i want to light up 3-5 LED with mobile batt. u know thar bat. r 3.6 -5 V and 500-900 ma,
what i have to put for light up 3/5 LED with this batteries?
if any diagram pls forward I'll be grateful.

thanx in advance. -[mamun2a at gmail.com]

LED = Normal White Flash LED, Current req. 20-25 ma, 3.6 v generally.
specifically i wanna know what calculation i have to do for the registor using with light up 3/5 LED by a 3.6 v 600 ma or 5 V 700 ma (mobile battery).

thanks

If the battery voltage drops down to 3.6V then a 3.6v LED will be extremely dim or won't light, when it has a current-limiting resistor designed to give it 25mA with a 5V supply.

You could use a voltage stepup IC circuit to give you 18V at 25ma to drive the 5 LEDs in series like in mobile phones. It is a fairly complicated circuit.

That link is broken.

Originally Posted by Hero999

That link is broken.

It won't work as shown, since it has an extra ".pdf" on the end. I copied and pasted it into my browser's address bar, then deleted the extra ".pdf". The datasheet loaded no problem. Like this ... http://www.zetex.com/3.0/pdf/ZXSC310.pdf
Jeff

But a simple series resistor isn't much use if you want to power an LED from the same voltage source as it's voltage drop - it won't allow a 3.6V battery to power an LED with a voltage drop of 3.6V.

Thanks all for help.
i have collect the solves from searhin and these are

"The next easiest is a simple resistor. The resistor does consume power, though, but is usually needed since an 'ideal' 3.6 volt source is rarely available. Use Ohms law (Resistance(R)=Voltage(E)/Current(I)) to calculate the value and wattage needed: (R=E/I)
Each white LED gives a voltage drop of 3.6 volts. As an example, for a 12 volt light, you can run a maximum of 3 white LEDs in series at full power (3.6 x 3 = 10.8 volts drop). Subtract this from your supply voltage of 12 volts to get the additional voltage that must be dropped (in this case, 12 - 10.8 = 1.2 volts of additional drop needed). In this case, 1.2 volts of additional drop / .025 amps (25 ma) = 48 ohms. Use the next highest value of resistor available, 50 ohms. You must also be sure the resistor can handle enough current. Volts x Amps = Watts; resistors are rated in watts. So in this case, 1.2 volts x .025 amps = 0.03 watts. A 1/4 watt resistor will work fine, but if you run a second string of 3 LEDs in parallel, each string would need its own 50 ohm resistor. It's important that each string has its own resistor....putting them in parallel with a single resistor is bad practice."

from http://www.otherpower.com/

Thank u all

dear all
i have facing a new prob. the ampere of battery is 600ma + so the LED r burn out.
Can u tell me how do i reduce the amp from 650 to 25?

It seems like you don't understand what we've been talking about throughout this thread, I'll try to summarise it for you.

Look up ohm's law on Google.



Current is limited by resistance - a 1hm: resistor across a 1V supply will only draw 1A regardless of how much current the supply is capable of supplying.

LEDs aren't resistive and have a similar voltage drop regardless of the current.

You need to add a resistor in series with the LED to limit the current to the required level.

This also means that there is a minimum voltage that an LED will operate on, for example a typical white LED has a forward voltage drop of 3.5V, therefore it will not operate satisfactorily below this voltage. Your battery voltage needs to be significantly higher than 3.5V throughout the discharge cycle in order to have a decent battery life for example three AA cells might start at 4.5V but they'll quickly drop to 3V so I'd recommend using six cells for a white LED. This means that unless you use a voltage booster circuit you can forget about running a white LED from a 3.6V battery.

You also seem to be confusing battery capacity with current, your battery has a capacity of 600mAh, that means (when fully charged) it can supply 600mA for an hour before it's discharged. Look up battery capacity and Ah in Google.

I understand that English is not your first language, have you tried using Google translate on this page?

thank u very much.
yes sir our first language is bloddy Bangla, and i'm afraid to find bangla translator in somewhere else in the World till now.
thanks again, i'll check.

but if my battery current cant reduce then the LED is blow out 3/4 minit later,
i undrstnd dat 3.6 v is not enough for lighting the LED, ok i use 6 V 4 AH battery, now what resistor i use to drop current from 4000 mah to 25 ma?

mamun, dhaka, BD (Bangladesh)

A fully charged "6V" lead-acid battery is 6.9V. If the LED voltage is 3.6V then 6.9V - 3.6V= 3.3V will be across the current-limiting resistor.
Using Ohm's Law, the resistance is 3.3V/25mA= 132 ohms. 120 ohms will result in a current of 27.5mA.
When the battery voltage drops to 6.0V then the LED current will be 20.0mA.