# Basic Driver Questions

C_K, is this the one you’re thinking of?

Yup thats it, thanks David. OP that should have enough info to keep you reading till 2015 or so.

I never understood why brightness has to change on low or medium modes.

I don’t think I understand this question. Is it in relation to a specific light/driver/etc? EG can you give an example of a light which has this thing you are trying to avoid?

You are right, I didn’t pose my question properly.

I was just basically asking about constant brightness lights and how their drivers work. I am puzzled as to why there are so very few constant brightness lights around. It seems like such a handy feature. Light doesn’t dim till battery is almost exhausted.

Ah, I think that’s a simple answer. (Good news, right?) Let’s restrict our discussion to approximately constant brightness unless you really want this to get painful.

At a given set of conditions (temp, drive current, etc) an LED has an arbitrary Forward Voltage, Vf. Linear, DD, and Buck drivers all require that the Vf be higher lower than the battery voltage (which is lower under load than at rest). … and that’s the whole explanation, really. Typically we’re talking about a Vf in the 3.0v to 4.0v range and a battery in the 3.0v to 4.0v range. The fresh battery produces about 4.0v to 4.2v under load, as it’s depleted battery voltage falls but the required Vf remains the same.

More information: Note that Linear and Buck drivers also have a voltage called the “drop out” voltage. You must add this to LED Vf, if the total is below battery voltage then your driver will “drop out” of regulation; brightness will be reduced.

[EDIT: oops! see strikethrough text]

Awesome!!!

Very very very informative

Thanks

Thank You!

You’re welcome. There are many interesting ramifications once you understand that basic concept: low Vf matters, low drop out matters, high voltage batteries matter, etc.

HKJ’s battery comparator suddenly becomes crucial.

So do these graphs from djozz (post #51).

… and here I skim over some more complex issues, but if we assume that the Vf of the LED is actually tied to the maximum drive level in most applications (due to PWM in most drivers)… then we realize that the lower the maximum drive current the driver is physically set up for, the longer the light will stay in regulation [at the same output level!].

Thanks again for the info, here’s HKJ’s article. The link is from the thread DavidEF posted a link to :

http://lygte-info.dk/info/DriverTypes%20UK.html

I have a light (actually several ) that the current goes down as the battery drains. So that is a direct drive light?

If I had a 8*7135 driver, I would get 2.8A (assuming 350ma)? But it would dim, because the voltage is going down, even though the current is constant?

But if the Vf is higher than battery voltage, I thought you wouldn’t have light?

Isn’t Vf on leds more like 3-3.2, (2.9-3.25 on XP-G) with batteries mostly 3.6-4.2?

Let’s tackle this after some other things.

The way Vf is presented in whatever education you have had (possibly just reading a spec, or maybe classes) has probably poisoned you a little bit. It’s an oversimplification to say that there is a voltage below which there will be no light. Please take a look at djozz’s graphs I linked to in post #13, they are VERY important and should help with your understanding of this subject.

1. Once you’ve scrutinized that graph, remember that current depends on voltage as far as the LED is concerned. Take another look at the graph with that in mind. So, if we have a power supply with infinite current and we adjust the voltage in the range shown on the graph… we’ll see the LED pull more or less current.
2. Batteries have voltage sag, presumably you’ve seen HKJ’s battery graphs - if not, go look. The more current you pull from them, the lower the voltage will be (measured instantaneously, I’m not talking about the drop over time as the battery discharges). The voltage will also fall over time as the battery is discharged.
3. Now model the interaction of these things in your mind: the battery voltage will fall due to various factors (note that current draw is one of them!), and the LED will draw less current the lower the voltage you present it with.

How are you doing now?

After considering these things please tell me what you think the answer to your last question in post #14 is. (Hint: remember that LED voltage is in nearly complete control of current draw for the LED)

[EDIT: see my EDIT back in post #10 to find the source of this confusion, or my clarification in post #22]

Well, if I’m reading the graphs right, blue on djozz’s graph is Vf. Which is 3.5-3.6 V at 3 amps. Looking at HKJ’s graph of the 2000ma AW, it doesn’t fall to 3.5-3.6V at 3A for about the first 750ma .

So the V on the battery is more than Vf on the emitter with a fresh 18650

Probably, I can’t look right now. I don’t see where you are going with this though. Are those the specifics of a light you have which dims during the discharge?

FYI 7135’s are supposed to have a 0.12v drop-out.

You said the drivers require that the Vf is higher than the battery, but this seems to contradict that.

OK , so 7135’s drop out at .12v. So does that mean that the V is lowered by 0.12v each?

The XP-G has a minimum Vf of 2.9V i think, just to keep it simple lets say that is true, i guess they vary individually. 350ma current.

So what happens when I run 2.7V @ 350ma?

I mean, I put my 4.22V 18650 into my 8*7135+XM-L light. What happens to the current, V, etc? especially as battery drops below what the driver puts out ( 3.6V?)

What happens to V, etc, when you get down around the Vf? 2.9-3.25V?

Very simple answer- it drops out of regulation, it’s no longer able to maintain the requested current and will direct drive the LED at a level limited by current battery voltage.

Another way to put it (note I do not have a graph in front of me, these are just random number’s picked out of the air)
Let’s say you’re using an emitter than has a vF of 2.85v at 750mA but rises to 3.3v at 1A (once again randomly picked numbers) and you’re running 4*3175’s for a total current of 1.4A (and at 1.4A the LED’s vF is 3.8). Remember to figure in all the total added up drop out, now let’s say your batters drops down below the needed 3.8V for the full 1.4A of regulation, as you get lower and lower remaining voltage you can follow the graph down to see what the max possible current the LED can pull is, when you get down to 3.3v the LED is only able to pull 1A, at 2.85v the LED can only pull 750mA. It doesnt matter how many 7135’s you have, even if you have the LED wired straight to the cell, if there’s only 2.85v for it to use it can only pull a max of the 750mA (which we get from the graph)

Crap, what a serious error on my part. I miss-spoke when I said that Vf must be higher than battery voltage, how confusing! I will edit my posts with strike-through to show the correct text ASAP. In the meantime understand that I simply said it backwards: LED Vf must be LOWER than battery voltage.

(it actually must be lower than battery voltage minus dropout voltage of course)

Oh that’s ok that makes sense. I thought it was that way, I see V as pressure, and if you don’t have enough, won’t go, although Vf isn’t constant in leds like most stuff, I don’t think.

Ok I’m starting to get this. CK where did you get the 3.8V from? You add up the 4x 7135’sx 0.12v, then whatever else is on the board? So thats what you were talking about Wight, with low Vf and drop out?

What would happen between 3.6 and Vf, though? Wouldn’t you still be above Vf, so you get more current than datasheet recommends?