3.6V rechargeable batteries for 3.2V LEDs

starsend · Post by **starsend** » Fri Apr 20, 2012 8:22 am

Hey all, I've been working on my 1/1000 TOS E for the past few days and all is going smoothly. I will end up running quite a few LEDs when all is said and done and thought I would try to use 3 AAA rechargeable batteries instead of 2 AA. Now the AA batteries are rated at 1.5V and so run a bit lower than the LEDs are rated for (3.2/3.4) and I would like things to be a bit brighter. But, 3 rechargeable AAA rated at 1.2V each will give me 3.6. The reason I can't use resistors is that I already sealed up the saucer and there's no way of adding the resistors now. Besides, there's no room for them! here's a link to the build:

http://www.starshipmodeler.net/talk/vie ... hp?t=99767

So the question is, will it be safe to run the kit off 3 batteries giving me a total of 3.6V or will I burn the LEDs? The only led needing a resistor is the red one which has a 56 ohm one installed. And that's been sealed up too. I have tried to very briefly connect the 3.6V battery pack and everything works but I am afraid I will be driving the LEDs too hot.

Thanks for your advice!

Joseph Osborn · Post by **Joseph Osborn** » Fri Apr 20, 2012 8:39 am

If your batteries are outside of the model, can't you just add the resistor(s) in-line with the power supply wiring going into the model?

I understand that you always put a resistor with LEDs to keep them from getting too much current (not voltage). It has been my experience that over-driving a LED, especially without a resistor, will result in an LED that glows very brightly, then soon starts dimming... and dimming... and then winks out.

starsend · Post by **starsend** » Fri Apr 20, 2012 9:18 am

But how do I figure out what resistor to use? I have all these leds in parallel that in theory would require each their own resistor. Hope this makes sense. Do you happen to know the formula to find the appropriate resistor to use in such a situation? Thanks!

Joseph Osborn · Post by **Joseph Osborn** » Fri Apr 20, 2012 10:35 am

The various online calculators should provide the answers. I have used the calculators here often: http://led.linear1.org/led.wiz

starsend · Post by **starsend** » Fri Apr 20, 2012 11:54 am

I know how to calculate resistors for a parallel or serial configuration, but I am not sure how to go about adding the resistor between the power supply and a parallel circuit already built. In parallel you want one resistor per led, but as I explained above, I can't do it. It will have to be one resistor for the whole thing, which I am not sure how to calculate...

Joseph Osborn · Post by **Joseph Osborn** » Fri Apr 20, 2012 12:49 pm

Okay, I think I see... you have the LEDs in your saucer tied to a distribution block of some kind inside the saucer itself and you only have a positive & negative wire coming out of the sealed saucer. There is a way to calculate the value of the resistor to use in this case, but I don't know how to do that. I would defer to someone who has more knowledge of electrical engineering than me

starsend · Post by **starsend** » Fri Apr 20, 2012 1:27 pm

Yes, you got it. Thanks anyway for trying to help1 If anyone knows, please feel free to illuminate us!

tetsujin · Post by **tetsujin** » Fri Apr 20, 2012 1:48 pm

If the circuit inside your sealed model is built to work at a specific voltage, and your supply doesn't provide that specific voltage level, get a voltage regulator. That'd be the best way.

starsend · Post by **starsend** » Fri Apr 20, 2012 1:57 pm

Well, the thing is that I have been using 2AA batteries for a total of 3V. That is a bit below the 3.2/3.4V suggested for my LEDs. But it works. The problem is that as I add more less I am noticing that I would like a bit more light. I don't have a 3.4V power supply and I would like to know if there is a way to bring my 3.6 supply (3 rechargeable AAA 1.2V each) down a bit with a transistor between the power pack and the actual circuit. Or even better maybe someone could tell me that 3.6 is just fine and to just plug it in and it will not burn my whole ship!

steam235 · Post by **steam235** » Fri Apr 20, 2012 2:16 pm

Yes the answer here it's to add one large resistor to the negative post of the battery box. This will drop the voltage from 3.6 to 3.2 . Then the resistor on the red led will drop the current for that led to what you Calculated before. To figure out the resistors value add all currents for all the Leds in the parrelell circuit then use a online resistor calculator substituting your entire circuits values for a single led.

tetsujin · Post by **tetsujin** » Fri Apr 20, 2012 2:22 pm

starsend wrote:Well, the thing is that I have been using 2AA batteries for a total of 3V. That is a bit below the 3.2/3.4V suggested for my LEDs. But it works. The problem is that as I add more less I am noticing that I would like a bit more light. I don't have a 3.4V power supply and I would like to know if there is a way to bring my 3.6 supply (3 rechargeable AAA 1.2V each) down a bit with a transistor between the power pack and the actual circuit. Or even better maybe someone could tell me that 3.6 is just fine and to just plug it in and it will not burn my whole ship!

