Driving a LED

[Overclocked]

I know you can drive an LED with just a resistor hooked up directly to the source, but which way would be more efficient in driving a LED; PMW (ie, 555 timer) or a Constant Current Source (A Transistor; mosfet, regulator..etc) when it comes to saving battery life while getting the most brightness?

 

[justDIY]

pulsed uses less power than continuous wave, if by brightness, you mean apparent brightness, as it applies to the human vision model ... then a 50% PWM with a freq of 150 to 1000 hz would do nicely. ps, you still need the resistor!

btw, what do you mean by efficient?

if your goal is to drive a single LED from a fixed voltage source, the most efficient thing to do is match the led's forward voltage to that of the source ... in other words, you can run two red leds off a 5v rail for the same 'energy cost' as it takes to run 1 red led off the same 5v rail, you're just dropping 1v in a resistor instead of 3v.

if matching Vf to Vcc is not possible, then a switching regulator will be your most efficient means of driving the LED, for example, high power LED taillights (think 1.5 to 3a), using a resistor (or worse, a linear reg) you end up wasting tens of watts of power, using a switcher you end up wasting only a few.

linear regulators (pass transistors and the like) are the worst in terms of efficiency, much worse than a resistor, however, they're also very flexible, so that is a trade-off to be considered.

 

[Overclocked]

pulsed uses less power than continuous wave, if by brightness, you mean apparent brightness, as it applies to the human vision model ... then a 50% PWM with a freq of 150 to 1000 hz would do nicely. ps, you still need the resistor!

btw, what do you mean by efficient?

if your goal is to drive a single LED from a fixed voltage source, the most efficient thing to do is match the led's forward voltage to that of the source ... in other words, you can run two red leds off a 5v rail for the same 'energy cost' as it takes to run 1 red led off the same 5v rail, you're just dropping 1v in a resistor instead of 3v.

if matching Vf to Vcc is not possible, then a switching regulator will be your most efficient means of driving the LED, for example, high power LED taillights (think 1.5 to 3a), using a resistor (or worse, a linear reg) you end up wasting tens of watts of power, using a switcher you end up wasting only a few.

linear regulators (pass transistors and the like) are the worst in terms of efficiency, much worse than a resistor, however, they're also very flexible, so that is a trade-off to be considered.

Well I had an Idea of Using LEDs for Emergency Lighting, since My Room is in a basement It gets awfully dark if the power Goes out. I planned on Running the LEDs off of a Rechargable Battery (I got a couple nice ones from a Lappy Battery- The battery did work) I want to make the most of the Batteries Current by Lighting the LEDs. (I also want to make a flashlight, wheres

I mean bright as in if you Ran a LED from 12V with a 180 to 270 Ohm resistor (blue, white..etc LEDs). Bright as in you can read from a single LED. Ive read (somewhere..) that driving LEDs with Resistors is somewhat innefficient because of the resistor (maybe it was hackaday or Digg..). But I think you hit My question..Now all I have to do is find the circuit. I have a bunch of Inductors from RadioShack that I have to measure.

 

[Overclocked]

I built a 555 Timer Circuit that operates at 32kHz, along with a Transistor on the Output. On the base of the transistor is a 1k Ohm resistor, on the Collector is a 180 Ohm resistor, and on the emitter are 4 LEDs in parallel with each other.

I measure 25 mA Total Current with the battery being at 8 Volts. The Current Going through the Collector is 17.80mA. Each LED Is getting 2.78V. These are White LEDs (Info here:**broken link removed**)

Now for the calcs, Pin= 8V*25mA = .2W, But heres where Im confused. If I do 17.8mA * 2.78V I get 0.04 W. Did I use the right Voltage? Now to Calc % Efficiency (Pout/Pin *100) So, 0.04/2= 20%!!! How can that be? 20% Efficiency? Im questioning by Numbers, Theres no way that can be right. (Maybe it isnt, Since Its a pulsed Signal..)

Btw, I measure 120mA with just the LEDs and resistors, So I Guess its an Improvement.

