the interests of battery life a resistor should not be considered.
Cheers Bob

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you can get them in 50 Ma as well this would be an idea i thought…or you could use a bc639/640 with a zener as another idea …good for 1Aor just a 5 watt zener . on its own biased up if need be…Q.R.T.

Don’t quite understand the critique in Brain’s reply:

• The battery is a 18650 lithium battery
• The microcontroller is a nano 33 BLE sense
• The bulb is 2.4V 600mA
• I’ve already 3d printed out a sleeve and I have 5 mm to play with around the battery, and a cylinder of space above the battery about 25mm diameter 25 mm height after the microcontroller is inserted.

I can power the nano using a S9V11F5 step up controller and the LiPo.

I don’t need longevity, this is a proof of concept prototype to test a hypothesis. If (big if) the idea worked then a future iteration would probably have a custom chip. Laryngoscopes are typically used for 60s at a time, but it would be a very bad look if one failed in use.

I didn’t put all that information down because I thought it clouded the original question, but I’m a programmer/ academic not an electrical engineer. Which is, to repeat:

What is the best way to reduce the voltage from this battery to the bulb in the space I have?

Kenneth

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I’m sorry I don’t follow any of this…

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Hi Ke3nneth

That is a start.

You know how much space is available. Until now no one else had an accurate idea. A short time ago I put a couple of C cells on the table and an 18650 battery alongside. I did not have a cylinder that size but I could not see any way the Nano was going to fit alongside the battery in a cylinder the size of a C cell (25mm). I was apparently wrong but I don’t have a Nano to try anyway. I did this because I was going to question if you actually meant D instead of C but on close inspection it would be a pretty fine fit but apparently is OK.

This would have been useful information earlier. I thought you were making something that had to last.
If you can tolerate the heat use a resistor. It may be only dissipating about .75W but that is still heat in a confined space. You will have to try it. Get some of those little temperature stickers that change colour when a predefined temperature is reached and stick inside your enclosure. Use a few stickers of different temperatures and you will get a good idea of how hot it gets under actual operation.
Although he did not explain it very well Brian’s idea of 3 diodes in series is valid. About 0.6V per diode so 2 would be probably OK. But you would still have the same 0.75W in heat to get rid of. You can’t avoid that.
So it is really up to you. If the bit of heat or battery longevity is not a problem a resistor would be simplest. Otherwise some sort of DC-DC down converter will be the go. You may have to experiment a bit in the real world out on the table.
Let us know how it all goes.
Cheers Bob

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Brian
I think a 1N4004 diode is about 0.6V so 2 should reduce to 2.5V.
Cheers Bob

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exactly i dont know my math fails me you only need 2 diodes and once there is load then it could drop lower…so i am not 100 % shure of it… its not enough to fit a reg…but …also pin 2 on his nano is 3.3 volt ref you could feed this into a transistor with common collector im thinking…?? but use the 3.3 as a reference point…

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Diodes hold the forward voltage drop pretty well over a wide range of current.
Cheers Bob

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i think 2x 1n5408 diodes in series would cover all the bases,

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So just to be clear the heat generated from the diodes would be the same as the resistor, but you think it is a superior solution as the voltage would be more stable under the variable voltage of the battery as it discharged?

Kenneth

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