# regulating voltage from 2 9v batteries

Discussion in 'The Projects Forum' started by cmmig, Apr 14, 2006.

1. ### cmmig Thread Starter New Member

Apr 14, 2006
6
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My Project: 3 series of 5 3.6v Leds, with a switch on each series; powered by 2 9V batteries, with a 27ohm resistor on each series

When I first tested the 2 9V bateries i got ~18.4 volts, more than i expected. I was hoping for exactly 18 volts since that would power 5 leds(5*3.6=18). Because of the extra voltage i put the 27ohm resistors in. Now my problem is that the batteries are putting out <18 volts, reducing the brightness of the LEDs, and this is exagerated by the resistors.

the LEDs i am using:
http://www.lsdiodes.com/shop/index.php?mai...1&products_id=1
http://www.lsdiodes.com/shop/index.php?mai...&products_id=15
http://www.lsdiodes.com/shop/index.php?mai...&products_id=19
They all draw 3.6v, but the white has a current draw of 25 instead of 20.

How can i regulate two 9v batteries to put out exactly 18V? This way i can leave out the resistors. And does the current draw matter?(i.e. 1,2, or 3 series could be on)

The only things i can think of are capacitors or the LM317T, but i have no idea how to apply either. (http://www.radioshack.com/sm-adjustable-voltage-regulator-lm317t--pi-2062601.html)
For the capacitor would i need one rated at 18v, and would it actually work? and how does one use the LM317T to adjust the output voltage to 18v?

Any help is appreciated.

2. ### Papabravo Expert

Feb 24, 2006
10,154
1,795
All batteries have a discharge curve. This is basically a plot of voltage versus time for some CONSTANT current flow. When a battery is brand new its voltage will be larger than the voltage printed on the package. 9V batteries are about 9.5 to 9.6 volts new. A 1.5V battery is 1.6 to 1.65. As the battery discharges the voltage drops. Most 9V batteries are "dead" when the voltage is below 8V and most 1.5 V batteries are dead at 1.0 to 1.1 Volts.

In your specific problem you want exactly 18V from two batteries that can output maybe 19.2 max. You want to use an LM317 voltage regulator. The problem is that 1.2 volts(19.2 - 18.0) is very nearly the point at which the regulator will drop out of regulation. The pretty good news is that as the regulator drops out of regulation the output follows the input down in a linear fashion.

You will have better luck with a so called LDO (Low Dropout Regulator). These don't stop regulating until the input to output voltage difference drops to 300 millivolts or so.

Hope this helps.

3. ### cmmig Thread Starter New Member

Apr 14, 2006
6
0
Thanks for the help.

So I need to find a LDO with a fixed output of 18V or a variable output capable of 18V. Sounds easy enough. Just to check though, the Vout will remain constant regardless of the load(5, 10, or 15 LEDs)?

Any advice on where to find something like this, radioshack doesn't seem to have much like this(i think they have 5V and 12V, which is more common than 18V). Are there any trusted online stores that carry electrical equipment like the LDO and other components that Radioshack doesn't carry.

EDIT: some spelling errors

4. ### n9352527 AAC Fanatic!

Oct 14, 2005
1,198
4
May I ask why do you need exactly 18V? 18.4V would power the LEDs just fine, the current is regulated by the resistor anyway and less than 3% of extra voltage wouldn't make that much difference.

5. ### cmmig Thread Starter New Member

Apr 14, 2006
6
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I guess it doesn't really matter if the voltage is a little too high, especially once it is spread over 5 LEDs. Ideally there would be something i could put in the circuit to make the current 18v at all times, regardless of the battery draining bellow 18v.

My other question still stands though, is there a reputable online dealer that carries electronic equipment(capacitors, resistors, ICs, PCBs, etc).

6. ### paultwang Well-Known Member

Mar 8, 2006
80
0
Stepping up voltage is more complicated and less efficient than stepping down. If you want your voltage source to be constant 18v even if the source is below 18v, maybe you should consider a regulated AC adapter, or larger mAh batteries.

Some online parts dealers:
www.mouser.com
www.digikey.com
www.jameco.com

I have personally used Mouser, and it is quite fast. Digikey has a more advanced search filter.

7. ### paultwang Well-Known Member

Mar 8, 2006
80
0
Duplicate post.

Jan 10, 2006
613
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My solution... use 4 strings of 4 LEDs (thats 14.4 volts @ 24mA per string) and drop the extra 3.6 volts @ 24mA in each string with a 150 ohm resistor.

The LED's will still dim a little as the Batts go flat, but this is by far the easiest solution

or 5 strings of 3, with a 270 ohm resistor in each string..(better brightness at low batt volts)

If you want more precise current control... i.e. constant current/brightness down to about 16volts, then try an LM317 wired as a current regulator in each 4 LED string. The identical circuit will work on strings of less LED's with constant brightness's down to lower battery voltages.

9. ### cmmig Thread Starter New Member

Apr 14, 2006
6
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Wow thank you Gadget. That seems perfect.

