# How LED drivers work, and efficiency?



## ImGeo

I was using a LED/resistor calculator, and realized that resistors were very inefficient (http://led.linear1.org/led.wiz).

So it made me wonder how drivers, boost circuits, bucks, etc, work. Can anyone explain a little about their efficiency (when having to adjust voltage) etc, how the voltage and current are regulated, etc? 

Thanks!


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## SFG2Lman

i would love to know too!! schematics and descriptions of how and why to use MOSFETS, etc would be a great learning experience for me at least, but i can understand how painstaking that might be to describe, so perhaps i'll learn some electrical engineering first.:thumbsup:


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## rmteo

Try googling "SMPS" (switch mode power supply). Here are a few to start you off:

http://en.wikipedia.org/wiki/Switched-mode_power_supply
http://www.smps.us/
http://www.smps.com/


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## RelativeEng

ImGeo said:


> I was using a LED/resistor calculator, and realized that resistors were very inefficient (http://led.linear1.org/led.wiz).
> 
> So it made me wonder how drivers, boost circuits, bucks, etc, work. Can anyone explain a little about their efficiency (when having to adjust voltage) etc, how the voltage and current are regulated, etc?
> 
> Thanks!



The reason resistors are inefficient is because all of the energy that they are regulating is converted to heat. You can reduce this affect by putting multiple LEDs in series to multiply the ESR (Effective Series Resistance) but you will always be limited on efficiency.

You are not changing the voltage by adding a resistor, you are limiting the current. The voltage is unchanged in the circuit. If you have a 12V battery hooked up to the LED, it will remain 12V. You will, however, have a voltage drop over the LED (forward voltage Vf) just as you would a resistor. A diode just responds differently than a simple resistor. It is dynamic in that it's Vf changes with slight variations in heat and current.

A IC (Integrated Circuit) whether boost, buck, buck-boost, SEPIC, hysteretic, flyback, etc. controls the DC current through the series of LEDs. It uses a comparator circuit (reference voltage and sense resistor) to monitor the current through the circuit and adjust duty cycle (%on vs %off) of the LED. It does this at a fixed frequency which is oscillated by an external capacitor. This simple explanation relates to all circuits.
They are different though in that some circuits drop down the voltage, while others boost it, or do a combination of the two. This is accomplished by an inductor (a wire wound around an iron ferrite core that uses field collapse of a magnetic field to boost voltage).

Efficiency on any circuit is directly related to the total amount of energy changed or regulated. If you have a 12V supply driving 3 LEDs in series (3x 3.4V=10.2V), the loss is minimal. If you use the same source to drive 1 LED (1x3.4V=3.4V) you will be much more inefficient. Keep this in mind when you are selecting your LED and power source. You must also keep it in mind when you are considering your thermal management of the resistors, transistors, and ICs.

Hope this helps. I am an engineer, not a teacher so I apologize if it doesn't make sense.


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## blasterman

Wiki has a slick article (also linked above) on power supply types that is somewhat related to this. If you scroll down they have relative efficiency descriptions.

http://en.wikipedia.org/wiki/Switched-mode_power_supply

*ImGeo* has asked as question that I've been meaning to ask for awhile. While resistors are obviously inefficient because they create heat, does 100% of their calculated current load get wasted as heat?

Reason: From an application perspective, using poorly made and inefficient chinese current regulators can sometimes be less efficient than using radio shack power resistors on 12-volt - to a degree.


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## RelativeEng

blasterman said:


> Wiki has a slick article (also linked above) on power supply types that is somewhat related to this. If you scroll down they have relative efficiency descriptions.
> 
> http://en.wikipedia.org/wiki/Switched-mode_power_supply
> 
> *ImGeo* has asked as question that I've been meaning to ask for awhile. While resistors are obviously inefficient because they create heat, does 100% of their calculated current load get wasted as heat?
> 
> Reason: From an application perspective, using poorly made and inefficient chinese current regulators can sometimes be less efficient than using radio shack power resistors on 12-volt - to a degree.




"While resistors are obviously inefficient because they create heat, does 100% of their calculated current load get wasted as heat?"

Yes. They emit no light and they do not resonate under DC conditions. A resistor is one of the few electronic components that produce only heat. The only reason that you could have a less efficient driver circuit is if it is designed incorrectly. The driver IC and circuitry can suck extra energy by doing a number of things:
1. Transient state transistors
2. Improper operating frequency
3. Line resonance
4. EMI backfeeding
5. Inefficient shottky diode
6. Cheap inductor
I could go on forever. Switch mode power supplies are pretty complex. It is easy to get them wrong. Energy can be wasted in alot of ways. Putting a simple resistor in series is much easier and has only 1 variable...heat. All resistances change by heating.


