Which LED bulb replacement

[msknight]

This has turned out surprisingly harder than I thought.

I have four cheap camping lights that I bought some time ago. The incandescent bulbs are 4.8v 0.5a screw fit. Running from four 1.5v AA batteries in series.

I want to replace with LED bulbs, presumably something that will spread the light wide. I quickly concluded that finding a plug and play replacement was not going to be the right avenue to go down.

I've included the picture of the outside and inside. I've got a reasonable amount of room to play with, so I can presumably mount a ring assembly of LED's rather than a single LED. Hopefully also to slightly under-run the LED's, or get ones of slightly higher power... or somehow balance the power between them, so that they last longer and don't burn out.

I also contemplated running two parallel banks of two AA's in the hope of running 3V LED's, but for longer. And that's where I got completely lost in mathematics, on-line LED shops and what to do about the situation.

Grateful for any thoughts please. - I'm in the UK, if that helps identify sources of components/kits.

[james_s]

How bright do you need it to be? Even just an ordinary 20mA LED with a resistor will be more efficient than the incandescent bulb. If you want a brighter light you could use a driver board meant for a flashlight, those are cheap and widely available, I've converted a few similar lights using that approach. Another option is lithium ion batteries in the same form factor as the AAs.

[tunk]

First remember that a new alkaline cell is ~1.6V, and that the
voltage will decline as you use it. End voltage will depend on
usage scenario, maybe 1V in this case. You can e.g. find some
discharge curves here:
https://lygte-info.dk/review/batteries2012/Ikea%20AA%20UK.html

The original incandescent bulb is 2.5W, and if you assume that
LEDs is five times more efficient, you need 0.5W of LEDs.

You could use a 1W led (plus heatsink) in series with a resistor.
Assuming a 3V LED voltage drop and that you want 0.5W from the LED
at 6V battery voltage, then use a 1W 18ohm resistor (R=U^2/W).
Or use 15-20 5mm LEDs in parallel, running each at ~10mA.
In both cases the light output will decline as the battery is
discharged.

[msknight]

I think a driver board for a flashlight as it might enable me to put small reflectors around and bounce some light sideways rather than having a single LED which is blocked by the joints at either end. At least, that's my thinking. Always open to learning!

[james_s]

You can get LEDs with different emission patterns. You can also put a cap or diffuser of some sort over the LED. Years ago I went looking for something like that and someone gave me some ink cups used by tattoo artists and those worked quite well. Not being into the tattoo scene it was not something I was aware even existed.

[msknight]

OK - I think my target will be a 1W LED, with heatsink, running from 3V. Just got to calculate the resistance and work out which LED to buy, and where from!

[msknight]

I'm looking at these...
https://www.switchelectronics.co.uk/white-10mm-led-diffused-5000mcd-40

Firstly, max forward voltage of 3.6v so should handle fresh batteries. Should hopefully also not require a resistor or heat sink....

... or am I understanding everything completely wrong.

(I am concerned about the 40 degree viewing angle - but at £0.13 each for 100, I can easily get a few and put them in parallel... yes?)

[mariush]

Well, first you have to see what you have to work with.

You have 4 AA batteries, so depending on what batteries you put there you may have :

* rechargeable ni-mh : 1.1v ... 1.35v x 4 = 4.4v ... 5.4v but 1.2v..1.25v per cell is the average voltage
* non-rechargeable :  1v ... 1.65v =  4 .. 6.6v  but 1.4v..1.5v per cell is the average voltage

However note that the bulb will be a significant enough load that the output voltage of the batteries will drop a bit, so it's unlikely to have 1.35v or 1.65v for any significant amount of time.

So let's just say your input voltage will be between 4.5v and 5.5v

Your incandescent bulb with nominal voltage of 4.8v works because that 0.5A of current is enough to lower the voltage of the batteries a bit and the filament can handle a bit of over voltage and can work just fine with less voltage (it will simply not shine as bright)

If you go with leds, the white leds typically have a forward voltage of 3v..3.25v typically so with your input voltage of around 4.5v ...5.5v you can't really put two leds in series because you'll need at least 6v...6.5v to make the leds shine bright.

You can put multiple leds in parallel or you can use a single led, but you must limit the current going through the led or leds otherwise they'll blow up - 4 aa batteries have enough of a current output and low internal resistance that they can do that.

The easiest way to limit the current would be a simple resistor ... you use the Ohm's law Voltage = Current x Resistance to figure out the value of the resistor you need to put in series.

Let's say you have a simple white led with 20mA  (0.02A) of current and a forward voltage of 3v and our input voltage is 4.5v ..5.5v  .. we can put the numbers in our formula :

Input voltage -  (number of leds in series x forward voltage ) =  Current x (number of leds in parallel)  x Resistance

5.5v - (1 led x 3v ) = 0.02A x Resistance  ->  Resistance = 2.5 / 0.02A  = 125 ohm , so a 120 ohm resistor could be used.

