Tweet
Some figures to think about.
Since Dave is temporarily out of service and unable to do any testing I decided to build a test stand for testing different kinds of coils. I have also just gotten a scope that will give me a readout of the rms or p-p or dc voltage levels. It will also give me the frequency of the signal.
So I ran a baseline test with no coils at all on my test setup. My motor was running on 12.4 volts and drawing 1 amp after the test setup had warmed up and the bearings had loosened up.
I then put one of my test coils on the test stand. The output of the test coil was connected to a 50 ohm load. The rms value was 9 volts and the frequency was 90 hz. 9 volts into a 50 ohm load means the load was consuming 1.62 watts.
The input to my motor went up by .1 amps. .1 amps times 12.4 volts means the motor was now consuming 1.24 watts more than before. So in other words for an increase in power input of 1.24 watts I was able to get 1.62 watts out. I realize that does not mean the overall efficiency is OU. But it does mean this particular coil is very efficient at converting mechanical energy into electrical energy.
You guys can argue all you want about what efficiency means but as far as I am concerned this is the best way to determine whether a coil is worth pursuing or should be scrapped and another idea tried.
Carroll
Since Dave is temporarily out of service and unable to do any testing I decided to build a test stand for testing different kinds of coils. I have also just gotten a scope that will give me a readout of the rms or p-p or dc voltage levels. It will also give me the frequency of the signal.
So I ran a baseline test with no coils at all on my test setup. My motor was running on 12.4 volts and drawing 1 amp after the test setup had warmed up and the bearings had loosened up.
I then put one of my test coils on the test stand. The output of the test coil was connected to a 50 ohm load. The rms value was 9 volts and the frequency was 90 hz. 9 volts into a 50 ohm load means the load was consuming 1.62 watts.
The input to my motor went up by .1 amps. .1 amps times 12.4 volts means the motor was now consuming 1.24 watts more than before. So in other words for an increase in power input of 1.24 watts I was able to get 1.62 watts out. I realize that does not mean the overall efficiency is OU. But it does mean this particular coil is very efficient at converting mechanical energy into electrical energy.
You guys can argue all you want about what efficiency means but as far as I am concerned this is the best way to determine whether a coil is worth pursuing or should be scrapped and another idea tried.
Carroll
Comment