Testing the TK Tar Baby

[picowatt]

TK,

Is your circuit "similar" or does it just "relate" to it?


PW

[TinselKoala]

Ahhh the current flows the opposite direction of what I thought it would flow alike...


Okay, I see, the DC current flow through the 10.3 Ohm resistor is from the main battery through Q2 back to the main battery ground.....
I thought this current would come from the 9 Volts battery, but this is not the case...!

Then just try to put a 100 nF cap in series, so it blocks the DC current and will only pass the AC.
ripple.
The 9 Volts battery should then not be discharged so fast...

Will it then still oscillate ?

The other new postings I have to catch up next week.

Regards; Stefan.

I think I  must still not be understanding you.

The "conventional" current, the one that works with calculations and right-hand-rules and all of that, is (thanks to Benjamin Franklin) assigned to be "out" of the positive pole of the source, around the circuit, and "into" the negative pole of the source.
Of course now that we understand things a bit better than Franklin or even Faraday, we know that what really happens is that _electrons_ carrying their negative charges bump into other electrons which bump into other electrons and so the _charge_ is transferred along the conductor from the Negative polarity to the Positive polarity. The electrons themselves bump along rather leisurely, but the _charge_  and whatever signal or power carried by it transfers at the speed of light in the conductor.
So in this video when I am describing the "conventional" and "anti-conventional" current directions, I am referring to the convention that Ben Franklin left us with, so that all the calculations make sense. 
It doesn't really matter, it's just a matter of sign, and so the convention remains with us and continues to confuse freshman EE students every September.

So I'm not sure what you are meaning about the current direction, and I'm still not clear on your use of "AC". Where is the AC in the circuit from the 9v battery (or power supply) to the gate input? It's all DC with a slight ripple on top. If I put a cap in there, nothing will get through, will it? The only place I see true AC is across the CVR... that is, in the main circuit itself when it is oscillating strongly.

So I suppose I'm still not following your meaning.

[TinselKoala]

TK,

Is your circuit "similar" or does it just "relate" to it?


PW

That's right, and I'm not claiming similarity, I'm just claiming to have _measured_ similarity.




(By the way I've made a new video and I think it shows the scope traces you wanted to see. )
http://www.youtube.com/watch?v=2LMthOsvbVU

[picowatt]

I think I  must still not be understanding you.

The "conventional" current, the one that works with calculations and right-hand-rules and all of that, is (thanks to Benjamin Franklin) assigned to be "out" of the positive pole of the source, around the circuit, and "into" the negative pole of the source.
Of course now that we understand things a bit better than Franklin or even Faraday, we know that what really happens is that _electrons_ carrying their negative charges bump into other electrons which bump into other electrons and so the _charge_ is transferred along the conductor from the Negative polarity to the Positive polarity. The electrons themselves bump along rather leisurely, but the _charge_  and whatever signal or power carried by it transfers at the speed of light in the conductor.
So in this video when I am describing the "conventional" and "anti-conventional" current directions, I am referring to the convention that Ben Franklin left us with, so that all the calculations make sense. 
It doesn't really matter, it's just a matter of sign, and so the convention remains with us and continues to confuse freshman EE students every September.

So I'm not sure what you are meaning about the current direction, and I'm still not clear on your use of "AC". Where is the AC in the circuit from the 9v battery (or power supply) to the gate input? It's all DC with a slight ripple on top. If I put a cap in there, nothing will get through, will it? The only place I see true AC is across the CVR... that is, in the main circuit itself when it is oscillating strongly.

So I suppose I'm still not following your meaning.

TK,

I believe Stefan is thinking that by placing a cap in series with the 9volt bias source, the 9volt battery will not have to provide a DC current and therefore not discharge.

I believe the correct answer to his question is, "no".  The source leg of Q2 needs both a negative voltage AND a DC path back to the battery negative (or the CSR).

PW

[TinselKoala]

A data point: