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EDIT: Here is a much brief version, the meandering motivated version is probably subsumed in the parent question on general battery life optimizations.

How can I lower the current into a LED without wasting battery?

I think using a resistor to lower the current will use up battery life. Is this true? How does it use up the battery if the current is kept small? Is it the power dissipation?

I think using a diode (and smaller resistor) will still use up the battery (LadyAda said something like "any linear device to lower the voltage uses the same amount of power"). Is this true? Is it the same amount?

Can I lower the voltage "for free", that is, without wasting too much power? How can a regulator IC do what a resistor or diode cannot?

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  • \$\begingroup\$ You seem to be asking the question, how do I lower the current of an LED. If this is it, you could shorten the question a lot. People will point you in the right direction. \$\endgroup\$
    – Kortuk
    Oct 7, 2010 at 23:43
  • \$\begingroup\$ Radically shortened. I linked to the other question for context. \$\endgroup\$ Oct 8, 2010 at 0:15
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    \$\begingroup\$ You still have 8 questions in your question. I think the most interesting one is if increasing the resistance of your current limiting resistor will extend your battery life. \$\endgroup\$
    – mjh2007
    Oct 8, 2010 at 12:37
  • \$\begingroup\$ I don't have time right now, but tomorrow I'll see if I can answer your question "How can a regulator IC do what a resistor or diode cannot?" \$\endgroup\$ Oct 8, 2010 at 23:37
  • \$\begingroup\$ Linear regulators can't do anything that diodes or resistors can't do, in the sense that you mean. They still waste power as heat. Switching regulators can, however, because they either pass the current without resistance or block it completely, so nothing is wasted in the device itself. \$\endgroup\$
    – endolith
    Oct 9, 2010 at 19:10

7 Answers 7

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Use a low current LED. For example, this one consumes 2mA.

Use Pulse Width Modulation (PWM) to drive the LED. During the off period, it's not consuming power. 50% brightness might be acceptable.

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    \$\begingroup\$ Nice idea. I saw something about this for the IR transmitter leds (apparently they get hot if they are not PWM). Cool, this effectively lowers the "rms" voltage, right? Basically for free (in terms of battery life), other than a microcontroller or oscillator or something has to be running, right? \$\endgroup\$ Oct 8, 2010 at 0:03
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    \$\begingroup\$ Along the same line of thinking, consider blinking the LED. A 1% duty cycle (10 ms of on timer per second) is still visible and consumes only 1% of the power. \$\endgroup\$
    – semaj
    Oct 8, 2010 at 0:07
  • \$\begingroup\$ @Jack Yes, it's basically free. The power for an MCU should be far less than an LED. \$\endgroup\$ Oct 8, 2010 at 0:17
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    \$\begingroup\$ I think PWM is the way to go, saving energy in the same way that a class D amplifier does. You can use a smaller resistor and send brief pulses of large current, and the LED doesn't care? It only cares about the average power dissipation and the rate at which heat is removed? \$\endgroup\$
    – endolith
    Oct 8, 2010 at 14:25
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    \$\begingroup\$ LEDs are more tolerant of overcurrent if it is pulsed (heat dissipation, as you said.) But, it's probably best to not exceed the max current rating by too much, just the same. Use a current-limiting resistor to get your "on" current to close to max, then use a reduced duty cycle PWM to get average brightness where you want it. If the flicker is too much, try putting a cap in parallel with the LED (after the resistor.) The cap would also allow a lower-value resistor, because it would store some of the energy for the "off" portion of the PWM cycle. \$\endgroup\$
    – Jesse
    Oct 10, 2010 at 10:09
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I think using a resistor to lower the current will use up battery life. Is this true?

The first statement is hard to give a yes/no answer to. Technically it uses up some of the power on itself, but as the resistance goes up the power goes down (the current also decreases, which is what matters in a battery).

How does it use up the battery if the current is kept small?

This is a very very messy statement. "Small", that means something different to every person. If you mean small, as in negligible, then it can be ignored, but I doubt this is true with a circuit counting uA. For it to be negligible a good rule can be that it's average current is less then 1/10 of your overall system current (I am fields, 1/10 is enough for 1/infinity).

Is it the power dissipation?

Yes, in the shortest answer. A battery has a voltage associated with it, which you normally know up front. Most devices have a current that varies as the voltage is varied, but since you know the voltage you can tell its current draw. Since the battery's capacity is measured in AH or mAH then you are set. take capacity, in AH, and divide by how many A your device pulls and you have your lifetime in hours. Number big enough? you are set. Gonna die 3 weeks early? now you have to find a way to reduce your average current draw, or get bigger batteries.

I think using a diode (and smaller resistor) will still use up the battery (LadyAda said something like "any linear device to lower the voltage uses the same amount of power"). Is this true? Is it the same amount?

As stated before you effectively have a current budget. if your average current draw is increased then you have lost battery life.

Can I lower the voltage "for free", that is, without wasting too much power? How can a regulator IC do what a resistor or diode cannot?

