Muller Dynamo

[gyulasun]

Hi all
I wanted to see how to collect the backemf/recoil with a bidirectional mosfet circuit that also has pulswidth control to it too, this circuit using two pairs of bidirectional mosfets, and one pair is normallyOFF switchesON and the other is normallyON switchesOFF and each pair has its own halleffect, so the pulse widht is adjsuted by distance between halleffects...
Anyways I was wondering how to take the  backemf/recoil out when using this circuit - there being 4 seperate mosfets in it.... and turns out the only way to collect it, is through the mosfet "C" as shown in this drawing....logically you would think mosfet "D" being the last in line would be the one to "tap" but that one puts out nothing, and the diode and cap need to connect to mosfet "C" then the cap fills up super fast and the "fzzzzzt" sound from the motor coil dissapears too.

Hi Doug,

You show an interesting schematic.  Lets consider MOSFET A&B as a single switch and also MOSFET C&D as another single switch: it is sure that in order to connect the 12V battery to the motor coil, both switches should be closed and then to break the motor coil current one of switches must open, i.e. the switch MOSFET C&D must be opened once you connected the recovery diode and the puffer cap into the C&D switch.  In this moment the MOSFET A&B switch must still remain closed, otherwise the backemf cannot get to the recovery diode in this schematic and also it is interesting that you drive the backemf current through the 12V battery too while some part of it goes to the recovery capacitor too. (One should ponder on the polarities of the battery and that of the backemf if the driving-through of the backemf current gives any advantage or not...)
Now the question is how the circuit is still closed when MOSFET switch C&D should be open during the backemf collection?  Well, it is the built-in body diode in MOSFET D (between its drain and source pins) which closes the circuit and makes the backemf capture possible at all in this schematic and you found it as the only way, which is correct of course.  The body diode of MOSFET D represents a forward diode for the backemf voltage (hence current) coming from the coil via the closed MOSFET A&B because the blue wire of motor coil will become positive polarity (due to the backemf) the moment the 12V battery voltage is switched OFF.

rgds,  Gyula

[konehead]

Hi Gyula
I didnt realize that the backemf is going back through the 12V battery too, besides filling up the capapcitor in  that circuit - can you explain this again? Is it from mosfets Aand B??
Is it reason of mosfets being bidirectional  instead of two single mosfets to do the pulse width contorl?
I would rather put all the backemf power go into the cpa if possible since I dont thing the backemf spikes going back to battery si going to really help charge it up very well - maybe desulphate or raise volts is all .... so if it can fill cap a little faster that would be better - maybe solutions like diode somewhere?

[gyulasun]

Hi Gyula
I didnt realize that the backemf is going back through the 12V battery too, besides filling up the capapcitor in  that circuit - can you explain this again? Is it from mosfets Aand B??
Is it reason of mosfets being bidirectional  instead of two single mosfets to do the pulse width contorl?
I would rather put all the backemf power go into the cpa if possible since I dont thing the backemf spikes going back to battery si going to really help charge it up very well - maybe desulphate or raise volts is all .... so if it can fill cap a little faster that would be better - maybe solutions like diode somewhere?

Hi Doug,

I edited your schematic drawing to show the relevant circuit.  So it can be seen the 12V battery is in series with the motor coil, whatever direction you consider the current flow, meaning both the emf and the backemf current directions.  And when the motor coil current is switched OFF, the coil will behave as a second 'battery' in series with the 12V battery as if you had connected a second source polarity-wise correctly in series with it (the two sources add like two 'batteries' in series, + - + -),  this means the current driven by this second source i.e. the backemf current goes through also via the 12V battery when it also charges up the recovery capacitor and this charging current lasts till the backemf (voltage-wise) reduces to the 12V level .
I indicated in blue arrow the backemf (voltage) direction and in red arrow the normal 12V emf.   I am uncertain if the backemf current has a real charging effect for the 12V battery, (remember: the coil current does not change direction when you switch it OFF, only its amplitude start decreasing towards zero),  it would be good to see the current waveforms by a scope across a series 1 Ohm or 0.1 Ohm 'shunt' resistor to study the ON-OFF process.

The 'classical' backemf capture in your original schematic would include a diode bridge placed directly across the motor coil with its AC input side and the recovery capacitor would be connected to its DC output.  This way the two bidirectional switches would clearly serve as a variable duty cycle ON-OFF switch (but the 12V emf would be also as an input to the recovery cap, so you would not want to discharge it lower than 10-12V when utilizing the captured energy from it.  And if you do not wish the 12V input would be also pumped into the recovery capacitor at the motor coil switch-on time, then you may wish to use a single recovery diode with a capacitor as you drew the steering diode above but not at the MOSFET C drain of course but at the blue motor coil wire and the capacitor negative point would be connected to the red motor coil wire.

To answer your direct questions: yes, backemf current has a low resistance path via MOSFET switch A&B, of course this switch must be kept ON during the capture process;  and no it is not a reason the MOSFET switches are bidirectional when they control pulse width (the bidirectional MOSFET switch 'only' improves upon the single MOSFET switch by mutually 'neutralizing' the body diodes in the two MOSFETs, the two diodes connected oppositely facing each other,  in a single MOSFET switch the body diode is perpetually included between the drain and source legs, this way the diode is in parallel with the two points to be closed or opened and this maybe a big problem when the current direction changes in the switched circuit).

rgds,  Gyula

[crazycut06]

Here it is the other cool effect of the coil. With a small drop in rpm ( less than 30rpms) the coil rises the voltage on the light bulb from 0.92 to 1.26 and stays there.
http://www.youtube.com/watch?v=AH7YQ_gOWAY&feature=youtu.be

Hi Marius,
    Have you tried using a short ferrite core, and what would the effect be? also try shorting the output coil see what will happen.


keep it up!

[konehead]

Hi Gyula
Thanks alot for all t hat and the drawing let me read througit 4 or 4 times until it all sinks in and I willget back to you on it.