Quote from: poynt99 on June 19, 2013, 10:32:57 PM
PW,

This model at least has a current output (common base) amplifier, not voltage. So the 50 Ohms looking back into the output port is a resistor to ground, not in series.

Since it is a current output, surely the generator can tolerate a dead short on its output, otherwise it wouldn't be worth a damn to the professionals using it. Any FG for that matter that can not tolerate a dead short indefinitely at full output should be tossed in the garbage imo.

It does have output voltage protection as well. They call it "Excessive Voltage Protection" using a few diodes and a transistor for each half.

I have some 15V TVS diodes at work; I will install them as well as the 250mA fuse as a precaution.

.99,

If your PG does not have a 50R in series with its output what is limiting the Q2 current when you apply the PG's open circuit -10V to the Q2 source?     

With -10V at the source, Q2 would turn fully on.  In a traditional FG wth a series 50R and an open curcuit voltage of -10V,  the 50R and Vgs(on) determines the Q2 bias current.  If there was no 50R in the FG, Q2 would attempt to turn fully on and current would be limited by Rload, RDSon, and the FG's output stage output impedance (until its output likely popped).

I'll see if I can dig up an 8160A manual when I have the time.

Anyone have a link?

PW

Just whose circuit is it anyway? It appears more and more that Ainslie doesn't know how to operate her own apparatus, she can't explain the circuit and not only that... .she doesn't even know the actual operating parameters.  Look at her latest.

QuoteHere's the problem.  We show a 12 volt applied at the gate when the coupling is DC.  And I'm not sure that this setting is correct.  That's what I've tried to explain before when I said that voltage relates to that coupling.  But right now the apparatus is with a highly skilled member of our team and he's going to show the zero relationship throughout the whole circuit.  Frankly I don't think we ever apply more than 6 volts ... IF.  But nor am I competent to prove this.  What I do know is that MANY really skilled people - INCLUDING academics - have never turned at hair at this fact.  They're all skilled.  And YET?  It seems to entirely MONOPOLISE picowat's attention.  In truth it was an academic who SHOWED me that the 'off set' could entirely restrict the flow from the battery supply.  I assumed that this fact was well known.  NOR has there been ANY objection from any of the reviewers - on this point.  IF it were impossible then I'm reasonably sure that the IEEE reviewers would have thrown that document back at us.

What is she talking about in that last bit?  The IEEE editors rejected her submission every time she submitted it.

Google is your friend.
http://www.ko4bb.com/Manuals/11%29_Stuff_Not_Sorted/4_Miscelaneous/PULSE/8160/8160Aopm.pdf
http://www.ko4bb.com/Manuals/11%29_Stuff_Not_Sorted/4_Miscelaneous/PULSE/8160/8160AsmV1.pdf

And please let's not lose sight of the claim. Ainslie's claim is that she can produce high heat in the load, including "bringing water to boil", using the circuit operating as shown in Figure 3 or other similar scopeshots that show no current in Q1 even though it is receiving enough gate drive to turn on.

Simply reproducing the shot isn't enough! I did that already! It must be shown that the circuit does what she says it does, when it's operating like that, regardless of the reason!

I say that the mosfet is inoperative for some reason, probably blown open due to heat stress. But as Ainslie says.... that doesn't matter, IF it can be shown to be a "benefit" which results in more heat at the load than there should be. So any test reproducing that scopeshot has also got to reproduce, or attempt to reproduce, the high load heat, the "steam evident",  the tiny bubbles, the "bringing water to boil" -- all the claims about that condition that are listed in the paper. If those _effects_ cannot be produced with the Figure 3 -type scopetraces NO MATTER HOW THE TRACES WERE MADE.... then once again, the claims in the paper are bogus and must be retracted.

Quote from: picowatt on June 19, 2013, 11:36:57 PM
.99,

If your PG does not have a 50R in series with its output what is limiting the Q2 current when you apply the PG's open circuit -10V to the Q2 source?     

With -10V at the source, Q2 would turn fully on.  In a traditional FG wth a series 50R and an open curcuit voltage of -10V,  the 50R and Vgs(on) determines the Q2 bias current.  If there was no 50R in the FG, Q2 would attempt to turn fully on and current would be limited by Rload, RDSon, and the FG's output stage output impedance (until its output likely popped).

I'll see if I can dig up an 8160A manual when I have the time.

Anyone have a link?

PW

PW,

It is a current output, so for a setting of plus or minus 10V, the current is limited to 200mA. If you attach a 50 Ohm load across the PG, the output voltage will drop in half, and the output current (from the transistors) will still be 200mA. So the maximum current that can be contributed by the PG would be 200mA.

Now if you tie the PG output to +72V, there would be 1.44A, +/- 200mA flowing through its internal 50 Ohm resistor (but at least it won't flow through the output transistors). The EVP circuit however should quickly detect the over voltage and disconnect the PG output.