OK, another long post coming to challenge what I think is some mixed quality theory from others. (All intended to clarify understanding, please don't take offence.)
Quote:
Originally Posted by SV650rules
The rectifier bit ( 3 phase ) produces heat due to voltage drop across diodes, which is a about 1volt per diode for silicon diode, ...
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This is correct. The semi-conductor junction has a fixed volt drop.
Quote:
Originally Posted by SV650rules
... this is independent of current passing through the diode, so heat produced stays pretty much constant. ...
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This is not correct. Power consumed in a component is the product of voltage dropped and current through it. P=IV. Dropping 1V at 10A generates 10x the heat of dropping 1V at 1A. (If this is not the case, why are diodes manufactured and sold with different current/power ratings?)
Quote:
Originally Posted by SV650rules
... The output of the rectifier is unidirectional sine waves, not dc, because it has a ripple. ...
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Yes, technically correct, although the overlap effect of the 3 phases affects the low end shape of the ripple and the battery does a decent job of smoothing ripple due to its energy storage and release capability.
Quote:
Originally Posted by SV650rules
... The alternator output voltage rises as the load current drops, the regulators job is to try to keep voltage at around 14 volts ...
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Yes, although the voltage rise is only due to the internal resistance of the alternator winding, which is generally low compared to the load resistance.
Quote:
Originally Posted by SV650rules
... it does this by drawing current from alternator, the resistance of alternator coils drops the output voltage, this clips the unidirectional sine waves and reduces ripple and produce a low ripple unidirectional voltage. As the alternator rotation speed increase the regulator has to work harder and absorb more current to keep voltage stable, this current causes more heat ( watts ) which is proportional to square of current. ...
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Maybe, but none of this changes as a result of change in load, which is what the original assertion was about.
Quote:
Originally Posted by SV650rules
... The resistance of alternator coils is constant so it needs a constant amount of current drawn to keep output voltage at required value, the more of this current is being drawn by devices ( bulbs, heated grips etc. ) the less current the regulator has to absorb.
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Except the coil resistance is in series with the load, so the volt drop across the coil resistance varies according to load. Yes, this will affect the output voltage that the regulator has to deal with, resulting in higher voltage at lower load, but the effect is much reduced. The current the regulator has to absorb is a function of the over-voltage generated, not directly related to the load current.
Quote:
Originally Posted by SV650rules
...It is pretty crude but works.
As already posted a car system controls alternator voltage output and keeps it stable against load by reducing excitation of rotor coil, some bikes have this system but SV is not one of them
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Yes.
Quote:
Originally Posted by glang
perhaps the diagram from the Suzuki manual will help. The output voltage from alternator changes with revs (the manual says 60vAC at 5000rpm is correct with it unplugged) this is changed when its rectified but its still too high to use in the bikes electrical system.
The three transistors in the diagram to the right of the bridge rectifier are controlled by the regulator to open sufficiently to allow some current to flow back to the alternator. This drops the output voltage of the alternator as it can only produce so much power (=volts x amps) ...
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Those thyristors act as crude crowbars to short circuit across the generator, causing voltage drop because of the internal resistance of the alternator. They're on/off devices and connected parallel to the load, so they operate independently of load current.
Quote:
Originally Posted by glang
... and now more current is flowing. The transistors present a resistance and as Watts=amps x amps x resistance theres an increase in the power to dissipate as heat.
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Don't confuse regulation current with load current. It's more about the (over)voltage that has to be dealt with. The internal resistance of the coils complicates the theory because both currents are superimposed through that, but that does not mean that any load current not "used" is absorbed by the R/R.
Having said all that, I think I may need to experiment! I'm feeling a bit rusty on both my theory and application.