Testing the TK Tar Baby

[poynt99]

The 10.5W Pin is correct.

Here are the pics:

[picowatt]

The 10.5W Pin is correct.

Here are the pics:

.99,

I suspect that the measured/calculated Iavg may be a bit high due to CSR inductance. The hazard of using such a low value CSR (.1 ohm) is that just a few tens of nHy's can produce a very significant error.

I would suggest repeating the measurements using a 1 ohm non-inductive CSR and comparing the results. 

Looking at various non-inductive resistor manufacturers' data, many simply state inductance as less than 100nHy for their non-inductive resistors, which for a .1 ohm resistor can be a very significant error.  The best I could find was from Caddock, and they state their's as 10nHy when measured .2" from the package.  Even this small inductance can be very significant relative to the .1 ohm value depending on the frequencies involved.  Using a higher value CSR, such as 1R, reduces the percentage of error due to inductance.

PW

ADDED:  I had assumed that with Greg using a gate driver the transitions would be much cleaner and at least 70-80% efficiency acheived.  Having now seen the waveforms, possibly this is a bit optimistic and the measured 50% is indeed closer to reality.   

 

[TinselKoala]

Hmm thanks.... I'll mull over the scopeshot
But what about the Wiki, in the image I attached above, where the actual derivation is given, showing that for a pulsed DC signal, all positive, the Pavg=Irms x Vrms, and the Vrms is given by sqrt(duty cycle) x Vbatt? What am I missing.... the derivation is right there and it makes sense to me and it even explicitly says that this is how the equivalent to a steady DC is determined.....
Is the Wiki wrong.... It even agrees with the Ham .pdf.... so what am I missing, please walk me through it.

[TinselKoala]

@.99.... that scope shot is the signal from the mosfet Drain, with respect to the negative rail?? Or is that the signal "across the load", with the probe on the drain side and the reference on the battery side, or vice versa? I'm not getting it. And with a gate driver.... I am also surprised to see such a messy signal anywhere. Is it that the PG50 just can't keep up at the frequency chosen?

[poynt99]

Hmm thanks.... I'll mull over the scopeshot
But what about the Wiki, in the image I attached above, where the actual derivation is given, showing that for a pulsed DC signal, all positive, the Pavg=Irms x Vrms, and the Vrms is given by sqrt(duty cycle) x Vbatt? What am I missing.... the derivation is right there and it makes sense to me and it even explicitly says that this is how the equivalent to a steady DC is determined.....
Is the Wiki wrong.... It even agrees with the Ham .pdf.... so what am I missing, please walk me through it.

The power in question is the input power. If the circuit is a "black box", and the supply is DC, how would you go about determining Pin_AVG?