Matching to the impedance of an alternator.

Could i see the models you used to set that up? I have an EL = application.

?-)

Duh! It's a HUB dynamo :-) ...Jim Thompson

Thanks, Marcel! I know just enough German to make a fool of myself :-) But I think I can decipher the data. Thanks! ...Jim Thompson

(Normally

loads to

using

voltage.

few

Of course. We used to go up to the Radcliffe (now officially Harvard) dorms ;-) ...Jim Thompson

Your perception of "brightness" is logarithmic. You probably need twice as much as your "gut" thinks.

load

getting

suspect

would

series

load

here -

current

the

pair

allowed

(experiment

Output

subliminally

posted

pay

rectification

Not always.

Perhaps. Sourced from an inductance? ...Jim Thompson

load

getting

suspect

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pull

load

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current

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Output

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got 6V

rectification

can

can

Close enough to work. Nothing is "always."

The source is practically a constant current. Transformers work with currents.

John

Watts not lumens.

I tinkered for a few minutes and came up with this. I think it may be in the right direction.

Version 4 SHEET 1 880 680 WIRE 128 0 80 0 WIRE 336 0 208 0 WIRE 512 0 336 0 WIRE 80 32 80 0 WIRE 80 128 80 112 WIRE 480 176 400 176 WIRE 512 176 512 80 WIRE 512 176 480 176 WIRE 400 224 400 176 WIRE 480 224 480 176 WIRE 512 224 512 176 WIRE 80 336 80 208 WIRE 304 336 80 336 WIRE 400 336 400 288 WIRE 400 336 304 336 WIRE 480 336 480 288 WIRE 480 336 400 336 WIRE 512 336 512 304 WIRE 512 336 480 336 WIRE 304 368 304 336 =46LAG 304 368 0 =46LAG 336 0 Vo SYMBOL voltage 80 112 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value SINE(0 {10*phi} {20*phi}) SYMBOL ind 64 16 R0 SYMATTR InstName L1 SYMATTR Value 1 SYMBOL res 224 -16 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R1 SYMATTR Value 1 SYMBOL res 496 -16 R0 SYMATTR InstName R2 SYMATTR Value 5 SYMBOL res 496 208 R0 SYMATTR InstName R3 SYMATTR Value 5 SYMBOL schottky 384 224 R0 SYMATTR InstName D1 SYMATTR Value 1N5818 SYMATTR Description Diode SYMATTR Type diode SYMBOL schottky 496 288 R180 WINDOW 0 24 72 Left 0 WINDOW 3 24 0 Left 0 SYMATTR InstName D2 SYMATTR Value 1N5818 SYMATTR Description Diode SYMATTR Type diode TEXT 96 272 Left 0 !.STEP param phi 1 10 3 TEXT 46 346 Left 0 !.tran 0.5 TEXT 296 24 Left 0 !.IC V(Vo=3D0)

...

Not the vacuum pump lubrication?

Grant.

What?

No, John. You are just plain wrong in the case of polyhase motors. The windings create a true rotating field and the PM rotor must try to = follow. It creates real and absolute torque in the direction of the rotating field.

Did you really learn anything about salient pole synchronous motors either? I bet not. They start by them selves even for single phase.

How could you possibly misunderstand that so badly. You deserved F grades like your scores indicated.

can't)

power

Try it before making a bigger posting any more. You shoot your mouth off a lot about "try it".

Now, now, josephkk, don't you know the rules around here? ...

Rule #1: Larkin is always right.

Rule #2: When Larkin is wrong, you must reply per Rule #1 ;-) ...Jim Thompson

No, in a pure synchronous motor, if you just slam on the AC, the torque wobbles at 60 Hz. If the rotational inertia is very small, the poles will follow the rotating field and it will start instantly at full speed. In a more normal situation, the poles can't move that fast, they miss tracking the field, and fall off. There's rotational vibration but the shaft doesn't turn.

Some synchronous motors have induction-start add-ons or some other way to get starting torque. Some require a pony motor or some other mechanism to spin them up.

I don't know why you want to say silly things without doing a little research first.

and

and lots more.

Absurd.

John

I don't have a bicycle alternator. Somebody who does could try it.

But think about it: these things are run close to shorted anyhow, which is why the lamps don't burn out when you go faster. In the normal speed range, the current is determined by a large induced voltage dumped into a lot of inductive reactance. The current is almost 90 degrees out of phase with the excitation voltage, so the vector product is nearly zero... no real power. So the current is not associated with a torque. An external ohmic load absorbs true power and rotates the phase of the current towards the phase of the voltage. which *is* real power, which requires real torque.

John