 

[audioguru]

The LEDs are only 2.78V but the battery is 8V so only 35% of the battery voltage is used for the LEDs and the rest is wasted as heat. Additional power is wasted because the 555 isn't a low power Cmos 555, and the 1k resistor also wastes some power.
Instead if 2 LEDs are in series and 2 groups of them are in parallel (and their resistor values reduced so they have the same current as before) then the efficiency is doubled and the wasted battery power is halved. The transistor works better as a common-emitter switch instead of an emitter-follower as you described.

If you pulse it then it looks dimmer. Your vision isn't a peak and hold circuit like some people think. Brightness is sensed as the average power in a pulsed light. That is how a PWM light dimmer works.

 

[Overclocked]

The LEDs are only 2.78V but the battery is 8V so only 35% of the battery voltage is used for the LEDs and the rest is wasted as heat. Additional power is wasted because the 555 isn't a low power Cmos 555, and the 1k resistor also wastes some power.
Instead if 2 LEDs are in series and 2 groups of them are in parallel (and their resistor values reduced so they have the same current as before) then the efficiency is doubled and the wasted battery power is halved. The transistor works better as a common-emitter switch instead of an emitter-follower as you described.

If you pulse it then it looks dimmer. Your vision isn't a peak and hold circuit like some people think. Brightness is sensed as the average power in a pulsed light. That is how a PWM light dimmer works.

So then is the resistor on the Collector Really Needed? If I just hook 3 Of them in series with each other, would it mean less Power is wasted since the resistor is no longer there? Just a thought, Since Ic=Ie, Then if I used 2-3 LEDs in series on the collector, Couldnt I do that on the emitter?

In a common emitter circuit isnt the signal on the collector Inverted? I guess it wouldnt matter if it was inverted or not.

 

[audioguru]

So then is the resistor on the Collector Really Needed?

Of course you need a current-limiting resistor. The battery voltage changes during its life so you don't want the LEDs to burn out if the battery is new at 9.5V.

If I just hook 3 Of them in series with each other, would it mean less Power is wasted since the resistor is no longer there?

2.78 x 3= 8.34V. Then they will burn out when the battery is new at 9.5V and will turn off when the battery voltage quickly drops to 7.2V. A 9V battery is considered dead when its loaded voltage drops to 6.0V.

Then if I used 2-3 LEDs in series on the collector, Couldnt I do that on the emitter?

An emitter-follower wastes about 0.7V. A common-emitter switching transistor wastes only 0.1V. The emitter-follower needs its base voltage to go much higher than the output of a 555 can go. A common-emitter switching transistor doesn't.

 

[Overclocked]

Intresting- All this time I thought I was running a Regular 555 Timer (NE555...etc), But In fact, I am using GLC555 Timer. In the data sheet it claims it uses 100uA to 300uA. So In Fact It is a low power 555 Timer, But the large amount of current is probably due to the RRC Network.

Now it makes me wonder, Since I am getting appr. 1/2 the brightness I would Usually, Is it worth driving these with a pulse? Cant I just save power by using a bigger Resistor? My Math Proves me wrong.

180 Ohms (Which is the resistor I use from voltages >5V) *2=360Ohms. Since a White Led Takes 3.3V, and VCC=9V, Vr=5.7V. since R=360Ohms, 5.7/360 = 15mA. Now since (For experimental Purpose...) Im using 4 LEDs, 15mA* (15mA Per branch) 4 = 60mA, Half of the 120mA I was using before with 180Ohm resistors.

So, P=IE, 9V*60mA= =.54 W. So, In this case I still Come out better using a 555 Timer. Granted, It takes up Less current, But at the cost Of Brightness and (possibly) efficiency. But Is there really a loss of efficiency? If 5.7V is Accross the resistor, and theres a current of 15mA then, Pr=0.0855W * 4 =.342W. So then Each LED is dissipating .198W Hmm, Then It seems I may have proved something I already knew.

So With that said, Which is more efficient, Boost-Buck or PMW?

 

[justDIY]

leds in series = more efficient ... dropping more than 2 or 3 volt across a resistor is wasting that energy by heating the resistor when it could be lighting an LED

a switchmode supply is only gonna make things a LOT more complicated for very little return. all an smps is going to do is squeeze more power out of the battery before the leds go dark. check out the joule-thief for this concept.