Check the specs on your LEDs. That will tell you how much current will flow through 'em for a given voltage level, and whether it's a reasonable amount of current for the device.

Alternately, try the batteries, but with an amp-meter hooked up in line. Measure the total current through the circuit. If you leave it hooked up just long enough to get a reading with the amp-meter it's unlikely you'll do any damage to the LEDs. Then divide by the number of LEDs and you'll know how much current each one is getting. Personally, not knowing specifics about the LEDs I'd be wary of giving them more than 25mA for any extended period of time.

Remember, though, if you do that test, you're testing those batteries at that charge level. If the ship draws enough current, the batteries may not actually provide their full 3.6V. (This is an effect known as "internal resistance" - basically it means that there's a limit to how fast the battery can supply current, and as it approaches that limit it will exert less force to move the current - less voltage. But it's often modeled as "a resistor inside the battery" which is easy to understand.) Depending on how much current your ship draws, the "internal resistance" of the batteries may be enough that you don't need to worry about regulating the voltage down.

On the subject of regulators - 3.3V regulators are very common and inexpensive. Most likely 3.3V will be perfectly safe for your 3.2V LEDs. If you install such a regulator in your model's base, you can hook up almost any DC power supply as input and safely operate the model's lights.

Finally, it's important to understand that the voltage/current relationship on a LED is not linear - it's exponential. Raise the voltage from 3.2V to 3.6V and you could get 40mA instead of 20mA through the LED. This is part of why we waste power with resistors: it helps to linearize the relationship and stabilize it, make it less sensitive to minor changes in voltage.

tetsujin · Post by **tetsujin** » Fri Apr 20, 2012 2:38 pm

steam235 wrote:Yes the answer here it's to add one large resistor to the negative post of the battery box. This will drop the voltage from 3.6 to 3.2.

A problem with this approach is you're dropping a relatively small amount of voltage (0.4V) with a fairly large amount of current (20mA for each LED - I don't know how many LEDs there are, though... Let's suppose it's ten.)

So V=IR, 0.4V = (20mA * 10) * R, so R = 2 Ohms.

The power dissipated by the resistor isn't an issue (V * I = 0.08 Watts) but 2 Ohms is not a terribly common rating for a resistor.

If you used a 1.5 Ohm resistor, you'd get about 27mA through each LED (which would probably be OK, if a bit on the high side...)

If you used two 4.7 Ohm resistors in parallel (to get a 2.3 Ohm resistor) you'd get about 17mA through each LED (reasonable...)

Using such a small resistor would mean the voltage/current relationship remains very unstable, however. I don't know if the batteries, freshly-charged, might supply more than 1.2V each, but if they did, even a half a volt over (from the three cells in series - so overage of about 0.2V each cell) could be way too much.

Again, personally I'd just use a regulator. Maybe in conjunction with a fourth AA cell, which could keep the ship powered for a longer period of time on a charge (since the lights would continue operating until the battery supplies less than 3.3V, starting with 4.8V means you'd be able to keep operating even when the battery voltage starts dropping from discharge...) It's also a great "fire and forget" solution. Maybe some day you want to power this thing with a wall wart? Not a problem if you've got a regulator in the base...

starsend · Post by **starsend** » Fri Apr 20, 2012 2:57 pm

steam235 wrote:Yes the answer here it's to add one large resistor to the negative post of the battery box. This will drop the voltage from 3.6 to 3.2 . Then the resistor on the red led will drop the current for that led to what you Calculated before. To figure out the resistors value add all currents for all the Leds in the parrelell circuit then use a online resistor calculator substituting your entire circuits values for a single led.

Thanks, we are talking mA and not Voltage I assume. I'll try and post the result. But you've nailed my problem exactly! Thanks so much.

Tesujin, thanks for the great post. I'm very new to all this and I appreciate the generosity on this board very much. My very first kit was only a few months ago, a 1/1000 refit and I lighted it without knowing much about any of this. That was done with no resistors either, but with two AAA and it came out alright. But I will need to learn eventually

.

