# Constant brightness

Discussion in 'The Projects Forum' started by ns21, Dec 19, 2009.

1. ### ns21 Thread Starter New Member

Dec 19, 2009
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Hi, i am trynig to make a LED lamp conected to the 6V battery. As the battery get's discharged, the light intensity drops. Is there a circuit or comnopent that can help me build a lamp that will limit current to let's say 1Amp, so that i can use the most power availalbe in battery and have a constant brightnes?

2. ### Audioguru New Member

Dec 20, 2007
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A constant current circuit will keep the brightness the same if there is enough extra voltage to power it because it is in series with the LED.
An LM317 regulator plus one resistor makes a simple constant current circuit. It needs a total of at least 3.25V plus the voltage needed by the LED.

3. ### ns21 Thread Starter New Member

Dec 19, 2009
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Of i go to local shop on monday... I'll try that and let you know. Problem i had was that voltage on battery was droping and therefore brightnes as well. I thought, since i had no 'constant voltage reference' that i can't achive that at all. i have tryed with other voltage regulators, and i even tryed using 'forward voltage drop' on power diode as 0.5 refference, and it didn't work. I hope this regulator does magic... thank you... regards.

4. ### Audioguru New Member

Dec 20, 2007
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896
Do the math.
The LM317 constant current circuit needs at least 3.25V. Your unknown LED might need 3.5V. Then the minimum battery voltage is 6.75V but yours probably drops to less than 5V. The LM317 will work if your LED is a 1.75V red one or if your battery has a higher voltage.

5. ### SgtWookie Expert

Jul 17, 2007
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You do need some kind of current regulator/booster.
However, an LM317 current regulator won't really help you, as you have limited voltage to start off with.

Adding an LM317 to the mix will mean that you will simply run out of power twice as fast, as half of it will be expended across the regulator.

If you post the circuit diagram of what you have now, it will help us to help you arrive at a better solution.

6. ### ns21 Thread Starter New Member

Dec 19, 2009
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There is no circuit to show, so far. All i want is to turn elctricall table lamp into battry powered table lamp. I have removed power cord and instaled 6V 4.5Ah Lead-Acid Battery, soldered 10 LED's in parallel and used regulator to drop voltage. Then i realised that i need stable voltage reference. As i wrote, i thought forward voltage drop on power diode will do the trick, but even that voltage changes as the battery gets disharged. Thats why i think that i should concentrate on regulating current insted of voltage. What do you think?
Simple project turned to be a problem that i am trying to solve for a past month. I realy need any help i can get.

7. ### Audioguru New Member

Dec 20, 2007
9,411
896
A 6V lead-acid battery is actually 6.3V to 6.9V. When its voltage drops to 6V then it should be charged or it is ruined.

LEDs should never be connected directly in parallel unless they are tested and sorted so that their voltages are all exactly the same. LEDs set their own voltage. You feed them with a regulated current not a regulated voltage.

Each LED needs its own current-limiting circuit or resistor.
What is the typical forward voltage and current for each LED? If you use the max allowed current then with ten they will all get too hot.

8. ### ns21 Thread Starter New Member

Dec 19, 2009
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Find tips... thank you. i will start by using one LED and take it from there. Thank you SgtWookie as well.

9. ### Wendy Moderator

Mar 24, 2008
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There are other ways to make constant current source that will go down to low voltages. The circuit would need tweaked (the extra LEDs removed). Figure the emitter resistors drop around 0.6V, and calculate the resistors you need from there. If it is 20ma then they would be around 30Ω. I picked 36Ω to give it some safety room for the LEDs. I figure this circuit would go down to 4½ Volts.

Last edited: Dec 21, 2009
10. ### Audioguru New Member

Dec 20, 2007
9,411
896
This low dropout voltage constant current sink is also pretty simple:

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