When the voltage of the batteries goes down to 4.5v,  the current through the led will go down to   

4.5v - 1 led x 3v = Current x 120 ohm => current = 1.5 / 120 = 0.12A or 12 mA

You can also figure out how much power is wasted in the resistor : Power = Current x Current x Resistance  = 0.02 x 0.02 x 120 = 0.048w

So we have 0.048w wasted in resistor and we have 3v x 0.02A = 0.06w consumed by the led ... pretty poor efficiency. You'd be wasting around  half the battery in that resistor, making heat.

You can use the same formulas to figure out resistor for multiple leds in series... for example let's say 5 leds in parallel to get 100 mA :

5.5v -  (1 led in series x 3v forward voltage ) = 5 leds x 0.02a x Resistance -> Resistance = 2.5 / 0.1 = 25 ohm so maybe go for 24 ohm resistor.


The easiest way to make it more efficient would be to simply lower the number of batteries from 4 to 3 ... simply use a wire to connect the + and - of the 4th battery location to reduce the voltage to 3 x 1.2v = ~ 3.6v for rechargeable batteries or 3 x 1.5v = 4.5v

This way you can use much smaller resistors and you'll get less wasted energy in the resistor.

[mariush]

I'm looking at these...
https://www.switchelectronics.co.uk/white-10mm-led-diffused-5000mcd-40

Firstly, max forward voltage of 3.6v so should handle fresh batteries. Should hopefully also not require a resistor or heat sink....

... or am I understanding everything completely wrong.

(I am concerned about the 40 degree viewing angle - but at £0.13 each for 100, I can easily get a few and put them in parallel... yes?)

Those are only rated for 30mA ...

There's https://cpc.farnell.com  which should have cheap or free shipping in UK  (but the website is down right now)
There's also https://tme.eu which should ship to UK

white power leds sorted by price : https://www.tme.eu/en/katalog/white-power-leds-emiter_113366/?s_field=1000014&s_order=asc&limit=20&currency=USD&page=1
 
white surface mount leds sorted by price : https://www.tme.eu/en/katalog/smd-white-leds_113362/?s_field=1000014&s_order=asc&limit=20&currency=USD&page=1

 through hole leds (pick the diameter or shape from the left, then filter to see only white) https://www.tme.eu/en/katalog/tht-leds-super-flux_112904/

If you want around 1w, then you're looking at leds with around 1w / 3v = 300-400mA  or paralleling 2-3 leds with around 100mA current.

Here's some suggestions :

$0.04 each 2.8v .. 3.6v  max 150mA  50-58lm  3030 package  : https://www.tme.eu/en/details/pj2n-ffve/smd-white-leds/prolight-opto/
 
Run them at 100mA and they'll last a long time.  They're 120 degree, if you want flashlight you probably want narrower angle.

$0.06 each 2.9 .. 3.3v max 120ma  54-59lm like the leds used in lcd monitor backlights : https://www.tme.eu/en/details/ltw-5630azl50lo_h4/smd-white-leds/liteon/ltw-5630azl50-xx-lo-h4x/

$0.07 @ 25pcs 3v..3.4v 120mA 68lm  https://www.tme.eu/en/details/ltw-2835azl40/smd-white-leds/liteon/


if you can handle higher forward voltage at 5.8v ..6.6v you have leds 3x as much luminosity at around 104-135lm
0.08 each,  150mA : https://www.tme.eu/en/details/ltw-3030dzl40/smd-white-leds/liteon/

[msknight]

Thank you very much for all the detail. That'll definitely get me going.  I did look on Switch Electronics which had some warm white at 350ma - https://www.switchelectronics.co.uk/warm-white-1w-star-high-power-led-105lm-120 - but the cold white are out of stock everywhere it seems.

That's given me some wonderful maths to work through and learn from. Thank you. I dare say I'll still get a few things wrong, but that's part of the learning process.

[msknight]

So... given that warm white LED from Switch Electronics ... four AA batteries together, 6v... minus the forward voltage of 3v gives 3v. Forward current of 350mA ... 3 /0.35 = 8.57 ohm resistor. Doesn't seem like much. Also, what about the peak forward current? Is that where it goes to when it starts?

[mariush]

You have datasheet there on the page  : https://www.switchelectronics.co.uk/pub/media/pdf/TYHP1WWS.pdf

peak current = 1000ma , it says conditions on page 3 : Peak Forward Current(1/10 Duty Cycle 0.1ms Pulse Width)

Basically read it as "the led is capable of surviving running for 0.1 ms at 1000mA afterwards it must have some time to cool off."
Think of it like flash bulb on a digital camera.

worth checking graphs on page 4 .. see top right graph ... the forward voltage will vary according to current (and also temperature but to a lesser degree)

By that graph at 300ma the forward voltage will be around 3.3v .. so 3.3v x 0.3 = 1 watt

Keep in mind that you should aim to waste as little power as possible in resistors.
If you go with 4 AA batteries and 6v input voltage, you're gonna waste half the battery capacity in that resistor.

