Kapanadze Cousin - DALLY FREE ENERGY

[baroutologos]

Hello Verpies and Grum,

I made a vid presenting the Mazzili oscillator with BJT. In contrary to my previous note that voltage will go as much as x4 of the supplyu voltage inside the tank circuit and the obvious experimental data that proves that easily goes x8 (i have not surpass that yet) i investigated and found that the way switched the transistors considerably affects p-p tank circuit values as well as waveform.

here is the vid
http://www.youtube.com/watch?v=Nx9fFRoQF2I to take an idea.

I also found that (inside limits) the p-p value is not affected by capacitance or inductance of the tank circuit. Several other notes are made. (as oscillation can happen with just one tranistor switching, without any resistors at all - only capacitance, etc.)

In all those cases, if put under load the oscillator by inserting another winding and loading it, the more stable form is the 1k resistor + 10 nf caps in parallel.
Have fun 

[Grumage]

Hello Verpies and Grum,

I made a vid presenting the Mazzili oscillator with BJT. In contrary to my previous note that voltage will go as much as x4 of the supplyu voltage inside the tank circuit and the obvious experimental data that proves that easily goes x8 (i have not surpass that yet) i investigated and found that the way switched the transistors considerably affects p-p tank circuit values as well as waveform.

here is the vid
http://www.youtube.com/watch?v=Nx9fFRoQF2I to take an idea.

I also found that (inside limits) the p-p value is not affected by capacitance or inductance of the tank circuit. Several other notes are made. (as oscillation can happen with just one tranistor switching, without any resistors at all - only capacitance, etc.)

In all those cases, if put under load the oscillator by inserting another winding and loading it, the more stable form is the 1k resistor + 10 nf caps in parallel.
Have fun

Dear baroutologos.

Many thanks for taking the time of making that video. IMO the basics are useful for helping the beginner!!


I think that the base bias resistors at 1 K ohm are driving the transistors full on, hence the larger output. As the resistance is increased so the voltage applied to the base decreases and  the transistor will not switch on fully. An interesting thing may be to add  variable pot's in series with the 1 K ohm resistors to find an optimum bias level. This IMO would also reduce current consumption to a minimum value sustainable and also reduce the  transistors heating!!

Cheers Grum.

[baroutologos]

@Grum,

Normally i would agree, but i do not know. Why while having the 1K resistors in place by adding parallel capacitors the voltage in the LC drops (the more the capacitance the larger the decrease)? Capacitors allow more current going to transistors' bases unimpedent so a more complete switching and a larger output should be done that way. No the case.

Clearly there are other variables into play too.

[Grumage]

@Grum,

Normally i would agree, but i do not know. Why while having the 1K resistors in place by adding parallel capacitors the voltage in the LC drops (the more the capacitance the larger the decrease)? Capacitors allow more current going to transistors' bases unimpedent so a more complete switching and a larger output should be done that way. No the case.

Clearly there are other variables into play too.

Dear baroutologos.

My own thought is that the capacitors are able to feed some ac voltage through the semiconducting P N junction?? Where as a resistor naturally resists the current!  I am now getting out of my depth!! And will say no more as the hole I am digging is getting quite deep!! 

Dear Verpies will, I have no doubt, have the correct answer!!

Cheers Grum.

[d3x0r]

I did add a pot in series, and attempt to modify the gate resistance to see the effect, but increased resistance decreased output; has to be another way to get current to it. I ended uphaving to leave the pot near zero.  I think I'm using 470's for the high side pullup... others say 690 or something... technically it should be proportionate to the pull-down so the voltage division for your target power gives the right voltage to the switch...


It's a DC input on the other side of the base/cap; not much of a chance of spike passing the cap...


the cap probably does hold a BJT on longer, since there is a current flow through the BJT's gate; was wondering if changing from mosfets to darlingtons would work the same; or BJTs...


mosfets don't have a voltage drain from their gate....they're capacitive... 


the resistor still has to be small enough to be able to turn on the gate... the capacitance is going to add an LR time to the gate turn on time; as long as this is less than a target frequency.