60 leds on a battery

ronsimpson

Joined Oct 7, 2019
174
Battery?
1574044233469.png
Looks like you are using a 3V LED. What part number? What current?
How long do you want it to run?
--edited--
Do you want the light to be the same brightness as the battery runs down or can the light get less and less.
 
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dl324

Joined Mar 30, 2015
9,150
Welcome to AAC!

We need more information.
  1. What current do you plan to operate the LEDs at?
  2. Is the 9V battery a transistor ratio type?
  3. How long to you want the battery to last?
  4. What do you mean by "how to get from 9V to 3V with little space"?
  5. What size is "little" space?
 

ronsimpson

Joined Oct 7, 2019
174
Current, voltage, time for a CopperTop 9V battery. I think it is not going to do 60 LEDs unless they are very small/low current.
I hate to pull more than 100mA from a 9V battery.
1574049002526.png
 

dl324

Joined Mar 30, 2015
9,150
Here's discharge data for Energizer 9V alkaline:
1574049465451.png

The information from Energizer for their Gold alkaline was pitiful:
1574049541697.png

Curiously, both are from Energizer and both are the same chemistry...
 

Audioguru again

Joined Oct 21, 2019
278
Why do you say that an antique Carbon-Zinc battery has the same chemistry as a modern alkaline battery that produces almost twice the performance??
The Energizer Carbon-Zinc 9V battery is obsolete and might have been called Eveready. I have never seen the Energizer GOLD 9V alkaline battery.
 

Tonyr1084

Joined Sep 24, 2015
3,676
Did you know Duracell makes batteries for RadioShack? Had a friend who worked there and told me the RS batteries were held to a higher standard, meaning they typically out performed Duracell. And RS batteries are cheaper than Duracell. However, I still don't have a collection of RS batteries.

{DISCLAIMER} I said that this is what I heard. Don't know this for a fact. So if wrong - sorry. Too often we repeat something we've heard as though it were gospel truth without knowing the REAL truth.
 

SLK001

Joined Nov 29, 2011
1,532
Build yourself a 9 to 180V boost supply and tie all your LEDs in a series string. Connect your 180V supply to the string and viola, your LEDs will light for about 15 to 20 minutes.
 

Reloadron

Joined Jan 15, 2015
4,591
what circuit do I need to use to power 60 LEDs with a 9v battery?
"how to get from 9v to 3v with little space."
Less a data sheet for the LEDs I can't begin to even guess. You need the LED forward voltage and forward current to even begin to calculate the best way to go about it. Next and as several are posted, a data sheet for the battery you plan to use so the amp hour rating is known as well as discharge curves.

Ron
 
The color of an LED determines its voltage range. Red and orange are 1.8V to 2.0V. Yellow and dim old green are around 2.2V. Modern green, blue and white are 2.8V to 3.4V.
A little 9V alkaline "transistor radio" battery has a small amount of juice inside. 60 LEDS with 30 strings of 30 that are bright with 20mA will each draw 600mA which will drain the little battery in about 20 minutes. You will see the LEDs dimming as the battery is quickly killed.
If the 30 strings of LEDs are dim with 3mA each then the battery will last for about 90 minutes, dimming all the time.
If you use a cheap Chinese "super heavy duty" 9V battery then the LEDs either will not light or will light dimly for only a couple of minutes.

Little space? I have a Chinese flashlight with 24 LEDs all in parallel with one little resistor to limit the current from three AAA alkaline cells in series. The LEDs all have the same forward voltage so somebody spent days or weeks measuring thousands and grouping them according to their forward voltage.
 

Tonyr1084

Joined Sep 24, 2015
3,676
ASSUMING you have an LED with a forward voltage of 3.2 Vf (sixty of them) and ASSUMING you want to run them at 20 milli amps, you put two in series along with a resistor. So that 3.2 Vf two of them equal 6.4 Vf. 9V - 6.4 Vf = 2.6 volts. The resistor needs to drop 2.6 volts at 20 mA. So 2.6V ÷ 0.020A = 130Ω. You'd need to set up 30 groups of this very circuit then wire them in parallel.

So 30 sets of 2 LED's & 1 Resistor means 30 resistors with 60 LED's

20 mA X 30 = 600 mA draw.

Again, this is all from making general assumptions. And you know what happens when you ASSUME; right? You can make an ass of You and Me. So I'll save you the trouble.

600 mA is more than a half amp. A quick search for "How many amp hours in a 9 volt battery?" turned up some results. I clicked on the Duracell page and it says it can deliver 600 mAh. That's 600 milli amps per hour. In other words, at the load I ASSUMED the battery would last one hour. See THIS.

[edit] GEE THANKS AUDIOGURU! You beat me to the punch.

