Testing the TK Tar Baby

[Rosemary Ainslie]

Well - THAT's all taught me something.  I actually assumed that they'd see that error that would slap your average power engineer between the eyes.  But they DON'T.  So.  Guys, with or without respect - I think we can now confidently say that TK and his band of vigilantes DO NOT KNOW how to do elementary power analysis.  And you can discount any efforts that they expend in this direction.

20 watts indeed.  What rubbish.

Regards,
Rosemary

added

[TinselKoala]

And as for THIS!
As you have been doing through over 2000 posts.  I've LEARNED the trick from you Leon.  And the rule is this.  What's sauce for the gander is also sauce for the goose.

Rosie Pose

No, you liar, I am attacking YOUR competency. Nobody else has supported you, especially nobody in the industry, so YOUR COMPETENCY ALONE is being attacked here, and YOU HAVE NO DEFENSE because you are indeed incompetent.

How's this for sauce:
Ainslie said, "I DID NOT POST THAT VIDEO".

[Rosemary Ainslie]

And guys, I LOVE that 'scrawny' reference.  An earlier one was 'scrawny wench'.  LOL.  Where does TK come from?  It's so funny.  It's not unlike those rather amusing descriptions of democracy that were promoted by the Chinese - in the 70's.  Those 'running dog' references.   I've just watched a wonderful clip on this by Colbert.  He's a genius of commentary.  I'd love to hear him let rip on TK's rather heavy handed propaganda.

Scrawny liar. Cite a definite reference and let's see if I need to "learn" any skill with respect to triggering.

While you are at it, liar, explain why YOUR scope is set to trigger where it is, and also explain why THAT LECROY is not stable in the video, but is hilariously dancing back and forth. 

IT IS BECAUSE ITS TRIGGER IS IMPROPERLY SET, you ignorant liar.

I have said what I know ... now YOU TELL ME WHY I AM WRONG. WHY IS THE TRIGGER SET WHERE IT IS IN THAT SCOPE SHOT?

You cannot tell me because you have no coherent reason.

TK.  Reason and coherence are tautological.  You cannot have the one without the other.  You are wrong in that calculation.  HOPELESSLY.  And I am not scrawny.  I have precisely the same shape and size as I had when I was in my early twenties.  And I am slim.  Not thin. Not scrawny.

Rosie Pose

[picowatt]

Some "power measurements" of the portable negative mean power oscillator:

With the battery probe at the old familiar (but erroneous) measurement point as shown in the schema01, the mean battery power is -4.2W.

We can "improve" this already improved negative mean by moving the "battery" probe to the MOSFET Drain as shown in schema02. At this point the mean battery power is - 6.3W.

When the battery probe is moved to measure directly across the battery with a differential probe (not shown), the mean power computation is +0.133W. This is the "actual" battery power being delivered to the circuit. See last scope shot.

.99,

I have got to get into some sim software.  I really like your work and what you are able to do with the sim.

PW 

[TinselKoala]

Well - THAT's all taught me something.  I actually assumed that they'd see that error that would slap your average power engineer between the eyes.  But they DON'T.  So.  Guys, with or without respect - I think we can now confidently say that TK and his band of vigilantes DO NOT KNOW how to do elementary power analysis.  And you can discount any efforts that they expend in this direction.

Regards,
Rosemary

CORRECT IT THEN, you blowhard.  What is the channel setting? 500 mV per division. What is the offset above the center graticle marker? 20 mV. What, then, is the voltage indicated during the "on" part of the cycle? It is near 80 mV, isn't it?

ISN'T IT?

And what is the value of the "shunt"? It is 0.25 Ohms , ISN'T IT?

And what is the result of dividing 0.08 volts by 0.25 ohms? Ainslie, what does your calculator say?

IS THIS CURRENT FLOWING, OR NOT? 

Do you object to the "20 watts" figure I cited, thinking yourself that your load is only dissipating less than one Watt, by I^2R? Well, good for you. But don't you remember that YOU YOURSELF have often simply multiplied a voltage value by a current value in just the same manner? Do you think that the duty cycle time enters into the calculations somehow? Of course it does... unless YOU are doing the calculations. Go ahead, Ainslie.... show me how to do the math and how to correct what you think are my errors.

The gate drive signal you are providing is turning Q1 partially on. When it is fully on its minimum Rdss is 2.0 Ohms. When it is operated in the linear response region by less than a full gate charge, its internal drain-t0-source resistance is considerably higher... hence the circuit is only passing 320 mW. Since you have a 60 "something" volt supply, the total DC circuit resistance during this partially on linear mode is R=V/I or 60/0.32 or == about 190 Ohms, most of which is in the mosfet.  0.32 x 0.32 x 190 is.... just about 20 Watts. Your circuit is dissipating 20 Watts, most of it in the mosfet, and ALL of it provided by the battery.