 

[audioguru]

You can't calculate how much current is in the LEDs unless you measure the voltage drop across the current-limiting resistor. Your transistor isn't a common-emitter switch so as an emitter-follower it has about 0.7V or more wasted across it.

Some LEDs are more efficient at higher current but other LEDs are less efficient. Check the datasheet for your LEDs.
If the efficiency of your LEDs stays the same at any current then if you pulse them with a 50% duty cycle to save power then they will look half as bright.

 

[Overclocked]

Ive been brainstorming a bit on how I could reduce current in this thing.
1) Move LED from Emitter to Collector
2) Reduce ON time as Much as Possible (10% Duty Cycle-100Khz Maybe?)
3) Possibly Remove the Transistor. I think a 555 Can supply up to 200mA. Im consuming a fraction of that.

Another Idea I have-
1) Use CMOS or TTL (Low power) Gates to make a Simple Oscillator. Theoretically, It SHOULD consume very low power.

2) Instead of LEDs Use CCLs. They use Inverters So there would be no real reason to power them with a 555 Timer. DC would work Fine. I have to seem how much mA they consume though. The only problem is they are bit more expensive that LEDs (Not Much though..)

 

[audioguru]

You don't gain anything if you power an LED from an oscillator. The duty-cycle of the on-off time determines the brightness. The same amount of power is used if it is the same brightness as with a continuous DC current. The only thing an oscillator provides is PWM to dim the brightness efficiently.

If you use a very low power Cmos oscillator circuit then the few mA of a 555 won't be wasted, but the LED still needs current pulses of maybe 60mA. 60mA is not low power unless you use a car battery.

You need to use a battery that keeps its voltage up as it runs down. Then the voltage across the current-limiting resistor isn't high so it doesn't waste too much power.

 

[Overclocked]

You don't gain anything if you power an LED from an oscillator. The duty-cycle of the on-off time determines the brightness. The same amount of power is used if it is the same brightness as with a continuous DC current. The only thing an oscillator provides is PWM to dim the brightness efficiently.

If you use a very low power Cmos oscillator circuit then the few mA of a 555 won't be wasted, but the LED still needs current pulses of maybe 60mA. 60mA is not low power unless you use a car battery.

You need to use a battery that keeps its voltage up as it runs down. Then the voltage across the current-limiting resistor isn't high so it doesn't waste too much power.

So use a boost converter to keep the voltage constant if Vbatt Drops, Correct?

 

[audioguru]

You don't need to boost the voltage if you use a cell from the "lappy". It is probably Li-Ion and is about 3.6V so it is just right for powering 3.3V white LEDs through resistors.
The Li-Ion cell will be 4.2V just after charging so the current-limiting resistor for each LED is 4.2V-3.3V/30mA= 30 ohms. When the Li-Ion cell's voltage quickly drops to 3.6V then the current in each LED will be 3.6V-3.3V/30 ohms= 10mA.

Don't connect LEDs in parallel. Each one has a slightly different voltage so their currents will be different and maybe the one with the lowest voltage will hog all the current and burn out. Then the next will burn out, then the next.
Use a separate current-limiting resistor for each LED.

You will need a circuit to shut off the power from a Li-Ion cell when its voltage gets too low, or the cell will be ruined. I think the minimum voltage is about 3.0V. Even with only 3.0V, a 3.3V LED will make some light.

Be very careful when you charge Li-Ion cells. If they get overcharged then they catch on fire and make a very hot, white flame like magnesium.

 

[Overclocked]

You don't need to boost the voltage if you use a cell from the "lappy". It is probably Li-Ion and is about 3.6V so it is just right for powering 3.3V white LEDs through resistors.
The Li-Ion cell will be 4.2V just after charging so the current-limiting resistor for each LED is 4.2V-3.3V/30mA= 30 ohms. When the Li-Ion cell's voltage quickly drops to 3.6V then the current in each LED will be 3.6V-3.3V/30 ohms= 10mA.

Don't connect LEDs in parallel. Each one has a slightly different voltage so their currents will be different and maybe the one with the lowest voltage will hog all the current and burn out. Then the next will burn out, then the next.
Use a separate current-limiting resistor for each LED.

