Coil discharge current
Channel: AlternateFarmhand1
Date: 09/02/2014

Showing the current caused by the discharge of a coil complies with Ohms Law. The energy released from the coil builds a voltage against the resistance to cause the flow of current to continue. The higher resistance the higher the voltage the coil discharges into and so the less the intensity of the current is. I misspoke and said milli Amps instead of saying milli Volts when reading the scope but corrected myself.

The voltage is measured across 0.2 Ohms of resistance and was 180 mV peak current which is 0.18 A / 0.2 = 0.9 Amp peak current, I selected RMS voltage in the video when I probably should have selected average current to get the charge current value, I remeasured using Average Voltage selected and got 26 mV across the resistors when I got 55 mV when RMS was selected.

So using the average voltage of 26 mV across the 0.2 Ohms resistance I get current of 130 mA of charge current to the battery. If I take the 12.8 volts the coil is discharging into and multiply by the charge current I get 1.66 Watts of output power.

The input looked like 220 mA at 12.5 volts which would be 2.75 Watts. But the meters on the input were not so accurate and I was not trying to determine efficiency, just showing the impulse charge current wave form mainly.

A similar but better test by using smoothing and filtering and showing efficiency can be seen here http://www.youtube.com/watch?v=iaxLdexemtU

Also an efficiency test of a boost converter is here. http://www.youtube.com/watch?v=eLnNgfaha10

Cheers