Tesla's "COIL FOR ELECTRO-MAGNETS".

[conradelektro]

I just measured a strange result (see the attached circuit diagram):

I compared power output over the "100 Ohm load" with and without the parallel 10 µF cap. One measurement (calculation) was done after the "speed up effect" occurred. Then the cap was removed and the power supplied to the motor was regulated to get about the same output over the "100 Ohm load alone".

There is a phase shift of about 45° between Voltage and current if the cap is in place and absolutely no phase shift without the cap.

If one compares efficiency (power needed to drive the motor for approximately the same output over the 100 Ohm load), the circuit with the 10 µF wins easily. But the phase shift might falsify the result?

You can see the phase shift in this post http://www.overunity.com/13460/teslas-coil-for-electro-magnets/msg384371/#msg384371 (top right scope shot, speed up effect)

What influence does a phase shift between Voltage and current have on the calculated Watt?

Greetings, Conrad


P.S.: I looked it up, it is called power factor http://en.wikipedia.org/wiki/Power_factor.

Since the phase angle shift is 45°, the "active power" = 0.71 * "apparent power" , 0.71 is cos(45°).

So, the active output with the 10 µF in place is lower than calculated. Therefore the comparison is not valid.

Real comparison:

- 0.79 apparent Watt is only 0.79 * 0.71 active power, 0.56 active Watt at 134 HZ (motor consumption 7.3 V at 0.94 A) if 10 µF cap is in place

- 0.57 active Watt at 125 Hz (motor consumption 7.1 V at 1.07 A) without the cap (I just did the measurement)

So, the whole set up is slightly more efficient without the 10 µF cap, which is to be expected.

[gyulasun]

...
For Ferrit core tests I have to wind two new identical coils (one monofilar and the other bifilar) over a 10 mm Ferrite core, what I want to do anyway soon. I can not fit my 8 mm Ferrite core into the big coil I am using now. And a 6 mm Ferrite core would rattle (that is why I did not buy it).
...

Hi Conrad,

Perhaps you could use a small piece of polifoam to wrap up the 6 mm ferrite core against the rattle?  just a suggestion to save you from winding new coils? of course you do as the most appropiate for you.

Okay on the scope probe placements, unfortunately I did not check it on your drawing...
Will make further comments tomorrow.

Gyula

[MileHigh]

Hi Conrad:

You notice in your scope trace where there is just the pure capacitor load (with no resistor) there is a standard phase shift of 90 degrees between the voltage and the current.  The current leads the voltage by 90 degrees.

You also noticed when it's a pure resistive load that the phase shift is zero.

When you add a capacitor to the circuit then the phase shift falls somewhere between zero and 90 degrees.  How many degrees difference is dependent on the size of the capacitor, resistor, and the frequency.  The higher in frequency you go the capacitor starts to predominate and takes more current.  That makes the phase shift move towards 90  degrees.

These things can be visualized with rotating phasor diagrams.  Here is one for a pure capacitor load at a given frequency.

Imagine there is a source of light directly over the phasor diagram.  The actual voltages and currents you will see on your scope are the "shadows" of the phasors on the "ground."

[MileHigh]

Conrad:

What influence does a phase shift between Voltage and current have on the calculated Watt?

If you have a phase shift that indicates that you have a partially reactive load.  You can argue that reactive loads are bad for your efficiency measurements because some of the energy is "ringing" and oscillating back and forth between the capacitor and the coil and being burned off in the resistance of the coil.  So that's power that is being 'stolen' from the load resistor.

Here is a thought:  With the capacitor as a resonator with the coil you can get more energy circulating and as a result get more power going to the load resistor.  However, nobody is saying that you can't simply remove the capacitor and change the value of the load resistor so you get more power in the load resistor.  Do you see what I mean?  You can argue that that is the true basis of an "even playing field" for comparison.

In the next posting I will cover using your scope to explore the coil in a bit more detail.

MileHigh

[MileHigh]

Conrad:

In posting #765 you changed your scope probe configuration.  Your original configuration (on the left) is arguably better and I will explain why.

Let's take out the "big guns" and look at the currents when the switch is closed and do some basic circuit analysis.  You can see two current loops in the circuit.  You see the left rectangle in the schematic.  That's one current loop.  Let's call the first current loop I1.  For the right rectangle let's call it I2.  Let's also say that clockwise current is positive current for both loops.

Here is the key thing:  Look at the capacitor and imagine the two current loops flowing clockwise.  The current flowing in the capacitor is (I1-I2).  Can you see that?

You can see that channel 2 is monitoring I1.   You can see that channel one is monitoring I2 because it's across the 100 ohm resistor.  You don't have to worry about the capacitor, all information about I2 is provided by the voltage measurement across the 100-ohm resistor.

So you can see how the original scope connection allows you to monitor both of the currents I1 and I2.  So you can calculate the power being burnt in the coil and the power being burnt in the load resistor simply by looking at the two RMS voltages and crunching a few numbers.

Beyond that, the scope channel across the 100 ohm resistor monitoring I2 is also implicitly telling the you the current through the capacitor.  The current through the capacitor is proportional to the slope of the voltage waveform across the 100 ohm resistor.  (That's a big 'trick' about capacitors.  As long as you can see the voltage waveform across the capacitor, you don't have to measure the current, you can "eyeball" the current with very good accuracy.)  Also, the current through the capacitor is also (I1- I2) and your scope may be able scale down the channel 2 voltage measurement by a factor of 100 before it adds it to channel 1 and displays it on your display as a third computed trace.  That would be a second way to get the capacitor current.

I note that you changed the scope probe connections because the polarities of the signals was confusing.  That's what the "invert waveform" function on the scope is for.  Instead of changing the probe connections around simply invert the waveform on the display and everything will make sense.

MileHigh