Overunity.com Archives is Temporarily on Read Mode Only!



Free Energy will change the World - Free Energy will stop Climate Change - Free Energy will give us hope
and we will not surrender until free energy will be enabled all over the world, to power planes, cars, ships and trains.
Free energy will help the poor to become independent of needing expensive fuels.
So all in all Free energy will bring far more peace to the world than any other invention has already brought to the world.
Those beautiful words were written by Stefan Hartmann/Owner/Admin at overunity.com
Unfortunately now, Stefan Hartmann is very ill and He needs our help
Stefan wanted that I have all these massive data to get it back online
even being as ill as Stefan is, he transferred all databases and folders
that without his help, this Forum Archives would have never been published here
so, please, as the Webmaster and Creator of these Archives, I am asking that you help him
by making a donation on the Paypal Button above.
You can visit us or register at my main site at:
Overunity Machines Forum



Testing the TK Tar Baby

Started by TinselKoala, March 25, 2012, 05:11:53 PM

Previous topic - Next topic

0 Members and 21 Guests are viewing this topic.

picowatt

Quote from: poynt99 on April 19, 2012, 03:34:03 PM
Just to clarify one point;

"bias" means a certain amount of pos. DC VGS. This can be completely floating wrt the rest of the circuit.

So HOW you provide that bias, makes no difference whatsoever, it can be positive, negative, as long as VGS is slightly positive, say about 5V.

I have simulated both cases, where the bias is applied to the Source and the Gate. In other words, I have proven what I wrote above.

.99,

I agree with the comments above, however, the gate is a very high impedance node for DC, while the source is a low impedance node for DC.

In your sim work, if you insert a -12V battery in series with 50R in the Q2 source leg (using the grounded gate configuration and connecting the battery/resistor where the FG is normally connected), wil not the -12V battery have to provide an amount of current equal to whatever DC current is flowing thru Q2?

Personally, I have no problem with the alternate bias method.  Tie the gate to the first Batt+ through some decoupling and insert a 50R at the source.  I would indeed be impressed if that circuit "ran forever".

PW


 

poynt99

Quote from: picowatt on April 19, 2012, 04:10:15 PM
In your sim work, if you insert a -12V battery in series with 50R in the Q2 source leg (using the grounded gate configuration and connecting the battery/resistor where the FG is normally connected), wil not the -12V battery have to provide an amount of current equal to whatever DC current is flowing thru Q2?

Yes, whatever is in the SOURCE will have to pass the Source current. That goes without saying. However, that current has nothing explicitly to do with biasing the MOSFET in a partially ON state. In this instance, the FG (if you will) is providing two functions; VGS bias, AND AC/DC path to ground.
question everything, double check the facts, THEN decide your path...

Simple Cheap Low Power Oscillators V2.0
http://www.overunity.com/index.php?action=downloads;sa=view;down=248
Towards Realizing the TPU V1.4: http://www.overunity.com/index.php?action=downloads;sa=view;down=217
Capacitor Energy Transfer Experiments V1.0: http://www.overunity.com/index.php?action=downloads;sa=view;down=209

picowatt

Quote from: poynt99 on April 19, 2012, 03:34:03 PM
Just to clarify one point;

"bias" means a certain amount of pos. DC VGS. This can be completely floating wrt the rest of the circuit.

So HOW you provide that bias, makes no difference whatsoever, it can be positive, negative, as long as VGS is slightly positive, say about 5V.

I have simulated both cases, where the bias is applied to the Source and the Gate. In other words, I have proven what I wrote above.

.99,

To further clarify, yes, the bias voltage or Vbias is the voltage applied to the gate relative to the source, i.e., Vgs.  But, but the bias current, Ibias, is the quiescent DC operating current that Q2 passes when, in concert with its source resistance and Vds, Vbias is applied.     

PW

picowatt

Quote from: poynt99 on April 19, 2012, 04:17:18 PM
Yes, whatever is in the SOURCE will have to pass the Source current. That goes without saying. However, that current has nothing explicitly to do with biasing the MOSFET in a partially ON state. In this instance, the FG (if you will) is providing two functions; VGS bias, AND AC/DC path to ground.

.99,

I would rather say "two functions; Vgs bias, the only DC path to ground and a partial AC path to ground", the other AC path being Ciss.

Regarding the bias source in the Q2 source leg, if we connect a 15 volt battery in series with 50R between ground (CSR) and the source of Q2 (batt+ at CSR), the measured drop across the 50R will be around 10.5 to 10 volts (the 15volts minus Vgs).  That means there is around 200ma flowing as Ibias thru Q2, and as well through the battery.  If the battery is a 200ma/Hr battery, does this not mean the battery will last only 1 hour?

PW

(corrected batt- to batt+)

TinselKoala

Quote from: picowatt on April 19, 2012, 03:08:36 PM
TK,

Hey, I thought you had "real work" to do today!

Your questions are good ones.  Regarding the bias source, that is why I asked what your 555 circuit's supply was drawing when it has Q2 oscillating.  If your Q2 bias is 150ma. for example, I would think your 555 supply will be showing a total current draw equal to the 150ma bias current plus 555 quiescent current.

For DC, the current path is through whatever is providing the turn on bias supply for Q2.  For AC, the path is through both the bias supply and the Ciss of the MOSFETs.  When the FG was used, its Rgen=50R was a lesser path for AC than the Ciss of the MOSFETs.  With your 555's 10R source, the 10R is probably close to an equal path for AC, as the 10R is close to the Ciss reactance at 1.5MHz.

PW
Thanks... I think I get it..... maybe..... So the bias power is "mixed in " with the main power from the batteries and must be dissipated in the load and mosfets (I suspect in the mosfets mostly). Hence the load heating and current viewing resistors are giving an inaccurate picture of the power flows in the circuit.... is that right?
(I know this has been covered before; I just want to hammer it home somehow if it is correct.)

I'm running so late that there's no point in making the drive. So here I am, stuck in Mobile with the Memphis blues again.

Anyway.... I tried the 50r/pot/cap/main battery thing and I can't get it to work.  And yes, I've confirmed that the pot is good and is adjusting the voltage properly, tried with three different valued pots.

This _does_ work just fine with the external battery (no decoupling cap), but using the running battery I can't get it to work. External 9v gives beautiful nearly perfect sinusoidal oscillations on the common drain when the 10 k (not 1 meg; I also tried 100 K, no luck) pot is set right, and the "voltage floor" still shows up: the negative voltage on the gate drive signal floors at about 4 volts negative indicated on the scope trace.

I think you really do need to provide a potential that is more negative than the negative battery terminal....

I just made another video... chaotic but the points come through I hope. It's uploading now.