[Closed] Electronics question

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

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.

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...

the leds are XPE red ones not XRE .

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

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" 🙂

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

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

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.

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

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

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

...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.

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

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... 😯

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

then the dynamo will happily output 500mA at 12V

if you pedal fast enough!

Give me some credit

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.

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

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.

were expletives used ?

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

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 [b]is[/b] daring 🙂

I'd been going in quite hard.

This always requires consideration.

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.

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.
[img] [/img]

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

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

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 🙁

I took the buckpuck out, the three LEDs are running in series and seem about as bright as before, so it seems the buckpuck [i]was[/i] 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

I'm sorry, but this is twaddle

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).

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 🙁

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.

[s]equally[/s]I am senseless

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.

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

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

I only invoke the Dr title when the good laws of physics are at stake. Now that's cleared up, it's just phil 🙂

OK, as it's you sfb, I know you're capable of understanding, so clearly I'm not explaining properly.

Let's work backwards - in the latest case http://www.singletrackworld.com/forum/topic/electronics-question?replies=34#post-2288170 what I'm saying is that you have a perfect voltage source, given a small enough resistance as a load it can supply as much current as you like. Now given a fixed resistance, the voltage source will drive a certain amount of current through it - it can't drive any more as that would require a higher voltage. Ohm's law at its most basic.

Get that? If so I'll work backwards to the next step.

Ohm's law at its most basic

and how is this relevant to a circuit made up of an inductor and various non-linear semiconductors ?

for a start, ignoring the forward voltage drops of the rectifiers (1.4V for silicon), the open circuit voltage is V = abs( X omega sin( theta ) )
where [b]X[/b] is some numeric multiplier,
[b]omega[/b] is the rotational velocity in radians per second and [b]theta[/b] the angle of the coil to the magnetic field in the dynamo,
so the voltage is proportional to the speed, but varying from zero to max n times a second. [n = (2*pi)/omega ]
As soon as you connect it up so current flows there's a back EMF v = L dI/dt (ie the inductance of the coil times the rate of change of current)

I may be shaky on the details as it's 40 years since I learned this stuff...

You're going into far too much unnecessary detail, sfb. I can't remember all that stuff, but it doesn't matter - you don't have to understand all the details in order to get the larger picture. It's the fact the circuit is made up of an inductor and non-linear semi-conductors that's irrelevant (even you are simplifying by ignoring the rectifiers - they're non-linear semis, do they not count?)

Try indulging me, and letting me know if you actually understand my last post - I'm working backwards from very basic principles to make my point. We can come onto exponential transfer functions for LEDs later.

yes I know Ohm's law.

Okay, let's flip this around to current sources then. Consider a perfect current source - given a large enough load it can provide as much voltage as you like. Now given a fixed resistance, the current source will provide a certain amount of voltage - it can't provide any more as that would require a higher current. Still Ohm's law (indulge me - I'll get rid of the awkward stuff in a minute).

zzzz - uh, wot ?

If you're feeling bored, then never mind - I really can't be bothered wasting my time trying to educate people who aren't interested.

I really can't be bothered wasting my time trying to educate people who aren't interested.

no, I'm fascinated, but not enjoying the sluggish stepwise approach! If you can educate me I'll be happy 🙂

no enlightenment [b]aracer [/b]??

Sorry - missed that you'd replied.

OK, so you want to jump in the deep end? Let's rewind to where you first seemed to have a problem before I started trying to simplify it:

"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

"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

First of all, lets ignore the non-linear nature of the LED - all that matters is that as you increase the current into the entire circuit (rectifier, driver and diode) the voltage increases and vice versa. Let's ignore the detailed nature of the dynamo entirely - instead modelling it as voltage limited (dependent on speed) current source. Get that - if not I'll have to revert to slow step-by-step stuff?

So, as the speed increases from stationary, the voltage of the dynamo will increase until it reaches a point the control circuit has enough voltage to start working, the FET has enough voltage across the gate to start turning on and the LED enough voltage across it to start passing current. With a little more speed and voltage you reach the point where the LED is shining brightly with 500mA flowing through it and the FET is hard(ish) on, since the driver control cirtcuitry senses not enough current so keeps it in direct drive. At this point the dynamo current source is saturated - the voltage of it can't rise any more, since to do so it would have to supply more current, and it can't do that. Hence "The voltage can't rise above this, as in order to do so it would need to supply more current" and "never allowing the dynamo voltage to rise enough to supply the required power."

Let me know if you need more simple steps in there to explain the latter two points you had a problem with before.

At this point the dynamo current source is saturated - the voltage of it can't rise any more, since to do so it would have to supply more current, and it can't do that

isn't this a circular argument - "It can't because it can't" ?

I guess at any particular speed there'll me a maximum current available where its back EMF + the load voltage will equal the open circuit voltage...

isn't this a circular argument - "It can't because it can't" ?

I guess at any particular speed there'll me a maximum current available where its back EMF + the load voltage will equal the open circuit voltage...

Hadn't realised the gap in your understanding was as basic as not getting that the dynamo can supply a max of 500mA - or have I not mentioned that figure before in this thread?

I dunno, but why is this the case (if it is)? And were it to be so, why were you rabbitting on about voltages ? What physical process limits the current ?

@simonfbarnes - the physical process which limits the current is the inductance of the dynamo coil. The faster the hub turns, the higher the frequency of the AC which is generated. The internal inductance of the coil means that the higher current which would be generated is reduced because the frequency is higher. This is generally A Good Thing as it stops the dynamo burning out itself or whatever is attached to it.

Each type of dynamo has a current at which the power transfer is highest. This varies a little with speed, but not a huge amount. More info and testing of a couple of dynamos here: http://bit.ly/eg4dj6