Rosemary,
When I read between the lines in your prose below, it looks like it never even occurred to you that current can actually flow through the two terminals of the function generator. That could either be into the signal terminal and out of the ground terminal, or out of the signal terminal and into the ground terminal. Well the short answer is that IT CAN, and so everything you have been saying recently is all wrong.
Poynt, if you're still there. I wonder if you could ask MileHigh to get his head out of those clouds and his feet on the ground. He's seriously proposing that upwards of 5 amps can flow into the ground rail of the probe - through all the circuitry of the signal supply source, nuke the most of those rectifiers, fry the delicate potentiometers, burn up most of that circuitry of that really sensitive instrument, that is decidedly NOT designed to take high amperage. And he then proposes that it can come out on the other side at the probe of the signal generator - to confront an applied negative signal at the Gate of Q1. It needs to reach Q1's source rail. So it IGNORES that signal? It simply overrides the applied charge and slips onto the source leg of Q1S. And then it flows unobstructed to the supply source or negative rail of the battery. That's unlikely.
If he's suggesting that the current from the battery can simply flow through the Q2 transistor at Q2's Drain through to Q2 Gate - AND THEN DIRECTLY ONTO THE CIRCUIT at it's source rail (or the negative battery terminal thing) then it would need to bypass it's own Source Q2S leg. Which means that we'd see a very visible arcing sparking flow of current in mid air, as it tries to find safe landing on a really slim landing site all of which is to managed while the current is in a kind of free fall. That's also unlikely. But both options are interesting on a speculative level. Especially as it would introduce some utterly exotic, if somewhat improbable, physics. And show him the schematic again. Here it is.
Q2s or the source leg of Q2 has NO CONNECTION AT ALL with the circuit battery negative. IT FLOATS. I really need a shot of this to show you guys. Hopefully soon.
As per normal for you, your technical discussion is all mangled up and your switch terminology on the fly. It doesn't matter, I understand what you are trying to say.
For starters, I am going to follow the naming convention for standard current flow.
He's seriously proposing that upwards of 5 amps can flow into the ground rail of the probe - through all the circuitry of the signal supply source, nuke the most of those rectifiers, fry the delicate potentiometers, burn up most of that circuitry of that really sensitive instrument, that is decidedly NOT designed to take high amperage.
By "probe" you mean the function generator signal and ground terminals, not to be confused with the scope probe. Beyond that I an not stating how much current can flow, all that I am stating is that current can and will flow through the function generator. The talk about "burning up circuitry" is just you revealing how little you know about electronics, which you freely admit, but then go onto draw erroneous conclusions anyways.
The only thing I can say right now is that since your function generator has an offset control, that means that it can both source and sink current when it is normally driving a load. The only thing that will be affected by having an external power source put current through the function generator is it's output stage. I am assuming that the output impedance is 50 ohms, and that is only a clue as to how the output stage will react to an external source of current flow. However, no matter what you think or say, current does flow through the function generator and that completes the circuit when Q2 is switched on.
And he then proposes that it can come out on the other side at the probe of the signal generator - to confront an applied negative signal at the Gate of Q1. It needs to reach Q1's source rail. So it IGNORES that signal? It simply overrides the applied charge and slips onto the source leg of Q1S. And then it flows unobstructed to the supply source or negative rail of the battery. That's unlikely.
I have no idea why you are talking about Q1 because it was established more than a year ago that Q1 is always off, which I also stated in my previous posting.
LET ME REPEAT IT AGAIN: CURRENT FROM THE BATTERY DRAIN RAIL (POSITIVE) FLOWS THROUGH RL1 THEN INTO THE DRAIN PIN OF Q2, THEN OUT OF THE SOURCE PIN OF Q2 THEN INTO THE SIGNAL TERMINAL OF THE FUNCTION GENERATOR, THEN OUT OF THE GROUND TERMINAL OF THE FUNCTION GENERATOR, THEN THROUGH THE CURRENT SENSING RESISTOR THEN BACK TO THE BATTERY SOURCE RAIL (NEGATIVE).
I hope that you get that Rosemary.
If he's suggesting that the current from the battery can simply flow through the Q2 transistor at Q2's Drain through to Q2 Gate - AND THEN DIRECTLY ONTO THE CIRCUIT at it's source rail (or the negative battery terminal thing) then it would need to bypass it's own Source Q2S leg.
The above quote from you is more mangled nonsensical talk. Everybody is supposed to know that the gate of a MOSFET transistor does not pass DC current. See the all-capitals description of the current flow in my prose above.
Which means that we'd see a very visible arcing sparking flow of current in mid air, as it tries to find safe landing on a really slim landing site all of which is to managed while the current is in a kind of free fall. That's also unlikely. But both options are interesting on a speculative level. Especially as it would introduce some utterly exotic, if somewhat improbable, physics. And show him the schematic again. Here it is.
The above quote from you is more mangled nonsensical talk. Even I can't figure out what you are trying to say there.
Q2s or the source leg of Q2 has NO CONNECTION AT ALL with the circuit battery negative. IT FLOATS. I really need a shot of this to show you guys. Hopefully soon.
Here you are starting to make some sense but you are dead wrong. You are alleging that "it floats" because it never occurred to you that the signal generator itself completes the circuit and allows current to flow. The bulk of the voltage that creates that current flow is from the battery set, and therefore the battery set is providing power that gets burned off in the RL1 inductive resistor.
Current flows through the function generator itself to complete the loop. Now that you know, perhaps you will manage to understand how your circuit actually works. All of your talk about "current flowing when the batteries are disconnected" is dead wrong. So your theory about the material itself that makes up the physical mass of the inductive resistor supplies the "extra" power or that "Dark Energy" is responsible for the "extra" power is completely and utterly wrong.
Your circuit is 100% conventional and no over unity barriers have been breached. The dam still holds firm.
MileHigh
Author
Topic: The Rosemary Ainslie Circuit (Read 477132 times)
