Alternator forensics

[ducwiz]

Jordan,

it seems that you made the tests without a load connected to the windings. I suppose you will receive a much more "sinusoidal" waveform when the nominal load resistance is applied (following the rectifiers).

cheers Hans

[DewCatTea-Bob]

By: Jordan...
" here's the setup. "

____ Very quite nice !


" Note the peaky rectified "sine" wave."

____ Well THAT particular view is to be expected to be pretty-much as is depicted when the voltage-scale to time ratio is set as you seem to have it...
If you were to re-adjust that ratio for the output-readout so that only one cycle were to displayed, then the shape of the displayed DC.curve-pulse wouldn't be as cramped-together (time wise) and thus-then appear displayed more as expected.


" some said that a true sine wave is difficult to achieve, and that power stations try hard to do it. So, all the textbook pictures are just theoretical it seems. "

____ Actually, when the load-system is PURELY 'resistive', then the common textbook-like example of a sine-wave, is kept fairly normal-appearing,, (once the displaying o.scope has been adequately adjusted to present what is considered to be a normal-appearing readout/depiction-display of such).


Fun-Cheers,
-Bob

[Jordan]

Jordan,

it seems that you made the tests without a load connected to the windings. I suppose you will receive a much more "sinusoidal" waveform when the nominal load resistance is applied (following the rectifiers).

cheers Hans

I had the 400 ohm resistor shown in one of the photos to get that waveform. You are right, it is more spiky without it. Should I use more resistance?
With the transformer connected to the house line power, I got a normal sine shape, even without a resistor.

Jordan

[Jordan]

" Note the peaky rectified "sine" wave."

____ Well THAT particular view is to be expected to be pretty-much as is depicted when the voltage-scale to time ratio is set as you seem to have it...
If you were to re-adjust that ratio for the output-readout so that only one cycle were to displayed, then the shape of the displayed DC.curve-pulse wouldn't be as cramped-together (time wise) and thus-then appear displayed more as expected.

I confess I'm no pro with the CRO. I tried every switch combination, and this is the most sine-like waveform I could get so far.
I'll try again when the probes arrive.

Jordan

[DewCatTea-Bob]

" I had the 400 ohm resistor shown in one of the photos to get that waveform. You are right, it is more spiky without it. "

____ That stands to good-reason, as the greater the resistance, the higher then will become the voltage-peak,, (as the o.scope's internal-resistance alone, ought be well-over 100,000-ohms, all by itself).


" Should I use more resistance? "

____ Actually, it's LESS 'resistance' (& more circuit-pathway/conduction) which is what you must really mean.
So-therefore you ought to try a much lower resistance-value -(higher current-conductivity), such-as merely-just 4-ohms.


" With the transformer connected to the house line power, I got a normal sine shape, even without a resistor. "

____ That's because the ratio of IT'S voltage-peak/base to frequency -(time), is much closer to that of the common textbook-example (as it's displayed), with your o.scope adjustment-settings set as they happen to be set at. ...
__ The voltage of your transformer (12v.?) vs. 60-herz is apparently at a ratio that's more suited to your o.scope's present input setting-adjustments, where-as the alternator's ratio of voltage -(up to 100v.) vs. frequency -(at least 3000 up-to perhaps 27000 'herz'), in comparison, provides a much taller and greater-number of sine-wave cycles to be crammed-together on the face of your o.scope-display,, so naturally then, the sine-wave shape can't appear to be much like the common textbook-example, (unless the o.scope-display is adjusted to show it in that rather normal-appearing way).


Enlightened-Cheers,
-Bob

[ducwiz]

I had the 400 ohm resistor shown in one of the photos to get that waveform. You are right, it is more spiky without it. Should I use more resistance?
With the transformer connected to the house line power, I got a normal sine shape, even without a resistor.

Jordan

Jordan,

400 ohm is not the nominal load for the alternators, you should use much less!
To my knowledge, the 5-coil CDI type produces ~75Watt@5000rpm, the 6-coil version ~90Watt. In case of the dual-halfwave rectification connected to all 3 leads, and choosing 7Volts as the nominal voltage, you need ~10Amp resp. 13Amp for the rated power outputs. Using a fullwave bridge rectifier connected only to the yellow leads and 14Volts, ~5Amp resp. 7Amp are necessary. As power equals voltage * current and resistance equals voltage / current, in case of 7 Volt you need roughly .7/.5 ohms load resistance, and 2.8/2.0 ohms for 14 Volt. You can construct those resistances by burdening the stator with several 6V or 12V bulbs with the proper power ratings, which will simply sum up due to the parallel connection. Please let as know of your results ...

Hans