Electronics question « Singletrack Forum

Bike Forum

[closed]

Electronics question

51 posts & 10 voices | Started 1 year ago by Mowgli | Latest reply from DoctorRad

Tags:

No tags yet.

Pages: 12 Next »

Mowgli - Member

I need to run two Cree XR-E LEDs (in series) off a 1000mA driver - they're being used as a tail light in series with a front light, and for reasons to complicated to explain they can't have their own driver circuit, nor can be put in parallel.

Problem is XR-E's max current is 700mA. If I put a resistor in parallel with the LEDs, it'll take off some of the current (and waste a bit of power). To halve the current in the LEDs I need to match their impedance. So what is the impedance of two XR-Es at, say 500mA? I don't have a voltmeter unfortunately.

Any pointers?!

Many thanks

Mowgli

Posted 1 year ago #
Mowgli - Member

My (probablu flawed) reasoning is thus:

500mA through the LEDs and 500mA through the resistor. Expect about 2.5v drop per resistor, so 5v overall. V=IR -> R=5/0.5=10 ohms. This seems like not very much.

Power loss through that resistor P=IV=2.5W. This is too much.

Anybody got any better ideas? Ideally I'd re-wire the LEDs to be in parallel with each other but I don't think that's possible.

There is a single Seoul 1000mA upstream of the rear light, in series.

Ta

Posted 1 year ago #
snaps - Member

XRE's are fine with 1A+ through them with good heat sinking but I think they'd be much to bright for rear lights even @ 700mA.
LED's don't have an impedance, they have a foward voltage (Vf) which needs to be applied across them to make them emit light, I think if you put a resistor in parallel with them, it'll could drop the voltage below the Vf & they will go out.
I'd keep the front & rear separate with there own drivers/dropper resistor.

Posted 1 year ago #
jond - Member

Err..kinda, but not quite.

The driver will try to drive current (ie 1A) at whatever voltage is necessary to drive that current (within it's designed operation). So what you're trying to do is dump the excess current (ie 300mA) through the resistor at whatever the controller winds up driving the leds at.

The drop across the leds is going to be something like 3-3.5V each (varies manufacture, current through them, temperature) so your resistor needs to sink 300mA with about 6-7V across it.

The worse case you need to work out is when the led drop is *low* - then if it happens to be high, the resistor will just sink relatively more current.
That's 1.8W in a 20 ohm resistor - that's gonna get a bit warm (and you might need to use a 5w rather than 2w to give a bit more margin).

You do realise you can buy a 350mA driver from BCT (see assorted lumi/led threads) for about 6 quid...

Posted 1 year ago #
trout - Member

the leds are XPE red ones not XRE .

vf if i remember right is about 2.5 v each and 700 ma max

Posted 1 year ago #
simonfbarnes - Member

they're being used as a tail light in series with a front light

do you actually mean this? Sounds a dumbass idea as a single point of failure takes out both lights Never mind the idea of dumping the "excess" power into a non-light generating load! Let's not dignify it with the name of electronics, it's "bodgism"

Posted 1 year ago #
Mowgli - Member

They're run off a dynamo which doesn't produce enough power to run two drivers, hence having to run them all off one.

Posted 1 year ago #
crogthomas - Member

A dynamo is pretty much a constant current device, so get rid of the drivers and just run the LED's directly from the rectified output. Works beautifully, apart from a little flicker at low speeds.
Maybe this site will help:
http://www.pilom.com/BicycleElectronics/DynamoCircuits.htm

Posted 1 year ago #
Mowgli - Member

Yeah, though sadly the front light needs 1000mA and the rear could do with 350-500mA, so can't run them directly. All the BCT drivers need 7-14V which is more than the dynamo gives (6V).

Found a low voltage 'bucktoot' which might do the job though, as it works from 5V. Quite expensive compared to BCT though, £17.

Posted 1 year ago #
aracer - Member

They're run off a dynamo which doesn't produce enough power to run two drivers

Wrong - it's not the drivers which need the power, it's the LEDs. By using a separate lower current driver for the rear you'll actually need less power from the dynamo.

All the BCT drivers need 7-14V which is more than the dynamo gives (6V)

Except the dynamo won't only give 6V - if you put a voltmeter across it with no load you'd find many hundreds of volts. Dynamos are current limited, not voltage limited - that 6V is only a nominal output voltage.

I'm sure it is possible, but it's a lot more complex than you seem to think.

