There is still a problem as the currents at the node above R42 do not sum to zero.

You noticed that too, huh.

 

All ammeters shows +1mA ??

Polarity shows direction of current. "+" from where to where. Where do resistances cancel each other? We still have 1 mA through R42. The current just comes from the input of the op amp on the right.
Yes, the input resistance is in parallel with R42, but not opposite in any way I can see. What did I get wrong?

 

Yes, this means that all the current is provided via second op amp.
View attachment 100824

From post #14 ...
I don't see what resistances are in parallel???
Current shown flows from +12 V, through the 200 ohm feedback resistor. At the input of the op amp we have 1 V. Current continues down through the load resistor what is in parallel?

(edited to add ...)
The output of the buffer may be where I get confused. I agree current at that point (inside the loop) is zero, or at least considerably less than the load. My observation confirms this.

 

Re post 14: the physical 200 ohm resistor goes from the left amp output to ground.

The gyrated negative 200 ohm resistor goes from the left amp output to ground.

Thus they are in parallel.

As Dick showed they form a singularity of infinite resistance.

One resistor consumes current to ground. The other sources current from ground. Thus the left amp neither sources nor sinks current from output to load; a small current does flow thru the bias network but not to the external components.

 

Re post 14: the physical 200 ohm resistor goes from the left amp output to ground.

The gyrated negative 200 ohm resistor goes from the left amp output to ground.

Thus they are in parallel.

As Dick showed they form a singularity of infinite resistance.

One resistor consumes current to ground. The other sources current from ground. Thus the left amp neither sources nor sinks current from output to load; a small current does flow thru the bias network but not to the external components.

Fascinating!
Thank you all.

 

Now I power the drill from a power supply with a negative output resistance that partially cancels the resistance in the armature and lead wires so that now the drill maintains nearly constant speed under load and operates much more smoothly than before

I've been trying for years to develop a good speed controller for a router that I own that works with 110VAC. I've tried phase control using triacs, while monitoring its RPM with a hall sensor, and using an MCU to adjust the phase accordingly, and the result was less than mediocre for my taste. I then tried using a high power mosfet with the same MCU-sensor technique, and while the result was a smoother one, its response time was still too slow.
Do you think that using a negative output resistance power supply would deliver better performance? How hard would it be to design one for this purpose?

 

Yes.

 

Do you think that using a negative output resistance power supply would deliver better performance? How hard would it be to design one for this purpose?

1. Perhaps.

2. Based on your experience I would say "very hard." your looking at designing an AC source for an unknown motor type, not that I am any expert on motors.

 

1. Perhaps.

2. Based on your experience I would say "very hard." your looking at designing an AC source for an unknown motor type, not that I am any expert on motors.

How much power is used to create a little negative resistance? We use 5 mA, maybe 10, to counter 1 mA in negative resistance?

 

U guess that negative resistance is another word for gain in the previous circuits. the only negative resistance circuits I have been aquainted with were an oscilator like the heathkit tunnel dipper using a tunnel diode as an oscilator, and an amplifier in a specialized reciever system in the army. the preamp for a 1 to 10,000 mhz reciever and analysis system.

 

Same power as anything: look at the output of the negative resistance circuit and find the volts and amps. Multiply for power... After you also include the current going back down the feedback resistors.

For the circuit of post 14 you have 1v at 5ma for 5mw, plus 2v at 1ma for 2mw in the feedback.

 

That depends upon the topology, but in the simplest control the current through the load is the current through the negative resistance (the negative resistance can be obtained by making the output voltage of the power supply increase with load).

Maybe a more revealing question would be "How much power would be dissipated in the negative resistance circuit?" The answer is that it depends almost completely upon the motor being controlled and the maximum motor load to be supported.

 

??? Did I do something wrong

No everything is correct for the gain of 2 input buffer.