Testing the TK Tar Baby

[picowatt]

TK,

I was thinking that was quite a bit of phase shift.  But, that is why one should always confirm measurements. 

Regarding the Ibatt versus Ibias, unless your DC ammeters are responding to the AC component, which some DC ammeters might might do with clipped or assymetrical AC waveforms, the DC currents should be the same.

The only DC path for Rload is by way of Q2's source and the bias supply/resistor (other than a very small gate leakage current).  At DC, the current should be the same measured anywhere in "the loop", unless there is a DC "sneak" path unaccounted for.

Have you ever attempted placing a cap across your Vbatt connections?

How do your dogs handle trips to the vet?

By the way, you sometimes do indeed provide a good chuckle...

PW

[Rosemary Ainslie]

Reposting what link? The one to your 25.6 million Joule calculation? The one to all the wrong things and lies you've made in the past three weeks?


How is my calculation wrong, Rosie Poser? Please enlighten us and SHOW YOUR WORK.

Do your own homework TK.  If you can't work it out one assumes picowatt would.  What's wrong with you all?  There's error after error after error and NOT ONE OF YOU ever notices.

Rosie

[TinselKoala]

@PW: They had fun. It's always an exciting outing and they usually get treats.


Apparently Rosemary doesn't yet understand how to measure current using a Current Viewing Resistor and an oscilloscope, even though I have posted the video showing exactly how to do it and comparing the reading with the Ohm's Law theoretical value as well as with two other meters.
Yet she refuses to say just exactly what's wrong with the method.

http://www.youtube.com/watch?v=fPVOkDQsXfs

The voltage drop across a current viewing resistor, Rosie Poser, is related to the current flowing through that resistor by Ohm's Law. I = V / R. So if you have a voltage across the resistor of 0.5 volts, and your resistor is 0.25 Ohms.... the current is I = 0.5 / 0.25 = 2.0 Amps. If you have a value of -2 volts, as you do in your scopeshot below, the current is -8 amps. That's the way it is, that's the way I calculated and once again you are wrong and easily refuted.

[picowatt]

TK,

I believe we are being reminded of the shunt's inductance.

PW

[TinselKoala]

Do your own homework TK.  If you can't work it out one assumes picowatt would.  What's wrong with you all?  There's error after error after error and NOT ONE OF YOU ever notices.

Rosie

Show us and explain, Rosie. You are talking to at least one active electronics bench professional, another who seems to be a retired one, and a few brilliant amateurs who all know a lot more about the subject than you do.

So enlighten us. Step by step. You  know I'd do it for you.