I'm pretty sure I'm not saturating my cores. How much field strength do you need to saturate a nylon bolt?

Yes, that's a great classic text all right, thanks for finding it. I found the Sams classic "Op Amp Cookbook" by Walter Jung in the used bookstore a few weeks ago and of course I scraped together my lunch money to buy it.

All op amps are not created equal. Many really do need the bipolar power supply to do their best work (like your analog power computer). Many can be used in single-supply mode as we have been doing. Some are more better for audio work, some are best for comparator applications, some are general purpose, some require dealing with more input and output options rather than just the two ins and one out of the opamps in the TL082 chip. Apparently I once again "aced" accidentally by choosing the TL082 for this application because of the very high impedance JFET input stage and the fast slew rate, and the fact that it works well on the single supply mode. And ease of use, low cost, and easy availability.

Here's something I found just a couple days ago, which made me very excited. This is another "classic" from a bygone era. The chips it talks about are still available, and many of them exist now in advanced versions that use very low power and are very sensitive. There are many many useful circuits and concepts covered in the TTL Cookbook.

ftp://apollo.ssl.berkeley.edu/pub/cinema/04.%20Science/TTL%20Cookbook_0672210355.pdf

It's a 12 MB pdf, an excellent scan, all 340 pages, clear diagrams and even the photos aren't too completely black.

ETA: I just checked EBay for sellers of TL082CN .... if you are willing to wait two weeks for a shipment from Thailand, you can get 10 for $2.79, free shipping. Or.... 50 for $9.99 from the same seller in Bangkok.
Twenty cents apiece! Delivered to your door! How can this even be possible at all?

I caught a mistake I made for the supercharged coil power analysis.   Note this is all about the resistive power losses in the coil, this is not about the power exported to the outside world by the stronger and faster-rising magnetic field pushing on the rotor.  I didn't even mention that which was another mistake.  That "export" power manifests itself as a (voltage drop inside the coil x the current flowing through the coil) and does not heat the coil.  You can't directly measure this "voltage drop due to power exported to the outside world," but you know it is there.

This may be counter-intuitive to some people but look at it like this:  You know the only form of power input to the coil is (voltage x current.)  You know the coil is exporting power to the outside world because because the rotor is turning.  That means that the export of power to the outside world has to "eat" a slice of the (voltage x current) input power.  This simply has to happen, and exactly the same process happens with a conventional electric motor.  The power to make the rotor spin is not coming from "nowhere."

So, back to the resistive power analysis that does indeed heat the coil.

So there is your supercharged drive coil:  You double the magnetic field strength of the drive coil if you split the coil into two halves and you decrease the energizing time constant by one half.  Nothing is stopping you from splitting it into three or even four if you want to.

Of course it goes without saying that when you do this you pay a price for this:  You double the current consumption and you double the power consumption of the drive coil also.

The resistive power dissipated in the original 100-turn coil:

We have one ampere and one unit of resistance so the power dissipated is = (one-squared x R) = R

The resistive power dissipated in the 50-turn coil:

We have two amperes and one-half unit of resistance so the power dissipated is = (two-squared x R/2) = 2R

We have two 50-turn coils so the total resistive power dissipated is 4R.

Conclusion:  When we split a 100-turn drive coil into two 50-turn drive coils the resistive power losses go up by a factor of four.

So let me finish this off by taking another look at the pure resistive losses in the coil vs. the power exported to the outside world to make the rotor spin.

Here is a thought experiment:

Setup #1:

Suppose that you have your 100-turn coil in your pulse motor and the rotor is locked so it can't turn.  You pulse the rotor and your sophisticated measurements tell you that the average power supplied to the drive coil is 10 watts.   In this case all of the 10 watts supplied to the drive coil will be converted into heat power that heats up the coil.

Setup #2:

Suppose that you have your 100-turn coil in your pulse motor and this time the rotor is free to spin.  You pulse the rotor and your sophisticated measurements tell you that the average power supplied to the drive coil is 10 watts.   In this case less than the full 10 watts of power supplied to the drive coil will be converted into heat power that heats up the coil.

It simply has to be due to the conservation of energy.  You could easily confirm this by monitoring the temperature rise of the coil over time.  And with some fancy footwork with your scope and provided that you knew what you were doing you could "see" this also.

The key issue here is to be able to visualize the flow of power.  If the drive coil is making the rotor spin then power is "flowing" out of the drive coil into the outside world.  If you really want to analyze what your pulse motor is truly doing then you have to be conscious of this.

That begs the question:  How much power is being dissipated in the rotor when it spins?  The assumption being that that power is coming from the drive coil, there is no other place!  That implies that you do a measurement of the moment of inertia of your rotor and then do some spin-down measurements.  That way you will know the rotor power dissipation as a function of RPM.  That's interesting information that may come in handy.  If you make very precise measurements, you should be able to see a non-linear curve which would tend to indicate the effects of air friction.

MileHigh

TK:

I will just comment on the analog computer circuit and some "weaknesses" in it, although that is too strong a term.  These are just some points for consideration.  They are not critical, just perhaps minor nuisances.

In a standard inverting amplifier configuration for an op-amp like we see twice in the circuit, the negative differential input pin is effectively a virtual ground because of the negative feedback servo amplifier.

When you look at the current sense input, you have a 1-ohm resistor effectively in parallel with a 2Kohm resistor.  So that's a 2000:1 ratio and you can pretty much ignore it.   The op-amp will not "disturb" the current measurement in any significant way at all.