6v - 3.3v = 0.3A x resistor  => r = 9 ohm  ... so you'll use either 8.2 ohm or 10 ohm

8.2 ohm will give you (6-3.3)/8.2 = 0.329A

When the batteries discharge to around 1.2v (or if you use rechargeable alkaline) you'll have 4.8v input voltage and then you're gonna have
4.8v-3.3v = current x 8.2 ohm  => current = 0.183A

Power dissipated in resistor is P = Current² x R
At 6v, you'll have p = 0.329x0.329x8.2 = 0.88w  which is too close to 1w so you're gonna have to use a 3w rated resistor to keep it cool.
But basically, you're gonna consume 1w on the led, and 0.88w on the resistor half the battery is wasted as heat in resistor.

Lower the batteries to 3 AA or a lithium battery and then you have 3.6v..4.5v or 3...4.2v for lithium - there's cheap 1-2$ boards on ebay with lithium battery chargers with usb port and you can buy lithium batteries.

[msknight]

So, with three batteries, say 4.5v ... that's 4.5-3.3 = 1.2  divided against 0.3 gives 3.66 ohms resistance ... so power is 0.33 and I use a 1w resistor. Yes?

If so, and quarter watt resistors is all I have, then four 4ohm resistors in parallel should do it for testing, rather than having to buy new resistors until I know it will work. Yes?

[msknight]

I've got some of those coming, but only the warm light was available. I'll try cutting off the yellow cover and see what happens.

But I've also decided to buy some of these.. I mean, at £0.11 pence each for 100, why not...
https://www.switchelectronics.co.uk/white-10mm-led-diffused-5000mcd-40

I'm intending to drive those from two banks of two AA, so 3V. At 30mA each, if I wire some in parallel I should get a good light and last longer and I shouldn't need a resistor in line... if I understand correctly. It will be interesting to see what happens.

[Wallace Gasiewicz]

For a more even light from the LEDs, use some fine sandpaper on the plastic LED. It will diffuse the light better. The light from the LED is only in one particular direction.
A frosted plastic envelope will help disperse it.

[tunk]

if I wire some in parallel I should get a good light and last longer

If you look at fig 1 in the datasheet, you'll see that it won't light
below 2.5V, i.e. only using ~50% of the capacity. It should be cheaper
to use three cells in series plus a resistor, i.e. 100% from the battery
minus ~1/3 lost in the resistor.

and I shouldn't need a resistor in line!

From the datasheet, the forward voltage can be 2.8-3.6V. If it's 3.6V,
the LED will barely light at all. At 2.8V you risk letting out the magic
smoke.

Also, I think the LEDs will last longer if you don't run them at
max current. They're also more efficient (fig 2).

[CaptDon]

Wow, that is a lot of calculations and assumptions etc. etc. There are many choices of 'flashlight' style LED's that include a control circuit on the wafer and the light will remain constant across a wide range of input voltage. Also the control 'assuming it's PWM or similar' circuit is WAY more efficient than a dropping resistor will ever be!!
Look at some of the specs that read "For 3 to 7 C or D sized cells". Flashlight retro-fit kit.

[tunk]

Or get something like this (the 1-mode type):
https://www.ebay.com/itm/313691735443?hash=item49097c0593:g:wWIAAOSwq05dTn5R
They have a 0.15ohm resistor. If this is for current sensing,
you probably could modify it to a lower wattage.

Or this:
https://www.ebay.com/itm/293662523306?epid=3008654957&hash=item445fa67faa:g:tGwAAOSwV1lfQ7I0
Since this adjustable voltage only, you should add a resistor
(maybe 0.5-1ohm) in series and adjust the voltage to a suitable
brightness.

Both these should be more efficient.

[CaptDon]

You do want some kind of current-sensing / current control circuit. A dropping resistor alone sucks because it wastes power and the light output will vary drastically. A voltage-only regulator is also a fail because LED's behave much like a zener diode, increase the voltage just a tiny bit above the turn-on threshold and the current skyrockets. I think it was Luxeon who made drop-in replacements with built-in current control to replace PR-2 type flashlight lamps and they also made screw-in replacements for screw versions of #44 / #47 bulbs. I used their products in scuba diving lights and was well pleased. No fear of burn-out in cave diving and huge extension to battery life!!!
 

[msknight]

One of the problems I've had, is that these bulbs don't have a "#" marking on them, so I can't easily find drop in replacements. I mean, searching for "#47 bulb" comes up with what I need, but searching for "led 4.8v 0.5a screw bulb" ... forget it.