Posted 1 year ago #
aracer - Member

...thinking about it a bit more, a dynamo is a current limited device. A standard "6V" dynamo is designed to drive a 3W bulb, hence 500mA current limited. Your Seoul at 1000mA will be over 3.5V, hence close to 4W in total including the driver. The only way the dynamo is going to provide that is by outputting 500mA at 8V. For a start, how exactly is it you're planning on driving the front LED for a start, as a typical DC driver circuit won't let the dynamo voltage get that high?

If you're looking at driving all 3 LEDs in series with a parallel resistor to dump some of the current (not that that would be a good idea in any context), then you'd have a over 10V across all the LEDs, hence 10W required, so the dynamo would have to be outputting 500mA at 20V! The dynamo output requirement is actually rather lower if you do something more sensible, but as mentioned above, it's all a lot more complicated than you think.

Posted 1 year ago #
simonfbarnes - Member

They're run off a dynamo which doesn't produce enough power to run two drivers

so throwing away some of the usable power is a really bad idea! Just fit a suitable zener to soak up any overvoltage!

if you put a voltmeter across it with no load you'd find many hundreds of volts

not strictly relevant as when it's not connected the voltage doesn't matter

Posted 1 year ago #
Mowgli - Member

Right for the sake of a fiver I've bought one of the BCT things, even though it needs a higher voltage than what I reckon the dynamo gives.

Maybe in the time it takes for delivery I can get my head around the theory here...

Posted 1 year ago #
aracer - Member

From your comments on here, I have to assume you don't actually know what you're doing. I'm afraid plugging that BCT driver straight into the dynamo simply isn't going to work.

Not strictly relevant as when it's not connected the voltage doesn't matter

Was just an example of the theory, sfb - if you want a better one, then if you put a 12V 6W (ie 500mA drive current) bulb across a "6V" dynamo, then the dynamo will happily output 500mA at 12V. The trouble is, an LED driver doesn't present as a resistive load, hence the dynamo won't automatically provide the required power.

Posted 1 year ago #
Mowgli - Member

Posted 1 year ago #
simonfbarnes - Member

then the dynamo will happily output 500mA at 12V

if you pedal fast enough!

Posted 1 year ago #
aracer - Member

Give me some credit

I was - by "straight into dynamo" I actually meant "straight into bridge rectifier" - had assumed you'd got that far. Care to give details of your current front light setup, and have you measured the current at the LED? From your description, I'm dubious it is actually running at 1000mA.

Trying to help here - I'm a professional electronic engineer (when I can escape from doing software), yet would have to go and look up lots of stuff to work out how to do this properly with a dynamo, so no shame not to understand it. I do know what won't work though.

Posted 1 year ago #
simonfbarnes - Member

were expletives used ? I see a blank post...

Posted 1 year ago #
Mowgli - Member

I'm not as thick as you might believe.

My front light have been working fine for months. I just modified it so it also gives a 5V USB output to charge my iphone, and now I'm adding in the rear light too.

The front has a 1000mA constant current buckpuck driving it, which can accept a 5-30v input. It's damn bright, but I don't have a way of measuring the current or voltage. As the buckpuck is 'constant current', I think we can assume it's delivering 1000mA.

Posted 1 year ago #
aracer - Member

were expletives used ?

No - seemed a perfectly reasonable post to me. If anything it was a bit angry directed at me defending his own competence, but I certainly didn't have a problem in the context that I'd been going in quite hard.

Posted 1 year ago #
aracer - Member

As the buckpuck is 'constant current', I think we can assume it's delivering 1000mA

Unfortunately no you can't, not given that the input is current limited, rather than a voltage source as is assumed by the buckpuck designer.

Posted 1 year ago #
simonfbarnes - Member

I'm not as thick as you might believe.

I think people might have been detecting some ignorance

I think we can assume it's delivering 1000mA.

again, if you're going fast enough...

My front light have been working fine for months. I just modified it so it also gives a 5V USB output to charge my iphone

that is daring

Posted 1 year ago #
TheBrick - Member

I'd been going in quite hard.

This always requires consideration.

Posted 1 year ago #
aracer - Member

My gut feeling was that your front light driver was just doing direct drive with the LED seeing the current limited by the dynamo - ie 500mA if it's a standard "6V" dynamo. Having had a proper think about it, I'm convinced that is the case - the buckpuck will measure the current using a sense resistor, and given it's seeing less current than it wants will increase the on duty cycle. As the on duty cycle is already 100% it can't do any more so will get stuck there, never allowing the dynamo voltage to rise enough to supply the required power.

