LED current question

I'm really good at building and modding but I'm not a theory guy... so I need someone educated to answer this. (Why I couldn't pass a simple exam for ham radio license)


I was told since time began that a device will only draw the current that it needs, IOW, say a device is spec'd out at a 2A draw at a given voltage. Let's say 3.7V for example. This device will only draw what it needs for its maximum whether your 3.7v is from a 2A or 100A supply.


As an experiment once I took a single XP-G, hooked up a 8X7135 driver to it, and it still only drew ~1.4A from that LED. (tailcap) So it wouldn't matter if there were 4 or a dozen chips on the board since the LED is only gonna draw ~1.5A, right? Yet guys have talked about driving these to 2.1A or more with a Shiningbeam driver.



With the triples I've built I've modded quite a few SB drivers to 4.2A by piggybacking 7135's on top of each other... the current drawn at the tailcap went up accordingly of course but reached a max and that was it. I did one with an XM-L and it still drew 2900-3000ma even with the 12 total chips on the board... see what I mean? (I was measuring with an IMR 18650)



So how can you "overdrive" LED's at a given voltage when it will draw a certain amount of amperage maximum unless you go direct drive? The only thing that's gonna fry it is overvoltage as opposed to overamperage as far as I can see.

You'd think I'd know this stuff, so be easy... feeling kinda dumb! Never went to school for any of this-I just build 'em.



Rich

You can not always assume that a device will draw what it need, the power supply (battery) must be able to deliver enough power at high enough voltage. Your problem is probably resistance, in wires and in your ammeter. Just 0.1 ohm will cost you 0.3 volt at 3 ampere.

Each 7135 will regulate the current to 350 mA, as long as there is sufficient voltage on the input to drive that current. So, suppose you put 8x7135 in parallel, the maximum current will be 8x350 mA = 2800 mA. But this is the maximum current, not the actual current. The actual current depends on your power supply, in this case whatever battery you are using. If your battery is not big enough to supply that current you won't get it, however many 7135's you bank up. At some point, your single lithium ion cell will not have enough voltage to drive the LED to the current you require, once you have allowed for voltage drop in the battery, resistance in the wiring, and for the drop-out voltage required by the 7135.

Another consideration is thermal management. When the 7135 is working, it gets hot, and if it gets too hot it will throttle back on the current to protect itself. So if you use a group of 7135 boards you might need to keep them well separated, or put a fan on them, or use extra heat sinking, to make sure the temperature doesn't get too high.

I'm only using the 7135 board as an example. There is no problem... the question is: How can you drive an LED over its rated current if it will only draw a given amperage whether you have a 1A or 100A supply? IOW, the device drawing (e.g) 2A max, even if you had a supply that will give you 100A?

On the same 7135 board modded to 4.2A I could wire a triple star and it will draw very close to 4A. Then I could remove the triple, wire one XP-G in its place and that device will only draw ~1.5A, even though the board is capable of 4.2A. See now? Then how could the one XP-G be overdriven?

I'm not good at putting it into words.

Rich

Leds does not have a fixed forward voltage (Vf), it depends on the current. Increase the current and you increase the Vf, this will prevent a high current as long as you are limited in voltage.
I will suggest you get an extra voltmeter and uses it to check voltage across all parts of you circuit (led, driver, wires, ammeter, battery), with current running. This will makes it possible for you to see where you looses voltage.

richpalm said:

I'm only using the 7135 board as an example. There is no problem... the question is: How can you drive an LED over its rated current if it will only draw a given amperage whether you have a 1A or 100A supply? IOW, the device drawing (e.g) 2A max, even if you had a supply that will give you 100A?

Click to expand...

This is not true. The LED will happily draw 100 A and vanish in a puff of smoke if you drive it with a supply capable of that. You are missing the capable part. Your supply is not capable of driving the LED that hard (read insufficient voltage), therefore it does not happen.

On the same 7135 board modded to 4.2A I could wire a triple star and it will draw very close to 4A. Then I could remove the triple, wire one XP-G in its place and that device will only draw ~1.5A, even though the board is capable of 4.2A. See now? Then how could the one XP-G be overdriven?

Click to expand...

This depends on the voltage characteristics of the LEDs. Your triple star has a lower voltage requirement than the XP-G. To overdrive the XP-G you need more voltage. Maybe you need more voltage than a single IMR 18650 cell can provide. As HKJ indicated, you may have some resistance limits. Have you made absolutely sure to have a low resistance path between the battery and the driver? Even a little extra resistance here could lower your current significantly. Even the resistance of your ammeter could do it when you try to measure the current.