[msknight]

...but even then, the question is, do I go for an LED replacement bulb which is 6v and hope that it operates when the voltage starts to drop, or do I go for  a 4.8v replacement and hope that it can handle over volting?

[wizard69]

I think people are over thinking this.   The first thing I would do is to find out what socket type you have and the current bulbs part number.   With that information go looking for LED replacement bulbs.   Dialight for example has the '586 VIP Series series tht fits an E10 / T3-1/4 socket and operates form 6 to 36 volts.   Now 650 mcd may not be enough, I have no idea how to translate that into the output of an old incandescent.   If that is not enough light, fixed voltage replacement bulbs often have far higher light output.

I realize that buying a bulb that is already setup to run may not be as much fun as a full blown electronics project.  I just see this as a reasonable approach, if you can find a "bulb" that makes you happy.   Anything different in my mind depends upon the design of the lantern.   How adaptable it might be is unknown to us.    Also there is the perfect forum for this: https://www.candlepowerforums.com/.   They are off the deep end when it comes to flashlights and similar.

[james_s]

Those Dialight bulbs look like just a standard 5mm LED inside a transparent plastic bulb. They're clearly intended as indicators rather than for general illumination. The rating in mcd suggests this too, it is a measure of luminous intensity rather than lumens which is what you usually want when speaking of illumination. A laser pointer has extremely high luminous intensity but is essentially worthless for illumination because the total amount of light is low.

[Rick Law]

I think a driver board for a flashlight as it might enable me to put small reflectors around and bounce some light sideways rather than having a single LED which is blocked by the joints at either end. At least, that's my thinking. Always open to learning!

I think this would be complicated, but a great choice to learn.  One issue is, your lamp is currently 4xAA.  4x1.5V is 6V and higher when brand new.  This is higher than the typical flashlight driver would be rated.  If memory serve, for the NANJG 105c, the AMC 7135 is rated at 6V and the MCU (ATTNY 13a) is 5.5V.  You will need to switch to NiMH recharge, that would be 1.2V per cell but coming off the charger it could be > 1.6V per cell (6.6V) or more for a moment.  The LED load should pull it down soon enough - just don't start with max brightness when batteries are fully charged.  Better yet, make yourself a "pass through" by shorting one of the battery slots so you run 3xAA at 4.5V.

A driver board like the NANJG 105c is even a platform you can modify hardware-wise and software as well. You can of course modify the native firmware with custom firmware.  That is a lot of learning you can do there.  I use an Arduino NANO as the ISP for my NANJG 105c's.

For NANJG 105c, you can change how many AMC7135 you put on it to control the total current (each gives you 350mA, you can have up to 8 on the driver).  The native firmware will give you brightness control.

Besides the AMC7135 count, you can even upgrade the MCU.  When you get really ambitious, you can swap the ATTINY 13 to an ATTINY 85 with more programming space.  The ATTINY 85 will fit on the NANJG 105c.  You need to bend the pins a little, but it will fit.  I have 1/2 dozen of them upgraded from ATTINY 13 to ATTINY 85.  All programmed with custom firmware. 

For LED, you can use multiple in parallel or choose a single LED like a Cree R5 T4 T5 T6 U2...  They have angle probably wide enough for camping or head lamp.  For single LED, expect to need additional heat sink for >350mA If you are not putting it into a flashlight with aluminum housing.  The tiny aluminum plate the emitter is soldered on is not adequate when without that flashlight housing as addition heat sink.

I've 3 running Cree XML2 (10W @ 169lm/Watt per spec) and some running XML T6.  Their XML-G (6W @ 199lm/W) and the XML-E2 (3W @ 137lm/W) series are lower power good single LED solution.  At least two are real Cree from Mouser, a few with a Chinese made compatibles (LightBright or some brand like that).  The real Crees are brighter (using a lux meter to measure), but not much, just about 10% to 20%.  So I use the much cheaper compatible ones for less demanding application (ie: the ones that doesn't leave the house and can double-up when I want it really bright).

If you want really wide, perhaps glue (with heat sink glue)  three or four R series (older cheaper) or four E2 series onto a single aluminum plate.

The learning opportunity there is big, but complicated getting there.  If you just want to start simple, string LEDs in series and put on a ballast resistor and start there.  Next step would be shorting out a battery slot (4xAA to 3xAA), modify the number of LEDs and change the ballast, now it is suitable for a native un-modified flashlight driver like the NANJG 105c.  Get two, now you can learn to program the other one while one serve as the working one...  Next round, you have your custom firmware.

[james_s]

I don't remember what driver I used but I have a lantern I converted to LED years ago that is run off of 4 AA cells and uses a single Cree XML emitter. Actually I have a smaller one with I think the same driver, the same 4 cell arrangement with a single blue LED and a remote phosphor dome. These both have proper buck regulator drivers.