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## Linger

RelativeEng said:


> Hope this helps.



Thank-you, and once again, thanx.

How is it that one model of driver can power led's with different vf's ?
Let's take a Souel P4, 'I' bin with a voltage forward of 3.25-3.50 volts.
A two-mode buck current regulating board will drive it at the output the user selects, for example 80ma, 700ma. Back to what you wrote in your post, "some circuits drop down the voltage," how is that set? for example, I know that I could use a different emitter with say a vf of 3.75-4.00v, and it would still work too, at the regulated current levels that the driver board lets me select (80ma, 700ma).
As most drivers regulate current, what happens to voltage? Both as the cells are depleted and voltage drops down from 4.2v to 3.5v
and, how does this change across emitters with different vf's.


(lastly, please confirm this: a driver regulates current by adjusting the 'duty cycle' this means the circuit from the cell to the emitter is completed, or power is drawn from the cell, in quickly cycling 'off / on' (the pulse width modulation) The chip has a separate circuit that remains charged so the chip remains 'on' and is able to monitor the current. So a driver could be over volted and fried (e.g., a max 6v driver connected to 20 IMR16340 in series), but a current regulating driver cannot be over amped simply by connecting a large power supply (e.g., same 6v driver connected to 20 IMR16340 in parrallel) as the driver regulates the current that is drawn from the power supply)

 that just came out. I've been reading a lot before assembling my first light from scratch.

-Linger


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## 2xTrinity

linger said:


> Thank-you, and once again, thanx.
> 
> How is it that one model of driver can power led's with different vf's ?
> Let's take a Souel P4, 'I' bin with a voltage forward of 3.25-3.50 volts.
> A two-mode buck current regulating board will drive it at the output the user selects, for example 80ma, 700ma. Back to what you wrote in your post, "some circuits drop down the voltage," how is that set? for example, I know that I could use a different emitter with say a vf of 3.75-4.00v, and it would still work too, at the regulated current levels that the driver board lets me select (80ma, 700ma).
> As most drivers regulate current, what happens to voltage? Both as the cells are depleted and voltage drops down from 4.2v to 3.5v
> and, how does this change across emitters with different vf's.


What any current regulator does is this: meausre the current through the deice, then vary the voltage supplied to the LED accordingly in order to maintain a particuilar constant current. If an LED has a lower Vf, the driver will supply a lower voltage in order to still see 80mA. If an LED heats up and CHANGES Vf, current will start to increase, the driver will sense this, then back off the voltage slighltly until the current settles back down to 80mA again.

In the case of a buck driver, this process continues to work so long as the input voltage is higher than the required output voltage. If the input voltage drops below the Vf of the LED, the driver can basically do nothing -- it requires a certain voltage "overhead" to operate. Some drivers on the other hand, such as buck-boost drivers, can actually continue to operate by stepping UP the input voltage.

From an exernal point of view, youi could expect to see soemthing like this:

LED with a Vf of 3V, and output current of 1000mA:

10Vin, 350mA in
6V in, 550mA in
4V in, 800mA in
3V in, 1080mA in
2V in, 1650mA in
1.5V in, 2400mA in

Input power (voltage * current) will always be close to output power. Howver, efficiency will tend to be better as the voltage input is closer to the voltage output, as the driver will have less "Work" to do. 

anytime there is a significant mismatch between input voltage and output voltage, that means a lot of energy will have to be stored in energy storing elements (such as inductor or capacitor) inside the driver. If input voltage is close to output voltage, the regular doesn't have to really do much, and can simply let the voltage "straight through" without storing up energy and converting the voltage.




> (lastly, please confirm this: a driver regulates current by adjusting the 'duty cycle' this means the circuit from the cell to the emitter is completed, or power is drawn from the cell, in quickly cycling 'off / on' (the pulse width modulation) The chip has a separate circuit that remains charged so the chip remains 'on' and is able to monitor the current. So a driver could be over volted and fried (e.g., a max 6v driver connected to 20 IMR16340 in series), but a current regulating driver cannot be over amped simply by connecting a large power supply (e.g., same 6v driver connected to 20 IMR16340 in parrallel) as the driver regulates the current that is drawn from the power supply)


Connecting more batteries in parallel doesn't necessarily "increase the current." It increases the amount of charge available (in mAh), and decreases the equivalent internal resistance of the battery. That means that if you were to connect 20 IMR batteries in parallel to a very demanding load (that is low resistance, or resistance comprable to the internal resistance of a single battery) -- such as a high power incandescent lamp -- the 20 IMR batteries would be capable of supplying more current than a single IMR battery. This is because for a single cell a lot of the energy will be absorbed by the cells own internal resistance. Running cells in parallel reduces this.

