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Bit of a newbie question here...

I want to build a battery powered LED light for use outdoors. So basically needs to be pretty bright and last a while.

If I have battery rated like this:

5.0Ah 6V
Discharge current 20 hr rate 250mA
Capacity:
20hr rate (0.25A) 5.0Ah
10hr rate (0.50A) 4.3Ah
5hr rate (1.00A) 3.8Ah
1hr rate (2.70A) 2.7Ah

An LED rated like this (lumens):

Forward Voltage: 3.7V
@ 350mA   : 100
@ 500mA   : 130
@ 700mA   : 185
@ 1,000mA : 250

So I want to create a circuit where I can switch between using 0.25A and 0.5A from the battery.

So, if I were to create a circuit with just the battery and LED, would the LED draw the maximum current it is able to? (lets say that's 1A, so it lasts 5h) Or would the battery dump all it's power into the LED since there is no resistance in the circuit?

To get the circuit running at 0.5A, by Ohms law, I would need to add in a 12 Ohm resistor (I think! - 6V / 0.5A)

Does the addition of the resistor reduce the power drain of the circuit?

Do the resistors regulate the current from the battery in a way that will extend the life of the battery?

Also, if the forward voltage for the LED is only 3.7V in a 6V circuit, how would I go about reducing the voltage for the LED?

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  • \$\begingroup\$ See LED simulator added at end of my answer. \$\endgroup\$ – Russell McMahon Nov 17 '11 at 9:48
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It is not very efficient using a resistor to drop the current of a high power LED.

In your example 3.7/6 of the power is used by the LED and 2.3/6 of the power will be consumed by the resistor which it will have to dissipate as heat.

Something like one of these components would do the job:

http://www.jaycar.com.au/productView.asp?ID=AA0593&form=CAT2&SUBCATID=976#1

It takes a variable battery voltage range, and has an input for a potentiometer so you could adjust the brightness.

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  • \$\begingroup\$ If Beno wants to build the circuit from scratch, he could use something like this part: maxim-ic.com/datasheet/index.mvp/id/5274 Just add an inductor, a MOSFET and some resistors/caps, and generate a PWM signal for the brightness. Unfortunately, there are only SMD packages available. \$\endgroup\$ – 0x6d64 Nov 17 '11 at 8:39
  • \$\begingroup\$ nice, i was looking for a part like this myself. \$\endgroup\$ – geometrikal Nov 17 '11 at 13:46
  • \$\begingroup\$ thanks for this, i'll check out this component. One thing I couldn't work out from the datasheet though... Input V is variable, output I is constant, output V is variable too. If I want a certain voltage (3.7V), how do i make certain of the output voltage? \$\endgroup\$ – Beno Nov 17 '11 at 13:59
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    \$\begingroup\$ The voltage of the LED changes with the amount of current going through it, and the amount of current is what sets the brightness. So I guess that is why output V is variable, just make sure it is within the range of your LED. Some parts can support a string of LEDs. This website ledsmagazine.com/features/4/8/1 has some nice graphs to illustrate this. \$\endgroup\$ – geometrikal Nov 17 '11 at 14:08
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    \$\begingroup\$ @Beno: You do not want a certain output voltage, but rather a current. The value of 3,7V is the result of the LEDs characteristics, not something you have to provide (you just make sure that those 500mA go through the LED). The only time you take into account the voltage drop over the LED is when you limit the current with a simple resistor. There you look at the LED as if it was a resistor (which is ok if you assume a fixed current) and calculate the dropper resistor accordingly. \$\endgroup\$ – 0x6d64 Nov 17 '11 at 14:15
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No, a LED doesn't limit the current by itself. With such a battery it would probably draw >10A and go up in smoke instantly.

Limiting the current with a resistor is possible, but for power LEDs it is not a good idea, because it wastes too much energy. In your case there would be about 2.3V on the resistor, so 2.3V/6V ~ 40% would be wasted.

Better use a buck converter, which uses a switched inductor to regulate the current. There are many ICs that do that, with efficiencies >90%, you can look for "led driver buck". But it is quite difficult to correctly build such a circuit, so it is probably better to simply buy a ready-made module as proposed by geometrikal.

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An LED driven at well above its typical forward voltage will draw very large amounts of current for very short periods. But only once.

Current for an LED with series resistor
I = V/R
I = V_across_Resistor / R
I = (Vbattery - Vf_LED) / R

So R = (Vbattery - Vf_LED) / R
Vf LEd changes with current but using 350 mA and 3.7V to start gives

R = V/I = (6-3.7) / 0.35 = 3.3/0.35 ~= 9.4 ohm
Use standard value of 10 ohms.

Ideally LEDs should be operated from a constant current source so that as Vbat changes or as Vf varies btween samples Reffecive can be adjusted to suit.

More anon.


SIMULATOR maybe

Try clicking on this LED simulation

If that does not run then first run this Diode simulation then right click the link above, select "Copy link address" then paste into address box on page with diode simulation.

You should get a diode plus resistor plus power supply. I have set the diode to Vf=3700m = 3700 mV at 1 A.

You can alter the supply voltage with the slider and hover over diode to see current.

Right click on resistor, choose edit, change resistor value.

Save any new model with File export.

enter image description here

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  • \$\begingroup\$ So, if I wanted to use 500mA in the LED, I would switch it out with a 6.6 ohm resistor? The other question is: Does the resistor regulate the current so that battery life is effectively increased when using a higher ohm resistor? \$\endgroup\$ – Beno Nov 17 '11 at 1:37
  • \$\begingroup\$ Close enough to yes. R = V/I = (6-3.7)/.5 = 6.6 Ohms. This assumes Vf = 3.7V at 500 mA. See data sheet for actual value. | Re "Does the resistor regulate ..."? - It's a matter of terminology. If you use more ohms them overall I will drop and battery will last longer. The resistor is oart of a lower current cicuit and increasings is value decreases current - you can say the resistor is "regulating " if you wish. "Controlling or "helping control" may be closer. \$\endgroup\$ – Russell McMahon Nov 17 '11 at 9:52
  • \$\begingroup\$ thanks, that simulator is awesome. I'll play with it a bit and try to learn! \$\endgroup\$ – Beno Nov 17 '11 at 13:53

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