The LEDs are from eBay and it says 3.4/3.2 at 20mA. I am not really sure about mA. I don't understand exactly how it increases in relation to the voltage. What I think I know is that as long as he voltage is at or below the forward voltage of the led there is no risk of the circuit drawing more than the necessary amperes. Is that right?

I'll get myself a meter, I think ill need one eventually. S basically you are saying to check for 1. That the mA with the 3.6V battery pack does not exceed 25mA, and 2. that the voltage is actually 3.6 as if it were 3.3 I would be safe. Am I correct?

As for the regulator, they are small components that look a bit like capacitors? I assume you just out it in the base between the power source and the two wires coming down from the stand? If I use that, could I move to a 3.7 or 4.5V power supply? I have a few in a box and it would be easier than using batteries.

Again thanks so much for your time and help!

starsend · Post by **starsend** » Fri Apr 20, 2012 3:12 pm

Steam, I tried your suggestion with 8 LEDs at 20mA each, 3.6 power supply and 3.2 forward voltage. I got as a result a 2.7 ohms resistor. Is that right? I don't know much but doesn't that mean that a resitor might not be necessary after all?

tetsujin · Post by **tetsujin** » Fri Apr 20, 2012 3:26 pm

starsend wrote:Steam, I tried your suggestion with 8 LEDs at 20mA each, 3.6 power supply and 3.2 forward voltage. I got as a result a 2.7 ohms resistor. Is that right? I don't know much but doesn't that mean that a resitor might not be necessary after all?

Whether you need a resistor depends on the "equivalent internal resistance" of the batteries. That is, if the battery still provides around 3.6V when you're draining 320mA from it, then operating without a resistor isn't safe.

Personally, I wouldn't chance it. How much work went into the model? What would you do if one or more of those LEDs burned out? Would you even be able to replace a LED without destroying the model? Reasoning like that is the reason why I err on the side of caution when it comes to choosing the power supply...

starsend · Post by **starsend** » Fri Apr 20, 2012 3:44 pm

tetsujin wrote:
starsend wrote:Steam, I tried your suggestion with 8 LEDs at 20mA each, 3.6 power supply and 3.2 forward voltage. I got as a result a 2.7 ohms resistor. Is that right? I don't know much but doesn't that mean that a resitor might not be necessary after all?
Whether you need a resistor depends on the "equivalent internal resistance" of the batteries. That is, if the battery still provides around 3.6V when you're draining 320mA from it, then operating without a resistor isn't safe.

Personally, I wouldn't chance it. How much work went into the model? What would you do if one or more of those LEDs burned out? Would you even be able to replace a LED without destroying the model? Reasoning like that is the reason why I err on the side of caution when it comes to choosing the power supply...

Oh I will! It was just to keep the conversation going and stealing more wisdom from you guys! If you get a chance to read my previous reply to your great post and have the time I would love to hear your replies to my questions. Again thanks a lot to both of you for the help!

Bellerophon · Post by **Bellerophon** » Fri Apr 20, 2012 5:54 pm

tetsujin wrote: If you install such a regulator in your model's base, you can hook up almost any DC power supply as input and safely operate the model's lights.

Good advice to mount the regulator outside the model, for a reason tetsujin doesn't mention here: voltage regulators can be really hot, depending on the load and how much it's dropping the voltage by. I have gotten some really nasty burns on my fingers, no big deal, they heal. But styrene doesn't.

starsend · Post by **starsend** » Fri Apr 20, 2012 6:12 pm

tetsujin wrote:
steam235 wrote:Yes the answer here it's to add one large resistor to the negative post of the battery box. This will drop the voltage from 3.6 to 3.2.
A problem with this approach is you're dropping a relatively small amount of voltage (0.4V) with a fairly large amount of current (20mA for each LED - I don't know how many LEDs there are, though... Let's suppose it's ten.)

So V=IR, 0.4V = (20mA * 10) * R, so R = 2 Ohms.

The power dissipated by the resistor isn't an issue (V * I = 0.08 Watts) but 2 Ohms is not a terribly common rating for a resistor.

If you used a 1.5 Ohm resistor, you'd get about 27mA through each LED (which would probably be OK, if a bit on the high side...)

If you used two 4.7 Ohm resistors in parallel (to get a 2.3 Ohm resistor) you'd get about 17mA through each LED (reasonable...)

Using such a small resistor would mean the voltage/current relationship remains very unstable, however. I don't know if the batteries, freshly-charged, might supply more than 1.2V each, but if they did, even a half a volt over (from the three cells in series - so overage of about 0.2V each cell) could be way too much.

