# How to measure the “efficiency” of a generator?

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?

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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. –  0x6d64 Nov 21 '11 at 9:07

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.

• 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

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.

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.

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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? –  JulioC Nov 22 '11 at 3:58
See addition to answer. –  Russell McMahon Nov 22 '11 at 4:13
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. –  Chris Stratton Nov 22 '11 at 7:46
@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. –  Russell McMahon Nov 22 '11 at 8:08