However of you were to connect first a single cell, then 20 cells in parallel to a relatively modest load (that is, a device whose resistance is much larger than the internal resistance of a single cell), the 20 batteries in parallel would provide the exact same amount of current as a single cell , but would just last 20 times longer.


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## Linger

First off, thank-you for the detailed reply.
The explainations of current draw with decreasing voltage was very illuminating. Now I feel I finally get why efficiency is highest when v in is closest to vf. I know i read Justincase writing this a dozen time, but now I get it, it seems simple to me.



2xTrinity said:


> It increases the amount of charge available (in mAh), and decreases the equivalent internal resistance of the battery.



I get this part (excellent diagrams on battery university).
My emphasis is directed towards the monitoring / current controlling function of the driver. Re-reading it I'm sure I've got it right; pwm controls current by turning the flow on / off (at a short intervals.)

Thanx again. I just wrote up a peice about direct drive and by the end i'd explained it to myself and deleted it. Thank-you, i've had a learning experience.
-Linger


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## LED-high

Don't a Constant current supply remove the need for a resistor in the circuit? As I understand these units have all the things needed in them and are designed to be used under sertain voltage ranges. As long as you opperate inside the limits you have close to no heat loss in the circuit, exept for the LEDs that is...


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## RelativeEng

LED-high said:


> Don't a Constant current supply remove the need for a resistor in the circuit? As I understand these units have all the things needed in them and are designed to be used under sertain voltage ranges. As long as you opperate inside the limits you have close to no heat loss in the circuit, exept for the LEDs that is...



No. Any current regulating power supply must have a resistor. Some times this can be internal to the IC but usually it is external so that the current can be adjusted. The only way that you can measure current in a circuit is by measuring the voltage drop over a resistance.


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## LED-high

RelativeEng said:


> No. Any current regulating power supply must have a resistor. Some times this can be internal to the IC but usually it is external so that the current can be adjusted. The only way that you can measure current in a circuit is by measuring the voltage drop over a resistance.


good to know, my crappy german is just as crappy as I thought...


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## Linger

LED-high said:


> is just as crappy as I thought...



... hrm?


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## ImGeo

WOW Thanks for all the info! Just have 3 more questions:

1. From what I understand, to change voltage and pass a current, inductors need a AC current. Does the alternation of on/off serve as the AC? Also, is the on/off alternation so fast that it provides a smooth light output?

2. Since the less voltage-change = more efficient, does that mean using 1xAA _must _be less efficient than 2xAA to power a 3.7 Vf lED?

3. *linger* was talking about variable Vf. If I understand correctly, the driver will adjust the Vf to maintain the current needed? For example, on a Fenix L2D at different brightnesses, the driver will always put out the proper voltage to match the Vf and current?

Thanks!


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## 2xTrinity

ImGeo said:


> WOW Thanks for all the info! Just have 3 more questions:
> 
> 1. From what I understand, to change voltage and pass a current, inductors need a AC current. Does the alternation of on/off serve as the AC? Also, is the on/off alternation so fast that it provides a smooth light output?


A changign current is required. Not necessarily an "alternating" current in the sense of a sine wave (like the wallplug power). "chopping" up DC using PWM will get the job done. In fact, this offers much greater control over the design. For exmaple, PWM in internal drivers tend to be very high frequency (megahertz) in order to keep the necessary capacitor and inductor values relatively small.

Also, just because PWM occurs inside the driver doesn't mean that the output flickers. This is very different from "PWM Drivers" used in some flashlights which run the LED at a fixed current only during the "on" portion of the cycle, and modulate the light output duty cycle to dim the light. In a regulted switching mode regulator, the output current to the LED is held constant, and the input current from the battery is modulated. In some cases, both the power from the battery, and the power TO the LED are filtered so that the whole PWM and transforming action only really happens "inside" the driver.



> 2. Since the less voltage-change = more efficient, does that mean using 1xAA _must _be less efficient than 2xAA to power a 3.7 Vf lED?


