4
\$\begingroup\$

I've built "linear generators" (a coil on a tube, with a free magnet inside), with different layouts and I want to measure which is one "is the best", or, at least, make some kind of profiling on them.

I know that it's not exactly "efficiency" (maybe potency?), but I want to know, how would I compare them to find which one is the best in a "quantitative way"?

Edit: my current idea is to drop the magnet inside the tube, from a fixed point above the coils. The magnet should have the same energy when passing the coil, so I could measure the efficiency. The problem is, how would I measure that? It would be the current on a resistor? The total charge on a capacitor?

\$\endgroup\$
  • 1
    \$\begingroup\$ Is this linerar generator supposed to be worn by a person or shaken by hand? It makes a difference! If you design this thing to be worn, efficiency is not the thing you want to measure, as the mechanical power that goes in is meaningless: If a person hauls around a brick, the "mechanical input" into that brick is 0. I would rather measure the absolute electrical power output and compare it to some other property like weight. \$\endgroup\$ – 0x6d64 Nov 21 '11 at 9:07
5
\$\begingroup\$

Build a two-at-once test jig:

  • Connect the two side by side separated by a convenient handle. This could be done as simply as by using a few pieces of wood or plastic and some tape. Do not use metal.

  • Ensure that there is enough separation that they do not interact. The width of a hand between them should be very adequate.

  • Arrange "handle" such that when handle is held each shaker is orientated in the same way that it would be if it was being held instead of the handle. This is so that the hand shaking motion shakes both at once in much the same way as it would if they were being held directly.

  • Wire outputs via wires and a rectifier to two identical loads. This could be a large capacitor (maybe a super cap, or a resistor with an oscillocope monitor or an LED or whatever.

Compare both at once:

  • Operate the pair in differing ways and note the outputs. This allow direct comparison without changes in shaking pattern, speed, stroke etc having to be standardised.

  • Use of a large cap as load with low leakage allows voltage to be slowly "pumped up" over time - a clear winner should emerge.


Question - Julio asked

  • If I could be sure that the same input energy would be used on each test, how could I use the oscilloscope to measure the efficiency

Answer:

Use a dual channel scope.

Connect one side of each generator to ground.

Connect a load resistor of the same value across each output.

Connect a scope probe / channel to each output.

Shake it all about.

Observe two waveforms on scope.
Make decisions about which looks "best".

Try different resistor values.

For DC comparisons, rectify to DC and use output capacitor perhaps 100 uF per channel.


My "large capacitor" charging test is arguable one of the best as
- it can be used to simulate battery charging,
- you can see the result at different voltages
- and its easy to measure with just a voltmeter (or two voltmeters).

You can go from eg
0 to 3 V
or 4V to 5V etc as suits you (to eg simulate battery charging).
You can use a power supply and battery to precharge the cap.


One at a time testing:

With a large enough capacitor you can test one shaker at a time with a reasonably good chance if comparison.

Say it takes about 1 minute to go from Vstart to Vend.

Test each in turn at the same level of shaking and see which is faster or which charges higher.

You should be able to match tests to within about 10%.
Try it several times on the same shaker to see.
Then anything outside that difference is probably due to relative performance.

\$\endgroup\$
  • \$\begingroup\$ If I could be sure that the same input energy would be used on each test, how could I use the oscilloscope to measure the efficiency? \$\endgroup\$ – JulioC Nov 22 '11 at 3:58
  • \$\begingroup\$ See addition to answer. \$\endgroup\$ – Russell McMahon Nov 22 '11 at 4:13
  • \$\begingroup\$ This answer makes a lot of sense for comparing inefficient things like shake generators where the work done just cycling them (and your hand) dwarfs the tiny amount of mechanical work that actually gets converted to electrical power. But for more efficient generators than are apparently being contemplated here, this idea won't work, as there's no way to tell how much of the mechanical load is due to each generator. \$\endgroup\$ – Chris Stratton Nov 22 '11 at 7:46
  • \$\begingroup\$ @ChrisStratton - Yes, but. The question was absolutely targeted to his home built shake generators so my answer is very much custom tailored to that question. It would be hard to extend my answer in a useful way to most other forms of generator. \$\endgroup\$ – Russell McMahon Nov 22 '11 at 8:08
5
\$\begingroup\$

Measure the mechanical work input and the electrical work output.

It sounds like you might be able to drive your generators mechanically using an electric motor with a crank and connecting rod to move the magnet like a piston. You could take comparative measurements by measuring the electrical power input (for example, current while running from a voltage regulated supply) and on the output measuring the voltage across a load resistor small enough in value to increase the current draw of the motor when connected, but physically large enough not to heat up (as its resistance might then change). This would be a comparative measurement only with which you could compare your different generators, as your calculated efficiency would be the product of your driver efficiency (which you wouldn't know) and the efficiency of your test generator. Arguably you need to run the driving motor at the same speed for all the tests, or have a plausible model for how its efficiency should vary with speed if your test generators have different speed requirements.

You could try doing something directly physical, for example a mass on a string winding or unwinding from a drum. In the real world this doesn't work as well as it does in physics textbooks though - inertia can overwhelm the continuous effect you are trying to measure, and the mass tends to want to swing side to side and bump into your knees or workbench, instead of just ascending/descending cooperatively.

Finally, consider that ultimate efficiency may be less important than effective utilization of the source of mechanical power available. For example, if you are building a shake flashlight, you probably want either the design that is the brightest, or the one that provides just enough light to see by for the least vigorous (or frequent, if you have caps in there) shaking.

\$\endgroup\$
  • \$\begingroup\$ Great answer. The initial idea was to drop the magnet from a fixed point and compare the measured "power" generated on the coil, but what would be that power (current, voltage, potency)? What method would I need to apply to get that? \$\endgroup\$ – JulioC Nov 22 '11 at 4:04
  • \$\begingroup\$ Current times voltage is power, integrated over (or if constant multiplied by) time gives work. Work would correspond to the energy given up by magnet falling through the coil, though likely it won't be slowed that much. If you had a pair of photogates with a timer you could measure the speed as it exists the coil and compare to what it should be after falling that far (or without the coil) to determine the actual mechanical power input. \$\endgroup\$ – Chris Stratton Nov 22 '11 at 7:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.