Quote from: MileHigh on May 14, 2012, 01:38:28 AM
Groundloop:


Now it seems very obvious.  I, and perhaps others, was fooled by my preconception that there should be no current flow.  I didn't even think about the diode in Q1.

So that was fun.  It shows the investigative process in action, and you figured out the mystery.

MileHigh

Yes you did.     Oh my. 

And yes it does.  Oh dear. 

Mags

Groundloop, TK:

What's interesting is that it's not trivial to calculate for the two unknown currents, i1 and i2,  in Groundloop's simplified schematic.

The 'smart' way to solve for the two unknown currents would be to do some simple linear network analysis.  You have two unknowns and provided you can develop two equations that describe the behaviour of the circuit (which you can) then you should be able to solve for i1 and i2.

The other way would be to use numerical analysis methods and iterate on the solution.

MileHigh

Quote from: MileHigh on May 14, 2012, 01:38:28 AM
Groundloop:

That looks excellent.  It looks like you have explained the very high current when the gate voltage on Q2 is high.  (Referencing your schematic.)

Now it seems very obvious.  I, and perhaps others, was fooled by my preconception that there should be no current flow.  I didn't even think about the diode in Q1.

Current from the +13-volt bias source flows through the 50-ohm resistor and then leaks through the diode in Q1 and then flows through Q2 and then back to the +13-volt bias voltage source.

This is a separate current loop independent of the main current loop driven by the +24-volt source.  This "extra" current loop causes a kind of negative 'feedback' on the main current loop and reduces the current flow in that loop very slightly.

So that was fun.  It shows the investigative process in action, and you figured out the mystery.

MileHigh

MileHigh,

I got almost the same result in simple simulation. So if .99 is reading this maybe he could do a Spice sim also?
I'm looking forward to see what TK gets with his measurement.

NB: A big storm is passing right now and my mains keep failing. So I do not know
how long I has the Internet connection up and running.

GL.

OK, please let's be careful about Q1 and Q2. It is the diode in my Q2 that seems important to me. The diode in whichever mosfet is NOT being turned on by the gate signal.

Here are my results. I had to use a 54.3  ohm resistor for the series resistor, so correct for that in the math. And also my ammeters are 1.8 ohm internal resistance.

And I am referencing the mosfets by the designations in the diagram below.

So I measured 2 conditions: Total main current and total bias current with Q2 IN, and total main current and total bias current with Q2 OUT.

So... with all Q2s OUT, and the hookup as below, I set the bias power supply to 13.02 volts. Total Main current through Q1 (inline ammeter at battery negative as usual) was 1.95 Amperes and the main battery voltage, running, was 23.8 V. And the bias current was zero, zip, zilch, nada, nichts, nichego. Zero, in other words, below the detectability of my ammeter. Mosfet warming, load heating well.

And...with one Q2 IN, I set the bias power supply to 12.92 volts. Total main current through... Q1 and ? was 1.78 Amperes and the battery voltage, running, was 23.8 V. And the bias current started at 90 mA and rose steadily to about 130-140 mA as ...something? heated up?... Q1 mosfet warming, load heating well.

And I tried all different mosfets in all positions, and there was a slight variation in this bias current but not significantly much.


I'm too tired to do the math tonight but I'd be happy to watch while somebody else does it.

The diagram showing the connections and my mosfet names. I think I'll call them Moses and Festus from now on.

Quote from: TinselKoala on May 14, 2012, 02:42:37 AM
OK, please let's be careful about Q1 and Q2. It is the diode in my Q2 that seems important to me. The diode in whichever mosfet is NOT being turned on by the gate signal.

Here are my results. I had to use a 54.3  ohm resistor for the series resistor, so correct for that in the math. And also my ammeters are 1.8 ohm internal resistance.

And I am referencing the mosfets by the designations in the diagram below.

So I measured 2 conditions: Total main current and total bias current with Q2 IN, and total main current and total bias current with Q2 OUT.

So... with all Q2s OUT, and the hookup as below, I set the bias power supply to 13.02 volts. Total Main current through Q1 (inline ammeter at battery negative as usual) was 1.95 Amperes and the main battery voltage, running, was 23.8 V. And the bias current was zero, zip, zilch, nada, nichts, nichego. Zero, in other words, below the detectability of my ammeter. Mosfet warming, load heating well.

And...with one Q2 IN, I set the bias power supply to 12.92 volts. Total main current through... Q1 and ? was 1.78 Amperes and the battery voltage, running, was 23.8 V. And the bias current started at 90 mA and rose steadily to about 130-140 mA as ...something? heated up?... Q1 mosfet warming, load heating well.

And I tried all different mosfets in all positions, and there was a slight variation in this bias current but not significantly much.


I'm too tired to do the math tonight but I'd be happy to watch while somebody else does it.

The diagram showing the connections and my mosfet names. I think I'll call them Moses and Festus from now on.

TK, thank you for taking time to do the bias current test. Yes, that confirms my findings. Without the other MOSFET
in the circuit there is no path for the bias current and you get zero current and a little higher total circuit current.

When you plug in the other MOSFET you get a path through the internal diode and you main circuit currents drops.
This is consistent with a negative feedback as MileHigh did say. You test 100% confirms my theory and also my
testing.

Thanks,
GL.