Simple answer: Yes. 1xAA is particularly bad for a few reasons. One is the fact that 1.2V just isn't that much. A lot of semiconductor devices, like most transistors for example have minimum switching threshold voltages to work. 

Stepping up 2.4V to 7.4 would actually be more efficient than stepping up 1.2 to 3.7, even though it's the same percentage increase. Stepping up 0.6V to 1.85V would be much worse. Most drivers don't even work on 0.6V.

The other thing is that with low input voltage like 1xAA, the current will necessarily be very high. However, internal resistance of the driver will tend to be some fixed amount no matter what. This means that losses due ot internal resistance in say, the inductor while may not be much for 2xAA, will be worse proportional to the square of the current for 1xAA. And to get the same output, 1xAA will have to be pushing MORE than double the input current as 2xAA. So that's a ~5x increase in resistance losses in the driver.



> 3. *linger* was talking about variable Vf. If I understand correctly, the driver will adjust the Vf to maintain the current needed? For example, on a Fenix L2D at different brightnesses, the driver will always put out the proper voltage to match the Vf and current?
> 
> Thanks!


The driver will adjust the voltage to keep the current at the desired output value at all times. Different brightnesses are usually accomplished by changing the reference signal, which causes the driver to settle at a different output value.


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## LED-high

LED-high said:


> good to know, my crappy german is just as crappy as I thought...


I must take this back, according to the supplior (www.leds.de) this unit does not require a resistor in the circuit. My german is adequate after all 

Is there a reason why they say so? Are there CCS's that ahs the resistor built into them?


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## rmteo

LED-high said:


> I must take this back, according to the supplior (www.leds.de) this unit does not require a resistor in the circuit. My german is adequate after all
> 
> Is there a reason why they say so? Are there CCS's that ahs the resistor built into them?



On a pre-built unit like the one you linked to, you do not need a resistor. The resistor is built into the driver board as you correctly surmised. Although I do not speak it, I'm pretty sure your german is perfect.


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## Lupino.86

2xTrinity said:


> Stepping up 2.4V to 7.4 would actually be more efficient than stepping up 1.2 to 3.7, even though it's the same percentage increase. Stepping up 0.6V to 1.85V would be much worse. Most drivers don't even work on 0.6V.
> 
> 
> 
> You means that a LED driver that has an operating range of 5 to 9V (2S battery pack)
> works better than a LED driver that has an operating range of 2.4 to 6V (1S battery pack)?
Click to expand...


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## AnAppleSnail

Lupino.86 said:


> You means that a LED driver that has an operating range of 5 to 9V (2S battery pack)
> works better than a LED driver that has an operating range of 2.4 to 6V (1S battery pack)?


It is likely to be more efficient - Buck regulators tend to have higher efficiencies than boost regulators, and are also easier on the battery. Slurping double current from a half-the-voltage pack causes the battery voltage to sag.


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## Lupino.86

AnAppleSnail said:


> It is likely to be more efficient - Buck regulators tend to have higher efficiencies than boost regulators, and are also easier on the battery. Slurping double current from a half-the-voltage pack causes the battery voltage to sag.



Thanks for reply.
I have to power a P7 LED, advise me a 1S or 2S battery pack? I will choose a drivers appropriate to the voltage, don't worry


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## HarryN

P7 power draw = approx 3.5 Vf x 3 amps = approx 10 watts

Driver losses from a quality driver = approx 20%, so 10 watts x 1.2 = approx 12 watts

For Li primary and Li Ion cells it is good to keep the discharge to no more than 2C (minimum 30 min run time) to avoid damaging them, so you need at least a pack with 6 watt - hrs

For primary CR123, 1 cell = approx 3 volts x 1.5 amp-hr = 4.5 watt - hrs - not enough

For R 123 cells, 1 cell = approx 3.5 volts x 1 amp-hr = 3.5 watt-hrs - not enough

For R 18650, 1 cell = approx 3.5 volts x 2 amp-hr = approx 7 watt-hrs - just barely enough

For R 26650, 1 cell = no problem.

Those are rough numbers but you get the general idea.

Your life will also be easier with 2S cells so that the Vbat does not cross over the LED Vf during discharge, or else you will need a driver that does buck + boost, which are harder to find.


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## Lupino.86

Thanks for the explanation. :thumbsup:
I will use 18650 batteries from 2200mAh in packages from 2 or 4 batteries. 
Following your discourse (and your suggestion) it suits me to use the 7.4V (2S) having so 2200mAH or 4400mAh, are OK this way?


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## Lupino.86

The discussion continue here :twothumbs


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