I only invoke the Dr title when the good laws of physics are at stake. Now that's cleared up, it's just phil 
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Electronics question
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Posted 1 year ago #
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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.
Posted 1 year ago # -
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 ?
Posted 1 year ago # -
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 X is some numeric multiplier,
omega is the rotational velocity in radians per second and theta 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...
Posted 1 year ago # -
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.
Posted 1 year ago # -
yes I know Ohm's law.
Posted 1 year ago # -
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).
Posted 1 year ago # -
zzzz - uh, wot ?
Posted 1 year ago # -
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.
Posted 1 year ago # -
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
Posted 1 year ago # -
no enlightenment aracer ??
Posted 1 year ago # -
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 twaddleFirst 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.
Posted 1 year ago # -
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...
Posted 1 year ago # -
isn't this a circular argument - "It can't because it can't" ?
No - the voltage can't rise because it's a current source which can supply a maximum of 500mA. It would be basic Ohm's law if things were linear (hence why I mentioned that). The fact the voltage/current curve for the load is non-linear doesn't change the fact that for a given current the voltage is fixed.
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...
Yes - 500mA for a "6V" dynamo at any speed. 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?
Posted 1 year ago # -
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 ?
Posted 1 year ago # -
@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
Posted 1 year ago #
Topic Closed
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