9volt & resistor??

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Robocop

Robocop

Mammoth Killer
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I am tinkering around with a standard 9volt battery and two 5mm LEDs from keychain lights. I can not find the information searching old threads on this mod and am hoping someone can help me now. I am trying to make something similiar to the pak-lite using two 5mm LEDs for long runtime off of the 9volt cell.

I have various resistors and need to know first should I wire the LEDs in series or parallel and can I use one simple resistor between the positive supply and positive lead of the LED? I would also like to know the value of the resistor to obtain something close to 25mA for decent runtime. The 5mm LEDs are all salvaged from various keychain lights and I assume they should all be pretty close in vf and specs.

I have several small smd resistors however do not know how to figure the formula to calculate the resistor....however I do pretty much know how to read the numbers on the smd resistors.

Seems like a pretty easy and fun mod for a long running night light and I have plenty of 5 mm LEDs and a few of the 9volt connectors. Thanks for any help on this.
 
It's best to wire the LEDs in series. Firstly it makes sure that exactly the same current goes through both LEDs, and secondly it makes more efficient use of the energy from the battery.

The resistor formula depends on the Vf of the LEDs. The Vf is about 3.2 V for white or blue LEDs and about 2 V for red or yellow LEDs.

Also remember the battery voltage will start out about 9 V but decrease to about 5 V when empty. So it is worth calculating for a slightly higher current when the battery is fresh so it averages downwards as the battery discharges.

Let's assume two LEDs in series giving a combined Vf of 6.4 V. We subtract that from the battery voltage (9 V) leaving 2.6 V across the resistor. To run the LEDs at 25 mA there is 25 mA flowing through the resistor, so by Ohm's Law the resistor value is:

R = V/I = 2.6 V / 0.025 A = 100 ohms

At 25 mA you will get about 10-20 hours of run time from a 9 V alkaline battery. Those batteries really don't have a lot of capacity. You could double it by reducing to 10 mA, and you might not notice much difference in the brightness.
 
Mr. Happy I somehow knew you would come to my rescue here and you surely do know your electronics...amazing actually.

I actually somehow believed I would get days of runtime from this combo however now realize I was wrong. Can you take a guess at what current the pak-lites are running to get their claim of 80 hours on high and 600 hours on low? Is it really just as simple as changing the resistor to a higher value?

So if I were to drop the mA to say 10 as you suggest would I simply double the resistor value to 200 ohms? Can you give me the standard numerical code on a smd resistor for 100 ohms? I assumed it would be 101-- meaning 10 plus the 1 on the end showing the number of zeros to add for 10+1 zero or 100 ohms????

Thanks again for your advice on this as you always seem to be able to get me to grasp certain things in my head with your knowlege on the subject.
 
25 mA is a rather high current for a 5 mm led, 5 to 10 mA is a much better value. To get a long runtime you can go much lower.

The general formula to calculate the resistor is:

resistor=(supply_voltage-led_voltage)/current

For 25 mA, a 9 volt battery and two led in series with 3.2 volt each:

resistor = (9-3.2-3.2)/0.025 -> 104 ohm, nearest stand value is 100 ohm

Going down to 5 mA:

(9-3.2-3.2)/0.005 -> 520 ohm, nearest standard is 470 or 560 or for better stocked suppliers 510 ohm.

I have done an electronic calculator that can do these kind of calculations and find a standard value, but it might be to advanced for you job.
 
Holy smoke HKJ that electronic calculator is most likely too complicated for a NASA scientist much less a weekend modder like myself. I do very much appreciate the resistor value and will most likely drop the value to try for 5 or 10 mA as some have suggested....thanks again.
 
Excellent and that does help me a little more...thank you for the link.

I was just playing aroung with different resistors and a single 5mm LED salvaged from an old Dorcy AAA circuit. I used a 1k resistor and attached it directly to a 9 volt cell and it seemed to my eyes to be pretty nice. I can not figure out how to obtain the probable mA with this resistor....can someone tell me about what the LED is pulling with a single 1K resistor.
 
Robocop said:
Holy smoke HKJ that electronic calculator is most likely too complicated for a NASA scientist much less a weekend modder like myself. I do very much appreciate the resistor value and will most likely drop the value to try for 5 or 10 mA as some have suggested....thanks again.
Click to expand...

My program probably look more complicated that it is, but the simple calculator mvyrmnd linked to is much better for your purpose.

With the correct selections, my program looks like this:
MiscelLed.png


Robocop said:
I was just playing aroung with different resistors and a single 5mm LED salvaged from an old Dorcy AAA circuit. I used a 1k resistor and attached it directly to a 9 volt cell and it seemed to my eyes to be pretty nice. I can not figure out how to obtain the probable mA with this resistor....can someone tell me about what the LED is pulling with a single 1K resistor.
Click to expand...

My formula (ohms law) can easily be changed for that:

current=(supply_voltage-led_voltage)/resistor

(9-3.2-3.2)/1000 -> 0.0026 -> 2.6 mA

With one led:
(9-3.2)/1000 -> 0.0058 -> 5.8 mA
 
Last edited:
Thanks for the information and I pretty much have it figured out for now. I may stick with the 1K resistor and 2 LEDs for a longer runtime. When I was playing around with this last night the 1K resistor version actually appeared plenty bright for an around the house light even at 5.8 mA as your graph shows......any idea as to the runtime with 2 LEDs at 2.6 mA and standard 9 volt cell??
 