In the RC filter stage though it's a different story.  You have a resistor divider network formed between the 1Kohm resistor and the 50Kohm resistor.   That's a 50:1 ratio and that means that the second-stage inverting amplifier is going to pull down the capacitor voltage a bit.  That's less than ideal, but it has been a long time that I have worked with op-amps and I was "playing it safe" in a manner of speaking.  I was worried about making the second-stage feedback resistor too high in value.

Here is a possible tweak:  Let's assume that you want to keep the RC time constant at one second.  Instead of 1Kohm + 1000 uF, you could change that to 500-ohms and 2000 uF.   The 500-ohm resistor is not an issue for the first op-amp at all.  Since it is a negative-feedback servo-amplifier the output impedance of the first amplifier is "zero" within certain constraints and you never go outside of those constraints.

Similarly, you might be able to change the 50Kohm and 200Kohm programming resistors for the second op-amp for 100Kohm and 400Kohm.   The assumption is that this is okay for the op-amp and it will work just fine.  Like I said I am not an op-amp guru so let's assume that this is true.  Obviously there is a limit for how high you can make the value of the feedback resistor and still have rock-solid signal integrity but I don't think a 400Kohm feedback resistor is anywhere near that limit.

If you try those changes, then the resistor divider that's associated with the RC filter is 500-ohms and 100Kohms.  So now we have a 200:1 ratio, which is a lot better than 50:1.  So with this modified configuration the second op-amp will be much less of a "drain" on the RC filter and "disturb" it much less.

The second issue is that there is no zero offset compensation on the non-inverting differential input pin for either op-amp.  If you look at the National Semiconductor applications guide you will note that most of the circuits do some kind of zero offset compensation.  I am making a reasonable assumption that any offset will be in the range of a few millivolts and can be ignored without it mucking the measurements up at all.  In other words, with no current going though the current sense resistor, and with the analog computer calibrated for your selected voltage source for your pulse motor, you might see a few millivolts at the output instead of zero.   Big deal, it's to make measurements on a pulse motor and the power draw for a pulse motor is never near zero.  So I am pretty confident that you can completely ignore this.

Note this is not meant to be a high-precision power measurement device.  It's intended to allow you to see the live power consumption of your pulse motor input or output as you make changes to various parameters.  For example, even moving the drive coil closer or further away from the rotor will likely show up as a power consumption change.  And like I said before, the delta power measurement should be deadly accurate.

MileHigh

Could be the start of a new fashion trend and therapeutic system!

TK:

Another thought or two about the issue of 1-ohm vs. 0.1-ohm for the current sensing resistor for the analog computer.

Using a 0.1-ohm resistor seems more "satisfying" as the design choice for the current sensing resistor because it disturbs the current flow through the circuit under test that much less.  I stated that I was "not comfortable" with the low voltages generated and the higher amplification required from the op-amp.  One more time I have to confess that I am not really sure if these are valid issues or not.  I have second thoughts about stating that the voltages wold be "low" for the op-amp differential input.  The whole point about the design of an op-amp is that the negative feedback reduces the differential voltage across the differential inputs to near-zero anyways.   For example, if the op-amp gain is one million, and the output is at six volts, then there is only a six microvolt potential difference across the inputs.

There is another biggie that I forgot to mention.  We know that the higher the gain for any amplifier, then the higher the noise is at the output.  Noise at the input gets amplified at the output plus the amplifier itself amplifies its own noise.  Supposing that there was a small amount of perceptible noise at the output of the first amplifier because of the high gain.  Well, it doesn't really matter!  The output from the first stage amplifier goes into a humungous RC filter that will squelch that noise to zero!  The second-stage amplifier will never see it!  Hence, 0.1-ohm may be the way to go.

Also, with a FET-input op-amp, this issue of bias compensation may be handled differently.  I was really talking about conventional op-amp inputs where you talk about a very tiny amount of bias compensation current.  There is no DC current flow for an FET input, and I don't know if they need bias compensation potentials.  I am also wondering if the silicon has been getting better and better over the years such that you simply don't have to worry about that issue at all.  I am not motivated to the point of going and reading the TL082 datasheet, sorry.

I suppose in a way that working on the bench is like riding a bike.  It would be fun for me to check every stage of that little analog computer and make sure it was working perfectly. Sigh....  It would be like old times!

One thing that I have never seen in my meanderings on YouTube is someone working with a digital logic analyzer.  Mind you, I have never searched on that.  The point being that you see a smidgen of work with digital logic by the free energy enthusiasts but never any mention of a logic analyzer.  For people that don't know, it's like a scope but instead for digital logic signals.  A typical "scope display" for a logic analyzer would display between eight and 20 digital logic signals.  They usually had two scope-equivalent analog inputs also.  When I used one you could transfer your waveforms onto a 3 1/2" floppy disk.  RJ-45 networking ports, USB, and USB flash drives were way way off in the distant future.

MileHigh

I've moved the sense coil to the underside/inside the rotor. Works great! I still am not seeing anything I can attribute to interaction between the sense and drive coils.

Later on I intend to test for interactions, without the rotor in place. I want to see if I can get feedback oscillations through coupling of the input (sense) and output (drive) coils.

@Tinselkoala,


                    Warp factor quantum leap! A N-S-N-S magnet arrangement would allow you to work the power coil to the inside also, to gain power from both coil poles, along with the sensor coil at 90 degrees. You'd have to cut that screw and tonge depressor off, build a skirt and dangle it.

Well... sort of. That's a couple steps ahead of where I'm at yet. Sticking with all one polarity magnets for now: how about this: Inside the rotor, a double ended drive coil, wound onto a tongue depressor with a central hole for the axle. The winding would be opposite for the two ends, so they would both present the same polarity on the ends. This would provide a drive pulse to two magnets instead of just to one.