I'll do 3 strings of 4 LEDs with an LM317 on each string.

Am i right in assuming the 50Ω resistor in your diagram will keep the Vout at 14.4v regardless of Vin? I.E. If i put 3 9V batteries (27V) would the Vout remain at 14.4v. Or more importantly is there an equation for Vout i can use so i don't have to keep bothering the nice people here for help. I only found this: http://www.tkk.fi/Misc/Electronics/circuits/lm317.html , but the circuit looks different(2 resistors, capacitors, and power sources).

Jan 10, 2006
613
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The LM317 is wired for Current regulation, and will provide a constant current of around 25mA regardless of the voltage across the LED's which will naturally go to their forward voltage of around 3.6 volts each (14.4 total). This safeguards the LEDs and will provide contant current/brightness so long as there is enough voltage being supplied to power the 317 and LEDs properly..(I think there is normally around 1.5 volts or more required across the 317 to make it work properly... a little more possibly because of the resistor.. so at least 16 or 17volts required.. and should work with up to 35volt across the 317, although the 317 and resistor may get a little too hot). The value of the resistor determines the current via formula I = 1.25/R.
If you added another 9 volt battery, I would advise you add another couple of LEDs to each string without any other modification...as the current will remain at 25mA so long as there is enough voltage to make that 317 work properly as a current regulator but not so much that heat would not be a problem either.....

11. ### cmmig Thread Starter New Member

Apr 14, 2006
6
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It looks like you got the Resistor value(50Ω) from .025=1.25/R (25mA=1.25V/50Ω).

So the 25mA means that each LED will get 25mA(rated for 20-30 or 25-30). The 1.25 is the drop from the LM317. And the R is the 50Ω resistor in your diagram.

OR should the V be 14.4 since thats what the LM317 should be outputting? That would make more sense to me. If so then the equation should be .025=14.4/R; R=576.

Sorry to keep bugging you, but I am interested in circuits and I am just trying to understand as much as I can.

12. ### windoze killa AAC Fanatic!

Feb 23, 2006
605
24
And just another swing on the idea. How about putting 3 9V batteries in series and using a bigger resistor. Start of with a potentiometer and adjust it until you have the brightness where you want it and then measure the value and replace it with a resistor of a similar value.

Jan 10, 2006
613
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Actually, that was my original suggestion, but I think cmmig likes the idea of a constant brightness even as the battery voltage slowly drops.

Forget about 14.4 volts. The circuit with the 317 is a Constant Current circuit. It will supply 25 mA regardless of the number of LEDs or the supply voltage as long as there is enough spare voltage to keep the 317 operating, and not so much that you damage the 317.
If you have 27 volts, and only 1 LED, the circuit will supply 25mA. As the forward voltage of the LED is 3.6 volts, the output of the 317 will drop to 3.6 volts, as it can't supply enough current to overload the LED and raise the forward voltage up above the LEDs normal forward voltage.
Put 2 LED's in there, still 25mA, but the voltage will only be pulled down to 7.2 volts.
3 LED's, voltage pulled down to 10.8
4 LED's voltage pulled down to 14.4
5 LED's voltage pulled down to 18 volts (still 25 mA)
6 LED's voltage pulled to 21.6
7 LEDs Voltage pulled to 25.2.... although it will be somewhere around or just above this point that the 317 may stop working correctly, and the current will drop off rapidly.

Just think of each LED as a 3.6 volt Zener diode that glows....

14. ### cmmig Thread Starter New Member

Apr 14, 2006
6
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Well i am ordering the LM317 and a 500pack of resistors from radio shack today so ill be getting all the parts within a week. And Gadget was right about me wanting constant brightness despite the battery voltage drop. Ill be using the circut that Gadget drew with the LM317 and 50Ω resistor. Thanks to all who posted for all your help.

15. ### PDubya Member

Mar 16, 2006
28
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Could I use the LM317 to control the output of a wall-wort, or should I go with something different?

Jan 10, 2006
613
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What, may I ask, is a "wall-wort".... Perhaps I know it by a different name..?????

If so, then yes, the 317 can be wired for it's normal function as a voltage regulator (instead of the current regulator described) and can regulate from 1.2 up to around 30 volts at up to an amp or so depending on heatsinking.

17. ### PDubya Member

Mar 16, 2006
28
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Trust me, I thought it was an odd name too! I assumed it was more specific than "plug in transformer" or the like. I had someone on another electronics forum give me a hard time about calling it the wrong thing <shrug>.

Are there any considerations I would have going from a 9v "wall adaptor" -> 9v output voltage from the regulator? My biggest issue now is that the adaptor is supposed to be 300ma, but is waaaay above that - plus when used on circuits with just simple groups of flashing LED's driven by 555's, is really irratic.

Thanks for the help!

18. ### PDubya Member

Mar 16, 2006
28
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Is the attached diagram close to what I'm thinking I'll need?

[attachmentid=1410]
(tis my first go with pcb123 - be gentle )

If I wanted a fixed 5v output I'd replace R2 with something like a 740ohm fixed resistor?