Yes, if you use a device like an LED Driver then it can do the trick. Most of these are Switching mode power supplies. These are a relatively advanced concept in electronics, do not beat yourself up trying to understand it. Just understand that it cheats. It will have an efficiency in the 80 or 90% range. You can calculate the power consumption of your device and then factor in efficiency to get the real power draw. This can then be divided by voltage and should correlate to your average current draw.

My advice

There is an easier way. As Joby was talking about, PWM. But i would not suggest using a 10% PWM, or even a 1%. Instead, blink your LED for short periods to let you know of important events.

If you want to check your device is sleeping, blink it every 1 second wakeup. If you are transmitting when rs232 is connected, blink it every time you TX a "packet". This can give you "on" times of milliseconds every few seconds. If you think of this as a basic PWM, then you are getting less than a .1%. if you are pulling 20mA to blink(quite bright) then you are pulling an average of 20uA. Go to a 2mA diode, and you are doing 2uA average current. if you blink every minute, you get to divide by 60 on that: 1/30uA.

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  • \$\begingroup\$ The specific type of power supply required is a buck regulator. Switch-mode PSUs are a very broad term. All that's really required to drive an LED is an inductor and a switch, nothing like the complex and expensive device shown on the Wikipedia page. \$\endgroup\$ Oct 8, 2010 at 18:02
  • \$\begingroup\$ I have helped someone whom needed high brightness to implement an LED Driver that was a buck-boost. They all fall under SMPS. The led driver would boost the voltage to get the required current if required. I consider it an advanced topic for someone whom is still learning about resistors and LEDs. We all start somewhere. \$\endgroup\$
    – Kortuk
    Oct 8, 2010 at 19:53
  • \$\begingroup\$ I do wonder why I received a downvote, if whomever downvoted me could expand I would like to improve my post. I attempted to answer each question with some explanation but avoiding writing 10 pages and getting lost in details. \$\endgroup\$
    – Kortuk
    Oct 8, 2010 at 19:55
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    \$\begingroup\$ I don't see what would be downvoted. You expanded the scope of possible solutions in such a way that much more efficient solutions could be considered. \$\endgroup\$
    – Jesse
    Oct 10, 2010 at 10:13
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Here's a fancy app note from Microchip about high-efficiency LED driving. http://ww1.microchip.com/downloads/en/AppNotes/91060b.pdf

The gist is that they use an inductor to drive the LED with no current limiting resistor. The inductor automagically drives the LEDs at the appropriate voltage. I don't know how practical it is, but it may be worth investigating.

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A resistor wastes power, as you observe. Any other analog device like a transistor in a regulator circuit, any kind of IC, a diode, or whatever, if it's conveying current from a voltage source to the LED, is going to waste power. There's a voltage drop across it, from the supply voltage to the forward voltage drop across the LED, with whatever LED current is flowing through. Multiply to find the power wasted as heat. No fancy regulator IC can escape this basic physics.

There are only two electrical elements that don't waste power: empty air and wire. Empty air, that is to say, any non-connection such as an open switch or turned-off transistor, has zero current. Wire, or any conductor such as a turned-on transistor (ignoring the small saturation voltage) has zero voltage drop. Low power circuits must avoid resistors, and use transistors as switches, always either on or off.

For the LED, needing some particular amount of current, you can drive a transistor with a rectangular wave. Very little power is wasted in the transistor, whether "on" or "off". Charge is efficiently transferred to the LED. Of course, you don't want to let too much current flow during the "on" phase, and toast the LED or cause it to age, so smooth the voltage at the load with a capacitor. (This is optional, depending on the LED's max current rating, frequency of switching, etc) The LED brightness can be varied by altering the rectangular wave's duty cycle, though you'd want to hold back from using a 100% on wave - then the full voltage supply is given to the LED.

This is a basic switched-mode power supply, a vital technology of recent and coming decades.

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  • \$\begingroup\$ While a wire doesn't have much voltage drop, it isn't zero (unless you have a superconducting wire, which has zero resistance). A wire often just has a really low resistance which = a low voltage drop. \$\endgroup\$
    – pfyon
    Oct 15, 2010 at 17:11
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If you increase the resistance of your current limiting resistor I would think your battery life would be extended. With a voltage source (such as a battery) using a larger resistor means less current will flow through the resistor and the diode. The only downside is you have a dimmer diode. However since the current through the whole circuit has been reduced your battery life rated in mAh will be increased.

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What you want is a current control circuit. These are commonly found in better LED flashlights (example), and devices like MAX6969 can help you control a bunch of LEDs from a microcontroller. That particular chip won't save you power unless you combine it with a lower voltage (perhaps supplied by a constant voltage buck regulator) to power the LEDs or PWM. Increasing a series resistor will indeed draw less current, and thus give you longer battery life, but the light will be reduced as well. With the constant current supplies you can eliminate the power converted to heat in the series resistors.

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try running the led with a step down transformer instead.. use the side with more coils in series only,with the led..its provides resistance by making many turns in the transformer coil before reaching the led..wont have much heat also unlike a resistor..wastage will be less

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