Again, personally I'd just use a regulator. Maybe in conjunction with a fourth AA cell, which could keep the ship powered for a longer period of time on a charge (since the lights would continue operating until the battery supplies less than 3.3V, starting with 4.8V means you'd be able to keep operating even when the battery voltage starts dropping from discharge...) It's also a great "fire and forget" solution. Maybe some day you want to power this thing with a wall wart? Not a problem if you've got a regulator in the base...

Sorry testujin, for some reason your post didn't show up on the page while I was replying. Regulator it is then, and I,ll go with a forth battery (or maybe use normal 1.5v batteries and use 3 of them for a total of 4.5V).

EDIT: is this what I am looking for?
http://www.diotronic.com/semiconductore ... 16776.aspx
It's great to know these even exist. One final question for future reference is the following: is it correct that as long as the voltage of the power supply is equal or less than the forward voltage of the led I won't need a resistor? I don't mind using them but when space is a consideration, it will be useful to know how safe it is not to use them. In this case for example, if the regulator brings the voltage down to 3.3, can I seal the model and not worry about the missing resistors in the saucer?

rayra · Post by **rayra** » Fri Apr 20, 2012 10:43 pm

starsend wrote:I know how to calculate resistors for a parallel or serial configuration, but I am not sure how to go about adding the resistor between the power supply and a parallel circuit already built. In parallel you want one resistor per led, but as I explained above, I can't do it. It will have to be one resistor for the whole thing, which I am not sure how to calculate...

The method doesn't matter so much as your problem forces you to treat the entire setup as a single source of power consumption. Add up the total load values and treat it as a single LED, then try some experiments with resistors on the very low end 50-150ohm and see that you maintain sufficient power that all your LEDs function. You could experiment and increase the resistance until a light fails to emit, then go back one step and be at the level where the least overvoltage is hitting your other lights, to do the best you can with the lifecycle issue.

starsend · Post by **starsend** » Sat Apr 21, 2012 1:14 pm

I went out and got all the supplies: a multimeter, TS1117 voltage regulator, etc.
At the store I was told to connect the positive from the battery pack to the left pin of the regulator, and to come out from the regulator at the right pin again with the positive to continue on. the middle pin according to him was for the negative lead. I tried that and in two seconds the regulator was very hot and while I could read 4.1V from the battery pack, I got 0.3V coming out. Something must be wrong as I'm supposed to get 3.3V coming out correct?

EDIT: I think I was shorting the circuit before so I tried red in and out, leaving the middle pin empty. Black straight to The meter... Now I'm reading 4V exactly as if I had no regulator in the chain... Not sure where I'm screwing up. Google not helping either.

Thanks Rayra, I think I might end up doing the resistor thing as it seems easier. Just for the saucer, as the rest of the ship still hasn't been built and I will use resistors for each led.

starsend · Post by **starsend** » Sat Apr 21, 2012 3:27 pm

It's working! I was simply given the wrong information about how to hook it up and was shorting all the time. In the end I figured it out and got the drop from 4.5 to 3.3v as you can see in the picture.

http://i1075.photobucket.com/albums/w43 ... 2b1909.jpg

What I'll do now is follow Rayra and steam's advice for the saucer and place a small resistor by summing the current to all the LEDs in it and inputing that into a calculator for a power supply of 3.3V. For the rest of the LEDs that I have yet to mount in the rest of the ship I will calculate resistors for all of them starting from the the 3.2V/20mA specs for each led (and of course a power supply of 3.3V as well).

Thank you so much all of you guys for helping a noob like myself out. I'll do the same if anyone ever asks for advice.

rayra · Post by **rayra** » Sat Apr 21, 2012 6:57 pm

good deal. I missed that the whole build wasnt' complete. Makes the rest easier and you can tailor everything to the same feed voltage and get all your parallel circuits running off the same power source.

steam235 · Post by **steam235** » Sun Apr 22, 2012 8:17 am

rayra wrote:good deal. I missed that the whole build wasnt' complete. Makes the rest easier and you can tailor everything to the same feed voltage and get all your parallel circuits running off the same power source.

Yes for most of mine I go to mouser.com and filter by voltage. Then you can find several sizes running on the same voltage. Then you can do a single resistor or just a regulated power supply and not worry about resistors on every LED. Then paint the nav lights with clear red, clear green or clear blue. Simple cheap easy. but no blinking or anything more fancy.