You will need a circuit to shut off the power from a Li-Ion cell when its voltage gets too low, or the cell will be ruined. I think the minimum voltage is about 3.0V. Even with only 3.0V, a 3.3V LED will make some light.

Be very careful when you charge Li-Ion cells. If they get overcharged then they catch on fire and make a very hot, white flame like magnesium.

Unfortunately, I dont have Lion Batteries, They are NiMH Type, But I do have a spare Lead Acid From a flash light (one of those 1million candle light type lights). Both worked when I scavanged them. Actually for now I'll just Use a 9V to keep things simple.

 

[Overclocked]

I made the following changes to my circuit:

1) Moved the LEDs to the collector, and put the resistor on the emitter
2) Changed the transistor to a single NPN (it was a darlington NPN before)

These 2 changes bring the current draw down to 22.60mA. The battery (oddly..) was at 8.65V. I shaved off appr. 3mA. I also measured the duty cycle to be 67%. 3mA is small, but theres really not much else I can do.

Oddly, there doesnt seem to be enough voltage to power 2 LEDs in series. The voltage across the LEDs in parallel is 2.76V. How come the LEDs dim when more are connected in parallel?

 

[audioguru]

The 555 oscillator is reducing the average battery current by dimming the LEDs. The same as using a higher value for a current-limiting resistor. You don't need a dimmer.

Two LEDs in series that are 2.76V each need 5.52V. If they have a 390 ohm current-limiting resistor in series with them then they draw 8.9mA from a 9.0V battery, 6.4mA from an 8.0V battery and 3.8mA from a 7.0V battery. When the baterry voltage drops more then they will be too dim.
You can add a second set of two LEDs in series and in series with another 390 ohm resistor for twice the current and light.

 

[Overclocked]

Heres the Finished Product. I made a flashlight using the circuit, what a PIA to get the circuit board to fit...

Technical Data:

Powered by a 9V battery
9 LEDs, 4 Blue (because I dont have anymore whites) and 5 White LEDs (They are all in parallel with each other)

"Doom 3 Flashlight" (low) Setting: 20.6mA
"I see you" Medium Setting: 47mA
"All Hail the sun" High Setting: 66mA

On the "low" setting you can still see, but its for Power Saving. Most of the time I really cant get to this mode, its because of the ultra small switch Im using. On "medium" power, it can get you around in a house, and On high, it practically lights up the room.

Note, I can Increase the number of LEDs without Increasing the current needed to operate the flashlight. Ie, I took measurements with 1 LED and then put 9 LEDs on the circuit and found that current only increased 1mA in most cases.

So now your wondering, "How long will this thing last?" (all numbers here are estimated based on graphs from batteries- Go take a look on Energizer.com under technical info)
https://www.electro-tech-online.com/custompdfs/2006/12/522.pdf

Low Power- Est Resistance: 450Ohms
Med Power- Est resistance: 191.48 Ohms
High Power- Est Resistance: 136.36 Ohms.

Low Power Longevity (hours): Appr 40 Hours If the battery was at 6V
Med. Power Longevity (hours): Appr 20 Hours.If the battery was at 6V
HighPower Longevity (hours): Appr 12 Hours If the battery was at 6V

It may last Longer or shorter if the battery was a 9V. Only one way to tell, Get a fresh 9V and let it run all night!

 

[audioguru]

Your flashlight doesn't need 9 LEDs since at the High Power setting they are each operating at only 66mA/9= 7.3mA. Their max rating is 30mA each. So two LEDs in series and in series with a current-limiting resistor could be doubled to 4 LEDs with two resistors and would be brighter and use half the current so the battery would last for more than twice as long.

 

[Overclocked]

Your flashlight doesn't need 9 LEDs since at the High Power setting they are each operating at only 66mA/9= 7.3mA. Their max rating is 30mA each. So two LEDs in series and in series with a current-limiting resistor could be doubled to 4 LEDs with two resistors and would be brighter and use half the current so the battery would last for more than twice as long.

They arent in series with each other, it seems there wasnt enough voltage to drive them in series. Each LED Requires 3.3V, BUT each is only getting around 2.67V. So each LED is getting Pulses of 66mA (They are in parallel of each other..)