I build different small PM handcrank generators for fun, Recently I built a single phase generator and printed myself a couple transmissions. Now as far as I know 3 things affect voltage, speed of rotation, coil size/#of windings, and magnetic strength. Where I am mentally stuck is that being the same generator (all things equal) when I ran it on my 1-250 transmission, i generated 9 volts with a given load, with the same load and generator, using the 1-400, I got the SAME output, even though I was going faster (i have an rpm reader). Up until now, all the generators I have bought, many of them wye 3 phase, the faster i go, the more voltage. I am certain I am not reaching the limit of my wire, as im using 22 gauge, and only pushing 2.5 amps max. While there is a possibility this was a testing error, I double checked and got the same results. Has anyone ever run into this Voltage wall before? as in no matter how fast you rotate, you hit a wall.

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    \$\begingroup\$ Have you considered the winding inductance? As you increase the frequency the impedance will increase. \$ Z = j \omega L \$. That means that there will be an increase in the internal impedance of your coils and they will act as a potential divider with the load. \$\endgroup\$
    – Transistor
    Dec 1 '18 at 17:17
  • \$\begingroup\$ As transistor points out, the basic inductance of your coil is an impedance in series with your out-feed and that impedance rises with frequency. Halving the turns can quarter the inductance (but only reduces output voltage by half). More tightly winding the turns but having fewer of them gives better high frequency performance under load conditions. \$\endgroup\$
    – Andy aka
    Dec 1 '18 at 17:35
  • \$\begingroup\$ @Andy, we're old enough to remember the Sturmey Archer dynohub which relied on this principle to give a fairly constant voltage to the bicycle lamp over a wide range of speeds. \$\endgroup\$
    – Transistor
    Dec 1 '18 at 19:09
  • \$\begingroup\$ You have discovered dynamic induction \$V_{Induced} = NB \ell v\$. where N = turns, B = flux density, \$\ell\$ length of coil and v = velocity. \$\endgroup\$ Dec 1 '18 at 19:37
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    \$\begingroup\$ @Stainless: I think the problem is that that's what he was expecting but didn't get - at least not at the winding terminals. \$\endgroup\$
    – Transistor
    Dec 1 '18 at 20:51

As discussed in the comments, you need to consider the winding inductance. As you increase the frequency the impedance will increase as given by the equation \$ Z=jωL = 2 \pi fL \$. That means that there will be an increase in the internal impedance of your coils and they will act as a potential divider with the load.

You can do make a reasonable measurement with your multimeter as follows:

  • Take an open circuit true RMS voltage measurement from your generator.
  • Connect up to your load and find what frequency causes the output to be half the open-circuit voltage.
  • Repeat step one at that frequency and check step two again to confirm that the AC reading is halved when on load.

Now you have the frequency where the impedance of the coil matches that of the load, R. From that we can say $$ R = 2 \pi f L $$ so $$ L = \frac {R}{2 \pi f} $$

For a more thourough explanation see Is it possible to measure inductance without LCR Meter/OSC/FG.

With the coil's inductance and resistance measurements you can do all sorts of calculations and predict the output to a given load at various speeds.


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