Like AG said, the LED's will start to dim rather quickly. As the voltage falls so will the current. So you're not going to like the outcome of 60 LED's on a 9V battery.

Little space? I have a walk-in closet that has "Little space". Then again, there's a box on my desk that has little space; about 27 cubic inches. Sounds like a lot but that's a 3 x 3 x 3 box. Is that LITTLE SPACE enough for you?
 

Tonyr1084

Joined Sep 24, 2015
3,676
@Audioguru again mentioned forward voltages and the need to keep them real close in order to power them in parallel through a single resistor. In the video below I did a test with a limit of around 45 mA (and that's a lot for an LED of this type). What would happen if I used several LED's in parallel and adjusted the current up to keep them all brightly lit? Suppose I have 10 LED's sharing current and (assuming) each draws 20 mA. The supply would have to be 200 mA. But what happens if one of those LED's fail. Now there's only 9 LED's sharing 200 mA. That means each (if equal sharing) will draw 22 mA. Not much of a big deal. But if one failed, it becomes more likely that another will fail. So lets knock out one more. We're now down to 8 LED's each drawing 25 mA. That's a little bigger jump. So now they're running hotter and more likely to fail. We lose another. 7 LED's share approximately 29 mA. Typical recommendation is not to run LED's higher than 30 mA. So now we're really close. But we've only lost three LED's so far. What happens next? 6 LED's run at 33 mA. Now we're really pushing them. 5 LED's run at 40 mA. They will begin to fail at such a high rate you won't be able to see which one goes out next because as each one fails the next will flash and burn quickly. 4 LED's? 50 mA. That's just about the death of nearly all LED's. But maybe not instantaneously. But your LED's are falling quickly now. 3 = 67 mA. 2 = 100 mA and finally the last LED is going to see the full 200 mA.

And now the video:

 
The "store" says that a Duracell 9V alkaline battery capacity is 600mAh. But its datasheet shows typically 125mAh at 250mA, 275mAh at 100mA, 500mAh at 10mA and 520mAh at 2mA. This is typical when its voltage has dropped to 7V, some are worse.
Energizer says their 9V alkaline battery has a little more mAh than a Duracell.
 

Yaakov

Joined Jan 27, 2019
1,604
Generally, the (V ("Transistor", 6LR61, 1604x) is a terrible choice for a power source. It was originally designed to provide power to "transistor radios" which needed the 9V, so it is a genuine battery, not a cell. It is made of 1.5V cells, all quite small, and so offer very low capacities.

Today, for space restricted use, LiPo will offer far more capacity, even as a multicell battery. The 9V batteries may seem convenient, but in the end they are a huge compromise for most applications and should probably be avoided.
 
I have some LEDs that are plenty bright @ 1mA, so assuming at least 2 in series with a resistor, that’s like 30mAs.
But their current is 1mA only when the battery is brand new. When the battery voltage drops to 7V then their current is 0.3mA and when the battery voltage drops to 6V the current is almost zero and the LEDs might produce no light.
You need to say when the LEDs are dim enough to replace the battery.
 
what circuit do I need to use to power 60 LEDs with a 9v battery?
"how to get from 9v to 3v with little space."
First of all make sure how are you going to connect them. Be very cautious about the current rating of the battery. Do not overload it. If there is an essential need to use 60 LEDs, you can use it after using a current booster such as darlington pair amplifier.

Hello, if you are really concerned about the space, use a capacitor divider network.


---------| |---------------------| |-------------
|<-------C1--------->|<-------- C2---------->|

When a series arrangement like this is connected across the battery you have, voltage gets divided and voltage divided depends on the value of capacitance C1 and C2.

Take the output across C2 and it is given by

Vout = VC2 = C1*Vin/(C1+C2).

Choose the values of capacitors such that you get 3V as output.

Resistor divider can also be used but it unnecessarily dissipates power as resistor network continuously draws current from the source.
 
Be very cautious about the current rating of the battery. Do not overload it. If there is an essential need to use 60 LEDs, you can use it after using a current booster such as darlington pair amplifier.
Absolutely NOT! A Darlington does not boost the current coming out of the weak little battery. Instead it simply has a base current that is less than an ordinary single transistor.

Hello, if you are really concerned about the space, use a capacitor divider network.
No, because a capacitor divider works only for AC, not for continuous DC.

Resistor divider can also be used but it unnecessarily dissipates power as resistor network continuously draws current from the source.
No, a resistor divider is not used with an LED because only a single series resistor is used to limit the current. The LED itself limits the voltage. You can use a single resistor in series with an LED to work perfectly if the DC voltage is 1000V or more. It draws current only when the LED is turned on.
 
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