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I’m a stubborn old cuss. I guess I look at things from a different perspective. I think you guys (not you floodrod) are missing the point.
You want calculations so here you go.
Capacity = 100uF, Voltage = 20, charge = 0.002 C, energy = 0.02 J
Capacity = 100uF, Voltage = 10, charge = 0.001 C, energy = 0.005 J
Capacity = 100uF, Voltage = 5, charge = 0.0005 C, energy = 0.00125 J
Discharging the source cap to the load uses half the stored charge. 0.001C has been sent to the load and 0.001C remains in the source. Each cap has exactly half of the original charge. We have lost nothing but potential energy in each cap. The caps representing that source and load are for measurement purposes of one pulse only. The real load has used the energy and it has dissipated to 0. Now there is an additional 0.0005 C in the BEMF cap. These coulombs were obtained for free.
If you continue to pulse the load with 20V and collect the BEMF into that cap you know very well the voltage can rise far above the 20V source.
Say we wait until that voltage is 30V.
Capacity = 100uF, Voltage = 30, charge = 0.003 C, energy = 0.045 J
At that point in time we have 0.045 J of free energy.
Say we wait until that voltage reaches 100V.
Capacity = 100uF, Voltage = 100, charge = 0.01 C, energy = 0.5 J
Now we have 0.5 J of free energy.
Isn’t that the goal we are striving for? Is that not free energy?
I mean, as an analogy, if I put 10 gallons of gas in my car and go for a long drive and while I’m driving an angel pours 5 more gallons of gas into my tank, I sure as heck would consider that free energy. And I wouldn’t complain that it’s not OU because I didn’t get more than I started with for free.
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'Tis better to try and fail than never try at all
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Author
Topic: Back-EMF Studies (Read 4863 times)
