pulse motor.mpg
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Uploader Comments (johncarl43447)
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You may be seeing more current, but I'll bet the voltage when drawing that extra current is very low than your input. So measure the voltage when reading the output current and calculate the power in and power out. Use Power in =Vin X Iin
Power out = Vout X Iout You will most certainly find the power out is less than power in.
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I will also have to test the current with an analogue meter also.The amperage from the source might not really be that low ,this circuit should be drawing close to 500 milliamps from the source.
johncarl43447 1 year ago
what really surprises me is that the main power source and the energy from diode both in which I am measuring the current of are both connected to the same place ,the positive side of the coil so I am thinking the voltage from the diode would have to be greater than the source voltage to cause a higher current to flow because they are both connected to the same place.But I will have to take some voltage measurements as you have said.I am just trying to understand and figure this out in my mind.
johncarl43447 1 year ago
I'll have to try that .What surprises me is that usually the voltage is higher coming from the diode but the amperage in my motors coming from the diode has always been less than the amperage drawn from the battery.Here it shows the amperage higher from the diode ,Very different than the norm for my circuits.
johncarl43447 1 year ago