Could a sh775 kill a stator?

[Gorminrider]

Current can't stop instantly, if it did V = L dI/dt (time derivative), voltage, would have to be infinite. ; where did you see that?

here...posSSRgraph.jpg and yes I did assume V would have to be theoretically infinite but in real life with wire in a field I expect it doesn't really happen.

[DohcBikes]

You guys get that rocket built yet?

[Gorminrider]

The voltage regulator is the short and yes that short might be a small voltage (less than 0.5V) but that is why the R/R gets hot

Well no, my kindergarten electronics don't see it that way when the shunt is the only load in the circuit ...It must have a higher vdc.



Simplistically,and with wild assumptions as to proportion, Pretend the ground potential is 14vdc and the shunt is shunting ALL the stators output.
My idea says the RR portion is say 6vdc@15 amps. Thats 90 watts consumed. (The stator's would be 8vdc@15amps) That would be warm to the touch with a pretty good heatsink !
Your idea seems to say the RR portion is 0.5vdc@15 amps. That's 9 watts. I can't think of much that could ever burn out with 9 watts if it was encased in an aluminum heat sink. I don't think it could be felt with a touch.

of course amperage varies per stator and what loads are also on the bike...headlight, battery charging etc etc

[posplayr]

here...[ATTACH=CONFIG]48473[/ATTACH] and yes I did assume V would have to be theoretically infinite but in real life with wire in a field I expect it doesn't really happen.

The light blue is the current(see the legend).There is no place where the current has a vertical segment terminating or starting from zero.

[Gorminrider]

The light blue is the current(see the legend).There is no place where the current has a vertical segment terminating or starting from zero.

I must be misreading it. I don't find it easy. Ideally it should be 0 according to the general take on series regulation . I assume this pdf attempts to show how it works in detail. Nothing is perfect... the blue line squiggles above and below the line. It is 0 whenever it does and the average is near enough 0 ...coincidentally when the yellow voltage line does take a nearly square" hike".
But I will go back to my first impression... "ringing" on cutoff.

[posplayr]

Well no, my kindergarten electronics don't see it that way when the shunt is the only load in the circuit ...It must have a higher vdc.



Simplistically,and with wild assumptions as to proportion, Pretend the ground potential is 14vdc and the shunt is shunting ALL the stators output.
My idea says the RR portion is say 6vdc@15 amps. Thats 90 watts consumed. (The stator's would be 8vdc@15amps) That would be warm to the touch with a pretty good heatsink !
Your idea seems to say the RR portion is 0.5vdc@15 amps. That's 9 watts. I can't think of much that could ever burn out with 9 watts if it was encased in an aluminum heat sink. I don't think it could be felt with a touch.

of course amperage varies per stator and what loads are also on the bike...headlight, battery charging etc etc

Sorry, I'm not even going to address you analysis without a schematic.

You apparently don't want to take my word for it, so I suggest you actually look a circuit diagram of the stator(see one below). For example each stator leg is a AC source ,with series resistance and inductance. There are three AC sources all 120 degree out of phase.

When you short two of those phases together you have a circuit of 6 elements (2AC source, 2 resistors, and two indicators) all in series. Assume the short is a simple resistor with about 0.05 ohms.

You know the DC resistance of the stator from the factory manual, and you can estimate the reactance of the inductance from the phase angle when not regulating.

This might help you understand why current keeps flowing in the stator despite the shorted R/R. You can see according to 1st principles of circuits, the two legs have to be equal and opposite by symmetry. If they are non zero then the neutral moves from ground and current keeps flowing driven by the two AC sources in each leg.

https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

his law is also called Kirchhoff's second law, Kirchhoff's loop (or mesh) rule, and Kirchhoff's second rule.
The principle of conservation of energy implies that
The directed sum of the electrical potential differences (voltage) around any closed network is zero, or:More simply, the sum of the emfs in any closed loop is equivalent to the sum of the potential drops in that loop, or:The algebraic sum of the products of the resistances of the conductors and the currents in them in a closed loop is equal to the total emf available in that loop.

