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Author Topic: Woopy's Shorting Coil Test  (Read 7869 times)
Group: Guest
This is to discuss Woopyjump's new setup for a shorting coil test.  He has done a nice solid build with good components and he has what appears to be a very decent DSO with math functions.


Woopy, here is the main issue in your clip:  When you switch on the MOSFET circuit, it is switched on for too long a time during the sine wave.  If we suppose that the MOSFET circuit is switched on 95% of the time, then you are shorting out the coil nearly all the time and not leaving any time for ringing.  You hear the rotor slow down because the pick-up coil is casing Lenz Law drag.

Here is the first thing I would do if I was you:  If I understand your Hall sensors, one switches on for north, and the other switches on for south.  As the rotor turns the two Hall sensors go like this:  ON-OFF, OFF-ON, ON-OFF, etc.  So one of the Hall sensors is always ON, and as a result the MOSFETs are always ON.

My suggestion is to use only one Hall sensor for now.  You can experiment with changing the position of the Hall sensor and looking at the scope display.  You can change the angle of the position of the Hall sensor to affect the timing.  You can also experiment with moving the Hall sensor close or far away from the spinning rotor, and that should increase or decrease the ON time.

I see that you fixed the Hall sensors in place with hot glue.  May I suggest that you hot glue one Hall sensor onto a small piece of wood.  Then you will have the flexibility that you want.

« Last Edit: 2011-02-27, 04:15:58 by MileHigh »
Group: Guest

I am going to give you some other suggestions for your setup and test procedures for your consideration.  Please don't feel overwhelmed by the suggestions, and it is of course your choice if you want to do them or not.

1.  Add a diode to your drive coil to prevent the PNP transistor from being damaged.

There is a missing diode in RomeroUK's circuit.  Please add an ordinary diode across the dive coil.  You connect the diode so that it would let current flow from the terminal that connected to the transistor emitter to the terminal that's connected to the power supply positive.  When you do this it will prevent your drive coil from generating damaging high voltage spikes.

2.  Improve scope display with separate triggering.

I noticed that you are triggering on the rising edge of your waveform at -3.6 volts.  There is some small ringing or glitches in your waveform that is causing extra triggering also which makes the display a bit confusing.

I am assuming that there is no electrical connection between your drive coil circuit and your pick-up coil circuit.

My suggestion is to use Channel 1 for the triggering, and use Channel 2 for displaying the sine wave.  You will connect Channel 1 to the drive coil circuit.  Try connecting Channel 1 between the drive coil power supply ground and the resistor on the Hall sensor.  You should see a small signal there and you should be able to adjust your trigger voltage level and trigger polarity to get a stable trigger on that signal.  (Notice that this also tells you about the amount of time you energize the drive coil).

Now you can connect Channel 2 across the pick-up coil and when the motor is on you should see a nice stable sine wave.   If you take my suggestion in the previous posting and have just a single movable Hall sensor for the pick-up coil circuit, you should see some interesting effects on the sine wave as you move the Hall sensor around.

Group: Guest
I am just posting a link to an application note about using MOSFETs in an AC switching application.  I am not an expert in MOSFETs and I am parking this application note here because it might be useful.

Jr. Member

Posts: 72
   Kone is doing basicly the same thing with the two fets too.

Will try this question one more time.
If the magnets passing the coils produce 3v @ 200 ma and you
short them. Will this kick the voltage up but lower the ma? This
is what I was trying to ask in my last question and my terminology
is layman and most here are EE trained.

Group: Guest
When the magnet passes by the coil, a current is induced in the coil.  Shorting the coil at the peak of that current will have the effect of retaining whatever energy is stored in the coil.  The inverse is true with a capacitor.  Once a charge is placed into the cap, opening its terminals will allow the longest retention of that energy.

I think the idea here by shorting the coil at its peak current is to save the energy in the coil until the "best time" to release it.

The amount of current in the coil (if it had zero resistance) would stay constant as long as it was shorted.  How many volts you get when the coil is later "opened" or put across a non-zero load is entirly dependent on the load, as I explained before.

If it is a high value resistance, you will get a very short duration high voltage spike.  If it a low value resistance, you will get a longer-lasting low voltage decaying pulse.  In either case, the energy dischaged from the coil is the same and is equal to 1/2 LI^2.

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