OK, that answers the question! I was pretty sure it was up to voltage but had to ask and clear it up. So it's only gonna be overdriven if you increase voltage.

I take all the precautions to lower resistance on the dropins, which go into L2P's. I use a copper stud on the driver, no spring, and I take the tailcap apart, clean everything, then I solder a copper braid to the board through the spring center, and solder a modified copper rivet to the spring and braid at the end so I get a nice solid, flat contact to battery negative. The rivet made most of the difference and perceivable brightness improved.

This is a poor man's operation so I'm not gonna dump $40 on McClickies for each light. The stock switches so far have done fine-in taking them apart, they all have silver contacts.

Don't think I missed anything! <g>

Rich

I think the confusion here is caused by a misunderstanding about how circuits work. To (over-)simplify, they have an operating point and a parameter that is key. In the cases where we'd say the supply amperage doesn't mater, if above the minimum, the rest of the circuit is responding to the fixed voltage. An example from another thread was 12V at 3A max. In that case any supply of 3A or greater providing exactly the required 12V would work. (Supplies providing less than 3A might "bog-down" and provide less than their spec'd 12V and affect the rest of the circuit. Plus probably overheat.)

LED's are very sensitive to voltage variations and are easier to think about as current regulated. Hence a 2A circuit might require 3.695V with one LED but 3.725V with a different one due to process variation. This is something we expect the driver circuit to regulate & control. BTW- In the case of the first LED sample it might draw 3.5A at 3.725V and produce smoke.... (Might. To be accurate you need to LED circuit characteristics, the circuit details, & circuit theory. See the spec sheets & an EE.)

richpalm said:

OK, that answers the question! I was pretty sure it was up to voltage but had to ask and clear it up. So it's only gonna be overdriven if you increase voltage.

Rich

Click to expand...

This isn't true for leds the way it might be for an object that has a fixed resistance.
I am not the best one to teach theory here, but you are looking at only 2 legs of a pyramid based on ohms law.

If you have a 1 ohm load and you feed it 1 volt, it will only draw 1 amp current. Period. If you want to increase the amount of current fed to that object, you can do it one of two ways. Yes, you could increase the voltage. 2V/1ohm = 2A

But, on the other leg if the resistance changes... 1V/.5ohms = 2A as well.

In the case of leds 2 things are at play. 1st, as pointed out, they are very sensitive to voltage changes, so while voltage may only increase by a tenth of a volt, the current would increase dis-proportionately. And second, as the led heats up, the resistance changes. When one uses a resistor as the current limiting device in a voltage controlled circuit, it is certainly better than nothing. But, it doesn't work well when the load changes.

Hope that made some sense.

I am a mechanical guy and whenever i have to visualise electronics i compare a circuit to water... let me put that to use in explaining a basic LED circuit:

We have a high tank (or lets call it a cell) of water with a tube (wire) on the bottom. Lets call the hight of the water level 'voltage' (making the actual amount of water the tank's capacity so a wider tank lasts longer). If we dont put a stop in the tube water will flow out of it, lets call this flow 'current'. The amount of flow is only determined by the pressure and the diameter of our tube.

This is exactly what happens if you short a cell, basically unlimited current will flow only held back by how good your wire and cell are.

Now lets put an impeller in our tube that drives a device. All of a sudden the water will now not flow as fast as it wants anymore, it now is held back by how much energy the impeller need (thats where the whole idea that a device draws only the current that it needs comes from). Now to make sure nobody connects our impeller the wrong way (making our device running backwards) we hook up a one way valve before the impeller ('diode'). As long as the valve is big enough it will not influence the flow of the water at all, in fact when we remove the impeller the water will start to flow unrestricted once again. Now if you imagine this valve having a tiny little light and dynamo built in to show how fast the water flows through we basically have a valve that 'makes' light ('light emitting diode'). The faster the water flows the brighter the light will be and if the water flows too fast the little light will pop!

And that's basically how a LED works, you have to keep the 'current' low enough not to pop the little light but high enough to produce the amount of light you want. You could do this by adding a resistor or load on the circuit (like our little impeller) or fancy other stuff like smart devices. If you connect a led directly unrestricted to a cell it will let huge currents through and destroy itself in the blink of an eye.

Got it now-thanks all! Very well written replies BTW. Much appreciated!

Maybe this should be a sticky.

Rich