[Gorminrider]

rectifier_scheme.jpgI am thinking of this...but part of my analysis would involve placing a finger on the heatsink of an operating shunt R/R

[posplayr]

[ATTACH=CONFIG]48487[/ATTACH]I am thinking of this...but part of my analysis would involve placing a finger on the heatsink of an operating shunt R/R

That is the same thing, I'm just showing a simplified R/R but expanded stator equivalent circuit.When the R/R shorts nothing goes to the battery.Of course there is still current flow in the 3rd phase but it is all unbalanced. Look at the way the current curve is bouncing around with several different harmonics/frequencies/.

Rule of thumb (V=IxR); if finite current is flowing there is voltage.

[Gorminrider]

.

Assume the short is a simple resistor with about 0.05 ohms

. I get that it can ideally be, but if not, it might be just something less than 6 ohms . Somewhere around full-conduction there are halfway stages that are making them hot. Why else invent a mosfet type? Your model of 0.05 ohms suggests that as your stator burns, a shunt type regulator is colder than the wires attaching it- not impossible; connections ARE often cited- (see 850combat's post?) but anecdotally, how many replace the stator while the R/R is perfectly fine? I think not many. They seem to fail as a pair, or the R/R fails first...
anyways, I've got to get back to the rocket I was building

[posplayr]

.. I get that it can ideally be, but if not, it might be just something less than 6 ohms . Somewhere around full-conduction there are halfway stages that are making them hot. Why else invent a mosfet type? Your model of 0.05 ohms suggests that as your stator burns, a shunt type regulator is colder than the wires attaching it- not impossible; connections ARE often cited- (see 850combat's post?) but anecdotally, how many replace the stator while the R/R is perfectly fine? I think not many. They seem to fail as a pair, or the R/R fails first...
anyways, I've got to get back to the rocket I was building

You tell me the resistance of as power mosfet can't possibly be 0.05 ohms(have you ever even looked at a spec sheet for one?) but it can be 6 ohms????!??
Does 6 ohms across 3 ohms of stator sound much like a short?
Even the bike's electrical system is R=V/I=14/14= 1 ohm.

I'm trying to be nice; so I'll just end this here.

[Gorminrider]

no, I'm not saying mosfet and I have read general specs of SCRs but we don't communicate very well so yes, we might as well "end this here"

[posplayr]

Just to put a bow on this for anybody reading the thread.

An SCR is a bipolar device and the relevant performance for the current analysis is its forward voltage drop when activated. I don't have any spec sheet in front of me but the number is going to be similar to a forward voltage drop of a diode which is about 0.5V for a high power device. The reality is there is a exponential Voltage to current profile, but the limiting voltage with ever increasing current (to the rated limit) is about 0.5V.

On the other hand a MOSFET is a Field effect Transistor that operates on a different principle. The relevant performance of this device is the equivalent resistance of the channel when fully activated. The higher the current rating the lower the resistance. It is also temperature dependent. A 25-50 amp part is going to be in the 0.01 to 0.050 ohm range. Again I'm not going to look up a specific part appropriate for this design but that is what they are. The SSPB for example is a 10 amp FET part in the 60-80 milli ohm range.

For simple circuit analysis , it is much easier to use the FET resistance number as it is a pure resistive parameter and can be compared to the stator resistance which is a couple of ohms. The SCR is not going to be quite as efficient and will use dissipate more power but is close enough for the purpose of the analysis.

In order to short teh terminals of a device that has 1-2 ohms of internal resistance, you have to apply a resistance across that device that is substantially lower resistance than the device itself(the stator). That means to short the stator the R/R equivalent resistance has to be measured in 10'S of milli-ohms and clearly not 6 OHMS!!!!

The proposition that there is 6 ohms of resistance in a shorted R/R demonstrates a lack of any real grasp of the basic operation of the circuit.

[DohcBikes]

^ you have a serious inferiority complex

[posplayr]

^ you have a serious inferiority complex

You don't even know what inferiority is.

[Grimly]

More proof that you are overcomplicating and overthinking this extremely simple system. And maybe obsessed.

The sh 775 goes bad at least just as often as any other reg rec and the gs issue is the suzuki regrec.

Stators don't care how much power they put out. They either can or can't keep up and they couldn't care less either way. The bulk of the power distribution and management is handled by the reg/rec.

shunt regrecs have been working since the time of cave men. except suzuki shunt regrecs.

Bye now.

I don't know where to start...