Testing the TK Tar Baby

[TinselKoala]

So I still think about 20 Watts is being dissipated by the circuit, given that 320 mA and the 62 volt battery. (And the 5 volt gate drive from the FG, of course.) The various resistances can be used then to calculate the individual power dissipation levels of the various parts. The 11.1 ohm Load is easy to figure: assume the whole 320 mA goes through it and use I^2R and obtain about 1 Watt. What about the mosfet, biased partially on by the weak positive gate drive? Again, use the whole 320 mA of course, and look at the transfer graph to see what a mosfet at 25 C and 5 v gate drive will conduct at 100 Vds. It looks to me about 1.2 Amps, and so at 100 Volts that gives a resistance of R=V/I == about 83 Ohms. The load is 11.1 Ohms, the CVR ( shunt) is 0.25 Ohms and the internal resistance of the FG is 50 ohms. So we are in the ballpark of the total circuit resistance calculated from the estimate of current, using the eyeballed estimate of mosfet resistance from the transfer graph.

(EDIT I removed the part about the FG being in series when Q1 is on... I'm not sure about that at this point, the situation is not a strict mirror image of when the FG is putting a negative at Q1 gate which does put the FG in series with the battery through the Q2 Rds.)

[TinselKoala]

No MileHigh.  You are not correct.  Not even close.  And NO.  I have no inclination to explain anything at all.  I'm entirely satisfied that just about everyone reading here is aware of the error.  And frankly this discussion is again getting rather tedious.  I've still got to do that ruddy post relating to TK's video - which he laughing calls 'educational'.  And RIGHT now I've got chores to finish before I can even get back here.

Rosie Pose.

In English, that means SHE CANNOT explain anything at all in coherent speech with standard terminology, and not only that, she can't find any substantial error in my calculations.

It also means that she ONCE AGAIN has completely ignored Stefan's wishes and is STILL trying to get herself banned.

[picowatt]

This scope shot is being discussed. Full shot and then a blowup of the region in question.

I don't believe the mosfet is fully on.

The LeCroy autodetects probe attenuation if it is used with LeCroy probes, as it is, and is using the correct atten, I believe.

TK,

So, now what's the problem?

If channel 1 is the shunt voltage, I agree with your off screen measurements and it looks like about 320ma is flowing when the FG output is positive.  320ma across the 11R load resistor means that the load resistor is dissipating around 1.13 watts.  If Vbatt is 60 volts, there is about 56.5 volts across Q1 so Q1 is getting hot as it dissipates approx 18 watts.  Total dissipation (ignoring the small dissipation in the csr) is just under 20 watts.

PW   

[TinselKoala]

Here, Ainslie, in case you lost the link. I can't wait to see what you come up with to "explain" my ruddy video in light of your claims.

Ainslie said,

What you are trying to do is to get me to believe that a function generator is able to pass current from a battery supply source via its terminal to its probe. Since I KNOW that is is impossible I'm afraid I'm not receptive to you trying to teach me or anyone else.  So NO.  I spare me your 'lessons'.

And TK replied,
http://www.youtube.com/watch?v=GuBWVmRmUtc

Come on, let's see what kind of laughingstock you can make of yourself now, Ainslie.

[picowatt]

So I still think about 20 Watts is being dissipated by the circuit, given that 320 mA and the 62 volt battery. (And the 5 volt gate drive from the FG, of course.) The various resistances can be used then to calculate the individual power dissipation levels of the various parts. The 11.1 ohm Load is easy to figure: assume the whole 320 mA goes through it and use I^2R and obtain about 1 Watt. What about the mosfet, biased partially on by the weak positive gate drive? Again, use the whole 320 mA of course, and look at the transfer graph to see what a mosfet at 25 C and 5 v gate drive will conduct at 100 Vds. It looks to me about 1.2 Amps, and so at 100 Volts that gives a resistance of R=V/I == about 83 Ohms. The load is 11.1 Ohms, the CVR ( shunt) is 0.25 Ohms and the internal resistance of the FG is 50 ohms. So we are in the ballpark of the total circuit resistance calculated from the estimate of current, using the eyeballed estimate of mosfet resistance from the transfer graph.

(EDIT I removed the part about the FG being in series when Q1 is on... I'm not sure about that at this point, the situation is not a strict mirror image of when the FG is putting a negative at Q1 gate which does put the FG in series with the battery through the Q2 Rds.)

TK,

Just figure the drop across the load resistor (as it is a known value) and subtract that drop from the supply voltage.  The remainder is the voltage across Q1 and that voltage times the 320ma gives you the dissipation at Q1. (ignoring small things like the dissipation at the CSR, wiring resistance, etc).

PW