K5 Pointy Headlight Connector Burning out

[Ruffy]

Originally Posted by aesmith

The quoted 60V AC is an open circuit figure, with the AC disconnected from the RR. I would guess it drops substantially when load is being drawn.

Originally Posted by Bibio

as far as i know it dont work like that. the AC will rise and fall slightly with engine speed but not load. i could be wrong though.

In the real world, it's a bit of both.

• Voltage will drop with load because of the undesirable internal resistance of the alternator coils (The workshop manual quotes 0.2-0.7ohm so I estimate 1-10V drop depending on actual load and coil resistance.)
• Voltage also varies because of the engine speed (the desired generator effect - the faster it spins through the magnetic field, the greater the voltage).

[Ruffy]

Originally Posted by glang

Dont forget though that theres far more forced cooling available on a moving bike especially as, handily, more speed equals more heat to be dissipated.....

I remain to be convinced how significant that will be. Theorectically correct,yes.

On that theme, it will also be theoretically better in winter due to lower ambient temperatures. Who says winter biking is all bad!

[glang]

I think the voltage changes quite a lot with speed (not as much as an old fashioned dynamo which is why they were replaced) and thats why the headlights get brighter if you increase the revs from idle.
Also I think the volts do drop with load where the load is made up of real load (lights, heated gloves etc) and the artificial controlling load of the R/R shorting the alternator via its transistors....

[SV650rules]

http://www.bandofriders.net/forums/s...ad.php?t=82950

This shows how to fit a Honda (CBR600) RR to SV650

[aesmith]

It would be interesting to measure the AC current at a certain engine speed, with head lights on or off. If the current was unchanged it would support the theory that the alternator works at a standard rate, and excess power is dumped in the RR.


Not sure about the theory behind voltage dropping from the O/C figure when current is drawn, I had always assumed it was the case as an analogy with a motor drawing more current as it's put under load. Is it purely down to actual resistive load, or is there something directly related to the fact that power is being drawn? I know that regulators for small wind turbines vary the current drawn to hit the optimum power point for any given conditions.

[aesmith]

Duplicate for some reason

[SV650rules]

Originally Posted by aesmith

I know that regulators for small wind turbines vary the current drawn to hit the optimum power point for any given conditions.

The wind turbines have to regulate the load otherwise there is a good chance the old propeller would stop rotating, they have to limit the load to the optimum for wind speed. The wind turbines probably have a wound rotor where the field strength can be adjusted, Funny thing is about wind turbines, they need to draw power from the grid to be able to generate power, maybe once they are rotating fast enough they can be self sufficient.

https://www.wemoto.com/info/regulator_rectifier

Also this https://bikesamag.co.za/bsa/how-a-mo...em-works/6384/

In the first article the regulator bit is described as a 'shunt regulator' and in electrical terms 'shunt' means 'an alternate / parallel path' - ( in the south african article it says 'shunts excess current to earth - 0V) so it looks like the regulator is in parallel with alternator and 'shunts' any unwanted current through itself and dissipates it as heat (all energy ends up as heat). The most basic voltage regulation system consists of a resistor and zener diode ( a zener diode has a low forward volt drop but when it is reverse biased (positive cathode) they are made to have a fixed and very sudden 'avalanche' breakover at a fairly accurate voltage - the breakover voltage is altered by different doping of the silicon). The variable supply voltage is connected through a resistor to the zener and the 'regulated' voltage is supplied to downstream circuit from straight across the zener diode. If the upstream voltage rises the zener will divert any extra current through itself until it is drawing enough current to drop the voltage through the resistor to its zener voltage, if the downstream load varies and the current through the resistor changes the zener will attempt to keep the downstream voltage constant. You need a big heatsink on the zener as it dumps a lot of heat, I figure that the regulator bit on the RR works the same way.

Originally Posted by aesmith

By the way those LED bulbs linked earlier are actually 24W each, 48 for the pair. That's a typical figure that I've seen for H4 replacements. LEDs will be constant current so that power draw will be fixed irrespective of battery voltage. In comparison standard lights are 60/55W at nominal 12.0V so a pair on full beam will be drawing probably over 140W in reality. So nearly 100W different with the LEDs.

That makes sense - about the same as my Phillips chipped ones (with no fan), I thought the fan may be drawing the other 48 watts LOL (big fan) - the Phillips ones draw about 20 watts but are 80watt equivalent filament light output.

[aesmith]

By the way those LED bulbs linked earlier are actually 24W each, 48 for the pair. That's a typical figure that I've seen for H4 replacements. LEDs will be constant current so that power draw will be fixed irrespective of battery voltage. In comparison standard lights are 60/55W at nominal 12.0V so a pair on full beam will be drawing probably over 140W in reality. So nearly 100W different with the LEDs.

[Barney626]

Haven't really had time to read through the whole thread since I posted and it seems to have veered off to a slightly different topic.

But for anyone wondering about the LED bulbs, they are noticeably brighter and more defined than the philips ones I had in before, also havent had a single issue with them burning out yet.

Happy so far...