Robocop said:
....any idea as to the runtime with 2 LEDs at 2.6 mA and standard 9 volt cell??
Click to expand...


Usual it is calculated as battery_capacity/current_draw. For a 9 volt battery it is around 600mAh, this would give: 600/2.6 -> 60 hours

But this would not be the real value, because the current draw is declining while depleting the battery, giving a much longer runtime (Depending on when you decide the brightness is to low), on the other hand, the specification for the battery says down to 4.8 volt and you need more like 6 volt for two leds in series, this would give a much shorter runtime.

For a real estimate you would need better specifications for the battery and need to integrate the current draw over time :sick2:. It is probably much easier to do a test!
 
when I use resistors on LEDs I use a variable resistor and both look at various light levels and current levels then when I find the light level I need I have both current and resistor value. What I would do is figure out the lowest useful light level you can use and at least double the current level that way as the battery depletes it will reach that current/output level when the battery is more depleted than starting at that level and depleting to an unuseful level quickly.
 
I am not looking for anything really bright as this is mostly just for the fun of making something. I was searching for information and saw a few videos of the pak-lite that claimed something incredible like 1200 hrs of runtime on low.

I believe the website quoted both standard runtimes as well as lithium 9 volt however their claims, and reviews, were pretty amazing. I am not even exactly sure of what these spare 5mm LEDs draw however most of them are salvaged from my old Dorcy AAA circuits.

If I can use it to simply navigate a dark bedroom or as a small night light I will be happy....just like the idea of having something that can go for days on a single battery and I like tinkering with this stuff when I can. Thanks again for the input and I am now looking in my parts drawer for some different value resistors to experiment with.
 
If you run a 600mah battery at an average of 0.5ma it should run about 1200 hours. This probably would mean starting it around 1ma and it depleting the batteries down to perhaps a barely useful 0.25ma level. I just never was interested in 9v lights. I only bought the 9v light at a dollar store so I could drain the used 9v batteries from my answering machine, clock radio, and smoke alarm every few years. I specifically bought another clock radio used because it has a 2032 lithium cell backup instead of a 9v one. I can get 2032 for 35 cents easily here 9v cost close to $2 and lithiums I do not worry about leaking while I have had unattended 9v batteries leak several times over the years they run so long at times you forget to check them.
 
Hi everyone ! Please help me figure out what resistor I need... I have an Osram Golden Plus LED, which in it's datasheet specifies that the forward voltage is 3.2-3.7V. It also says that the maximum current is 1A, which can be achieved at 3.9V(this value is represented in the -forward current relative to voltage- graphic). So, please help me with this, what resistor do I put in series with a 4.8V(2100mA CC source) to get this led to work at it's maximum light output at 1A ?(i have a proper heatsink, so this will not be a problem) Thankyou! Here is the led's datasheet if anyone wants to see it
 
bjcryss said:
2100mA CC source
Click to expand...

I don't think you mean what that says. What don't you tell us what the actual power source is and it will help to give the right answer?

In general a resistor is not the right solution for a high power LED running on battery power. It has too little control and it wastes too much power. You should normally use a proper LED driver circuit for this purpose.

However, if running from line power like a wall wart or something then a resistor may acceptable in a pinch, although it is still not ideal. It will get hot and will need to be big and expensive and mounted where it doesn't burn anything.
 
bjcryss said:
Hi everyone ! Please help me figure out what resistor I need... I have an Osram Golden Plus LED, which in it's datasheet specifies that the forward voltage is 3.2-3.7V. It also says that the maximum current is 1A, which can be achieved at 3.9V(this value is represented in the -forward current relative to voltage- graphic). So, please help me with this, what resistor do I put in series with a 4.8V(2100mA CC source) to get this led to work at it's maximum light output at 1A ?(i have a proper heatsink, so this will not be a problem) Thankyou! Here is the led's datasheet if anyone wants to see it
Click to expand...
If you have an initial 4.8v source and need 3.9v at 1A that would mean the resistor needs to drop 0.9v at 1A. Ohms Law says E=IR or R=E/I, 0.9/1 means you need about a 0.9 ohm resistor. I would recommend a 1.5 watt although theoretically a 1 watt resistor should suffice.
 
Mr Happy said:
I don't think you mean what that says. What don't you tell us what the actual power source is and it will help to give the right answer?
Click to expand...
Is that something hard to believe? :) I've got 4 x AA 1.2V 2100mA rechargeable Batteries put together is series, this gets me 4.8V @ 2100mA, and I'm thinking of going with 4 x AA 2500mA Sony Rechargeable batteries which I found at a good price.

Mr Happy said:
In general a resistor is not the right solution for a high power LED running on battery power. It has too little control and it wastes too much power. You should normally use a proper LED driver circuit for this purpose
Click to expand...
You are absolutely right, but for a led driver I'd have to pay 20euro extra in my country, so i'm looking for a much cheaper solution.
 
Lynx_Arc said:
If you have an initial 4.8v source and need 3.9v at 1A that would mean the resistor needs to drop 0.9v at 1A. Ohms Law says E=IR or R=E/I, 0.9/1 means you need about a 0.9 ohm resistor. I would recommend a 1.5 watt although theoretically a 1 watt resistor should suffice.
Click to expand...
Thankyou ! :D this helps a lot ! I think I'll go with your calculus and recommendation :D
 
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