Circuit setups for pulse motors

[gyulasun]

The energy is so little right now that I don't think the zener is eating to much of it. But thank you for the heads up, I was not aware of the heating issue. However, I would assume that the zener works like a diode and a resistor mixed.
Jason

Hi Jason,

Yes, ok.  I agree with your thoughts.  And in the meantime, after reading your Reply #146 it was clear you wished to charge up the cap at first to a certain voltage level,  then dump some charge from it into a battery through the Zener diode.  I was not fully in the picture as to your aim but the dissipation in the Zener still holds. Of course this loss is low as long as the output power need to charge a battery is really  small and loss will increase proportionally with battery charging current need.
To minimize this loss,  and this answers Peter's asking too,  it is possible to sense the voltage level in the capacitor and compare it to a reference voltage level with the help of a comparator  and this latter would fire a switch to dump charge from the cap into a battery (or a load). This way the discharge of the capacitor can be controlled more efficiently, especially at increasing load current demands where Zener diodes surely would dissipate increasingly.  The full circuit  needed to accomplish this needs power to operate of course,  so considering its usage should be judged both by the power level involved and the extra components cost. 
I am aware of such a circuit designed by Jean-Michel Cour in his TuneCharger devices (patented, see WO2005091462 or FR2868218). He runs a Yahoo mail group on TuneCharger ( http://tech.groups.yahoo.com/group/TuneCharger/ )   
I remember he made a schematics for experimenters too which is in his Files section if you are interested to see,  I cannot cross-post it here without his permission.

Gyula

[hoptoad]

@ hoptoad, thanks again for your excellent pages of juicy info, although I must admit the inclusion of mosfets etc on the third page is way out of my league for now! I wired up the pnp to your specs and found it to run, although it ran slower and amp draw sat at a constant level as it sped up. Perhaps I have wired it up wrong. I have gone back to the old config with the coil connected to the emitter, but I will try the hall hookup to a npn and see if I can improve it. On this old config the collector and emmiter read 200v spikes as speed increases, but I do not get anywhere near that when I insert a diode on both or even one, in any direction. I look forward to studying your info over the next few days.

Hi Ren,

If your motor is hooked up in the configuration I recommended, but you are not getting the results you expected, e.g. running slower, fairly high
constant current, etc, you may require a higher resistor value to ensure that your transistor is actually turning off completely.

One simple test you can do to see if your transistor is turning off properly, is to:

1. Connect your circuit.
2. Put a voltmeter across the collector and emitter. (doesn't matter whether its NPN or PNP or which leg you have your coil connected to.)
3. Stop the rotor from turning by hand, then manually turn it until the switching magnet is in front of the Hall IC to turn the Hall IC on.
4. Measure the voltage across the collector and emitter. - If the Hall IC is in the "on" mode, then you should only see .6 Volts across the 
   collector to emitter junction.
5. Now turn the rotor so that the switching magnet is away from the Hall IC and the Hall IC should now be in "off" mode.
6. Measure the voltage across the collector and emitter again - If the transistor is turning off properly, you should now see almost the full supply
   voltage minus a small drop due to the presence of the coil. If the voltage is not nearly that of the supply or very close to the supply voltage, then
   your transistor is not turning off completely. - Remedy put a variable resistor anywhere from 1K - 5K in series with the existing resistor and
   repeat steps 3+4 to test for "turn on" and 5 +6 to test for "turn off". Vary the pot until the transistor turns off. - You will know it's off because the voltage will read nearly that of the supply.

Hope this helps you to sort it out. Also, when reading your circuits without any load on the output with a meter and seeing very high voltage spikes
which seem to disappear when you connect the output to a load; this is normal. These very high spikes produced by collapsing EMF in the unloaded output state are what is commonly known as "floating voltage".
As the load is connected (capacitor or resistor or battery) it will absorb the very high voltage spikes. What voltage remains is commensurate with the voltage attainable for a given current flow into a given load resistance/capacitance. 

[Ren]

champion. Thanks for simple tips. Yes I have seen the voltage at collector and emmiter. It reads 12.1 v or around abouts. I will check to see if it drops during firing. I have some more simple diagrams that  I will post of my setup soon.

[hoptoad]

Hi again Ren,

I just changed the test steps a little bit due to minor error in statement.

Should have said : - repeat steps 3+4 to test for "turn on" and 5 + 6 to test for "turn off".

I have added the simple test statements to page 2 of my site, and have also added an extra diagram to page 3 to make it a little simpler to understand circuit Fig 7 : - http://www.totallyamped.net/adams/index.html

Cheers from the Toad who Hops

[hoptoad]

Hi again Ren,

I just changed the test steps a little bit due to minor error in statement.

Should have said : - repeat steps 3+4 to test for "turn on" and 5 + 6 to test for "turn off".

Also, if your transistor is turning off OK but you are still getting a fairly high constant current draw, then still try including a pot of 1k-5k in series with R1. If your coil has a really high number of windings, then its impedance is probably higher than the base circuit resistance during run time, and the constant current you are measuring is most likely the base to emitter junction resistance via R1. In another words your controlling base current is higher then your drive current.

I have added the simple test statements to page 2 of my site, and have also added an extra diagram to page 3 to make it a little simpler to understand circuit Fig 7 : - http://www.totallyamped.net/adams/index.html

Cheers from the Toad who Hops