Your LED might seem "damn bright", but then an LED driven at 1000mA vs 500mA is only 60% to 70% brighter, and given the logarithmic nature of vision, that only translates into a 25% or so increase in perceived brightness.

Posted 1 year ago #
Mowgli - Member

I'm off to bed. Below is the circuit that, today, charged my phone on the way to work and lit the street on the way home. Sorry for the dodgy MS Paint effort. The circuit fits inside the headtube - the old one was in a tatty saddlebag.

If the buckpuck's not giving 1000mA then it may as well be taken out I guess. I will try putting the front light directly to the rectified dynamo output.

Posted 1 year ago #
simonfbarnes - Member

never allowing the dynamo voltage to rise enough to supply the required power.

??? if it's stuck hard on it isn't stopping the dynamo from doing anything it can't do for lack of revs

Posted 1 year ago #
aracer - Member

if it's stuck hard on it isn't stopping the dynamo from doing anything it can't do for lack of revs

It is stopping the dynamo voltage from rising because of the dynamo current limit. We have a habit of thinking of a voltage driving current through a load because most sources are voltage sources (or at least closer to a perfect voltage source than they are to a perfect current source). A dynamo works the other way round - the voltage will rise enough to supply a fixed current. In this case with the converter working in direct mode, the dynamo voltage will rise until it is supplying the LED with 500mA - ie the LED Vf at 500mA plus RDS for the FET times 500mA. The voltage can't rise above this, as in order to do so it would need to supply more current, despite the fact that if supplied with a suitable load the dynamo would quite happily provide 500mA at 8V (ie 4W, sufficient to run the LED at 1000mA).

Posted 1 year ago #
simonfbarnes - Member

The voltage can't rise above this, as in order to do so it would need to supply more current

I'm sorry, but this is twaddle

Posted 1 year ago #
Mowgli - Member

I took the buckpuck out, the three LEDs are running in series and seem about as bright as before, so it seems the buckpuck was limited by in the incoming current. Good to know, and saved me a few quid (got a refund on the BCT before it was shipped). The lights still have to be in series though to get the 500mA. I might replace the front LED as at the moment the Seoul is only running at half it's capacity.
Cheers

Posted 1 year ago #
aracer - Member

I'm sorry, but this is twaddle

Would you have a problem with the statement "the current can't rise above this, as in order to do so it would need to be driven by a higher voltage"? If it helps to make it simpler consider a purely resistive load.

Posted 1 year ago #
aracer - Member

The only issue you then have with running 3 LEDs in series, as sfb keeps alluding to, is that you then have a minimum speed requirement to generate enough voltage. If you're going slowly then your lights will go out - in such a case you'd actually do better by switching the LEDs to parallel drive, or just switching off the rears using a bypass switch to short them out (you do have a backup LED as well given lights that go out when you stop?)

Regarding the Seoul LED, I'd leave alone. There's nothing you can replace it with which will provide more light at 500mA (well I suppose there is, but then the same argument would apply if driving at 1000mA - the lower drive current isn't of itself a reason for switching).

Posted 1 year ago #
simonfbarnes - Member

Would you have a problem with the statement "the current can't rise above this, as in order to do so it would need to be driven by a higher voltage"? If it helps to make it simpler consider a purely resistive load.

equally senseless

Posted 1 year ago #
goldfish24 - Member

I don't think there's space here to convincingly explain current sources and sinks, but I do want to pitch in and say that aracer is completely correct.
Most peoples basic understanding of electronics is based around voltage sources, I.e batteries, and resistive loads, I.e bulbs and resistors (bulbs not perfect in this sense). A diode, however, is a current sink, that's why we have to use a 'driver' which is basically a voltage limited current source for the led. The dynamo is also a voltage limited current source. Ideally we'd use a separate driver for the front and rear light, bit it's not clear how the current will be shared when the drivers are not saturated. In all probabilities one led will steal all the current.

Btw, a 1 amp current source will not supply 1 amp @ the threshold voltage of the led if there isn't enough power going in!

Regards Dr phil, electronic engineer.

Posted 1 year ago #
aracer - Member

~~equally~~I am senseless

FTFY

I though you understood Ohm's law and basic electronics?

Thanks, Dr phil - I agree the understanding issue is because of the current sources and sinks not being people's normal understanding of things.

Posted 1 year ago #
simonfbarnes - Member

I though you understood Ohm's law and basic electronics?

which is what leads me to believe that what you've written is tosh...

Posted 1 year ago #