Different ways to step down V?

[eebowler]

Good morning everyone. Is there any way, other than by using an energy wasting resistor to step down dc voltage? Do some applications use ICs for this purpose? If so, is the stepping down process any different from using a resistor? Is it more or less efficient?
Question 2) Power loss through a resistor is I*V. ie, current flowing through the resistor by voltage drop across the resistor. For a 1W Luxeon, I= 0.35A and the foward V is 3.42V Now, if I use 3 batteries, the V drop will be= 1.08V, therefore, power loss through resistor =0.378W.
If I use 4 batteries, V drop will be= 2.58V and the resulting power loss will be= 0.903W! That's more than twice the power loss just by the addition of one 1.5V battery. Here is the question: Assuming that in both cases the current through the LED is the same (350mA), why should anyone want to use a 6V source to power a Luxeon LED if the power loss through the resistor is greater than if they used a 4.5V source? What is the advantage of the extra battery? Does it keep the LED lit for a longer time?
Thank you all for sharing your knowledge with me.

 

[Doug Owen]

Yes, the basic idea is different. A resistor is like a narrow pipe while IC solutions are 'switchers', they are automatic switches that turn on and off very fast and include some sort of part (usually an inductor) to store a bit of energy to use during the 'off time'. The idea is the average is 'half on, half off' with each condition as close to 100% efficient as possible. It can be from say 60% to maybe 90% efficient so they may or may not be more efficient than a linear regulator (either resistor or active).

The 3 or 4 cell delema is interesting. What you say it initially true, but remember the battery voltage is dropping all the time. For alkaline cells, you still have half the current (more or less) at 1.2 Volts. This is the point where 3 cell solutions 'die' (with half the energy still available, but at lower voltage). With four cells, you can drain 'em to the nubs (.8 to .9 Volts each), depending you actually get more light out of a dozen cells 3 or 4 at a time.

So yes, the fourth cell keeps it lit longer. It may or may not be better overall.

NiMH is another matter, the more or less flat 1.2 Volts over discharge makes 3 cell solutions very efficient indeed. Better in some cases with linear regulators than IC switchers in fact.

Doug Owen

 

[eebowler]

[ QUOTE ]
Doug Owen said:
The 3 or 4 cell delema is interesting. What you say it initially true, but remember the battery voltage is dropping all the time. For alkaline cells, you still have half the current (more or less) at 1.2 Volts. This is the point where 3 cell solutions 'die' (with half the energy still available, but at lower voltage). With four cells, you can drain 'em to the nubs (.8 to .9 Volts each), depending you actually get more light out of a dozen cells 3 or 4 at a time.

So yes, the fourth cell keeps it lit longer. It may or may not be better overall.
Doug Owen

[/ QUOTE ]
Ok, now I am /ubbthreads/images/graemlins/confused.gif.I understand the voltage is dropping all the time and I understand that I can still get 1/2 of the initial current at 1.2V when using alkaline batteries. What I don't get is when you say the 3cell solution dies. I also don't understand why the 4 cells will be drained to .8V each while the 3cells wouldn't. My problem is that in both cases, there are resistors present to reduce the V going to the LED to 3.42V. The resistors are alwayse there and will alwayse drop the V a certain amount depending on the voltage input(4.5 or 6V).
The only way I can believe that the 4cells are better is if you tell me that the V drops at a slower rate. Other than that, I still don't get it.
Sorry for sounding broken up. I haven't had any /ubbthreads/images/graemlins/sleepy.gif since yesterday afternoon.
Thank for the help.

 

[INRETECH]

Don't think of LEDs or resistors as "voltage devices", think of them as current devices - a resistor limits the amount of current, the voltage across it will change with current and to a small degree - the temp

The best way to run a LED is in constant current mode, then the LED will always get the proper amount of current no matter what the voltage is, use a LM317 and a sense resistor

R=1.25/Desired current

 

[eebowler]

INRETECH I understand what you are saying and know that it is true, but, with respect to that fact, there must be a point where the V is not enough even though the current flowing through the LED is. How does someone determine that point and avoid it? Are you saying that for a 1W Luxeon, as long as the power flowing through the LED is arround 1W, the LED should be lighting close to its max?
Thanks for sending me to the LM317. I am checking it out now. It however may not be available in my country.

 

[INRETECH]

You should be able to find the LM317 most anywhere, if you can't I have over 100 of them - send me email

LEDs being diodes have a min voltage before they strart drawing any current, this depends on the construction of the junction - but for White and others its about 3v; until you reach that voltage - they device will simply not draw any current at all

The LED will tend to 'self-regulate' itself on voltage when a voltage and resistor are used; the LM317 in Constant-Current mode is a wonderfull device except for the fact that you lose 1.2v thru the sense resistor and about 2v thru the device itself

So, for low voltage applications - its just best to "guess" the resistor, and then pad it from there, good ole Ohms law will get you close

R=

(Input_voltage - LED_Voltage)
-------------------------------
LED_Current

And then go from there

A more eff way to get the right voltage - would be to use a switcher; that way the current will average out to the proper amount and you will not waste energy/heat in the dropping resistor

In order of eff - they are

) Cuk'
) Ripple mode
) Voltage mode
) Current mode

There are some really easy to use parts like the "Simple Switcher", that all you have to do is add a inductor/cap/recovery diode

 

[fivemega]

If voltage is main concern, why not to use voltage regulator? Even if you want to regulate current, it would be more acurate to use current regulate device after regulated voltage.

 

[INRETECH]

LEDs are current devices, not voltage - once you reach their initial voltage, the current goes up very quickly

For example, a LED might do:

3v-0ma
3.1v-10ma
3.2v-50ma
3.3v-100ma
3.4v-200ma

The numbers aren't specific for any LED, its just a example of the curve of a diode/LED

Therefor, current mode is the best

 

[idleprocess]

A voltage regulator will behave somewhat like a dynamic resistor - except it will do nothing to regulate current, so you still need a resistor or some other method of limiting current.

You'll hear people talking about "thermal runaway" in LEDs, but that's academic - failure can take milliseconds. If LEDs are overdriven hard enough, they tend to fail instantaneously.