[SOLVED] how to glow 30 LED on a single 9V battery????

I want to glow 30 LEDs on a single 9V battery plzz help me for this........telll if any ic can do this thing.

Search for led boost IC, you need to boost voltage to higher level corresponding to leds voltages in serie or parallel. In other hands you should know that standard 9V battery is not huge in capacity, therefore be taken into account working time of this device.

Do u knw any boost ic on whch i can work easily?

Few examples:

https://www.linear.com/product/LT3598
**broken link removed**
https://www.linear.com/product/LT3791
https://www.electroniq.net/led-drivers/isl97672a-power-led-driver-electronic-project.html
https://www.linear.com/product/LTM8042

I dont think that you need a boost converter as this will add complexity and consume.
You need to know at what current Id you will operate your leds and the voltage drop Vf of each one. That depends on the specific leds you want to use.
For example with Vf=2v Id=10mA you can put 4 leds all in series with a resistor of 100ohms. And repeat this circuit x7 to have 28 leds. That is a combination of series and parallel circuits. The two remaining leds go in series with a resistor of 500ohm.
Take into account that in this case the total current will be 8x10mA= 80mA. The battery may not last very long. With more powerful leds the battery could be depleted in short time.
This page will do everything for you:
**broken link removed**
Hope that this helps.

albert22 said:

I dont think that you need a boost converter as this will add complexity and consume.
You need to know at what current Id you will operate your leds and the voltage drop Vf of each one. That depends on the specific leds you want to use.
For example with Vf=2v Id=10mA you can put 4 leds all in series with a resistor of 100ohms. And repeat this circuit x7 to have 28 leds. That is a combination of series and parallel circuits. The two remaining leds go in series with a resistor of 500ohm.
Take into account that in this case the total current will be 8x10mA= 80mA. The battery may not last very long. With more powerful leds the battery could be depleted in short time.
This page will do everything for you:
**broken link removed**
Hope that this helps.

Click to expand...

thanks alot albert you solved my problem. But now the question is whether my battery is able to supply 80mA output current? I am using normal battery. Or if you know any battery please tell me i have to submit my project to my teacher as soon an possible. Ya always thanks....

What is a "normal" battery?
Is it old fashioned carbon-zinc which is also called "super heavy duty" with low capacity or is it modern alkaline with higher capacity?

All the details are on the datasheet of the battery that is available at the manufacturer's website. Energizer 9V alkaline batteries supply 75mA for 2 hours when the voltage has dropped to 6.8V. Since the voltage drops all the time then you will see the LEDs slowly dimming.

What leds you plan to use?

Thanks to all my friends..... I done with my project. Once more thanks. Oh i am very happy. I have done it as albert says..... Thanks albert

This is the preferred way to calculate LED series resistors when using batteries..

Red glow at 1.3V and are rated >=1.6V so if rated at 20mA the .3V rise means the internal ESR= .3V/.02A = 15 ohms
White glow at 2.6 and are rated >3.2V so if rated at 20mA, the .6V rise means the internal ESR= .6/.02= 30 ohms

But if using a high power Red rated at 2V @300mA , ESR is (2-1.3)/.3 =2.33 ohms

Keep in mind the battery also has an ESR so voltage drops as more current is drawn , deltaV/deltaI=ESR

Simple way is to match LEDs to battery and add up all the ESR's.

for RED 5mm LED rated at 20mA

consider 9V new alkaline battery as 9.5V with ESR = 1ohm

A series string of 7 LED's
9.5V - 7*1.3V - I(7*ESR(led))+ ESR(bat))=0 neglect ESRbat. in this case as it it much much lower,
I=0.4V/(7*15R) ~ 4mA which is dim (25%max)

So you can power 4 parallel strings to use up 28 LEDs and have 2 spare and no external series resistor the current will reduce to zero when the battery reaches 9.1V at which point it still has some 50% capacity.


So consider a string of 6.
again using the dim threshold for Red

9.5-6*1.3 = 1.7V for full 20mA you need a series resistance of 1.7V/0.02A=85 Ohm but recall the Red Leds are 15 Ohms ESR so 6*15=90 Ohm means again you don't need to add a series resistor unless your battery new is actually higher than 9.5V like 9.6, so some variance can be expected.

Now with 30LEDs you would use a 5P6S config. with no Rs in series.

Dont try anything less than a string of 5 unless you add a series resistor..