Circuit sj1. Terse and Technical only.

[lanenal]

Here I replaced the Schottkey diode with a scilicon diode.

[exnihiloest]

...

[poynt99]

Here is a simulation of the self-run circuit. Of course, it did not turn off the battery. Just a proof of concept.

Good job Lane. Looks like you have the frequency about right.

Have you considered trying the sim with a non-linear core rather than the simple linear coupling? How close are those waveforms to what the Professor is seeing?

.99

[lanenal]

Good job Lane. Looks like you have the frequency about right.

Have you considered trying the sim with a non-linear core rather than the simple linear coupling? How close are those waveforms to what the Professor is seeing?

.99

Thanks .99. This is not exactly like his model, but close, and the frequency in the simulation is of the same magnitude. I'm not sure how to work with a non-linear core yet, but that's a good idea! I'll see if there is anything available. I'll also try to replicate his model too. Just need a 151p cap and some battery...

[PhysicsProf]

  Interesting idea, laneal.  Evidently you are adding the third winding to use that to extract "output" energy from the system.  Not a bad idea, but note that this will certainly change the dynamics of the circuit.  Certainly worth a try.

  Let me emphasize a curiosity in the original schematic (attached again for convenience) -- which is that point 6 is the "low point" in the potential, as measured by a DMM and diode Do in the direction shown.  Right now, I'm running with a capacitor in place of Ro, with the NEGATIVE terminal of the cap at point 6, as shown in the diagram.  This charges the cap quite rapidly under good conditions to about 6 volts, with 2.5 volts input.

[lanenal]

  Interesting idea, laneal.  Evidently you are adding the third winding to use that to extract "output" energy from the system.  Not a bad idea, but note that this will certainly change the dynamics of the circuit.  Certainly worth a try.

  Let me emphasize a curiosity in the original schematic (attached again for convenience) -- which is that point 6 is the "low point" in the potential, as measured by a DMM and diode Do in the direction shown.  Right now, I'm running with a capacitor in place of Ro, with the NEGATIVE terminal of the cap at point 6, as shown in the diagram.  This charges the cap quite rapidly under good conditions to about 6 volts, with 2.5 volts input.

Thanks professor. Could you maybe give an exact specification for replication when you are more or less certain of the best values you have got in your experiments? There are many changes here and there now and I am not sure which is the best way to follow, especially when you might have got better configurations.

I agree that that's the right way to put a capacitor. But to loop the device, it is much harder. Using a third coil is one of the simplest way I can think of. You are right that will change the dynamics a bit, but won't be too much -- because the L0 and L1 will be equivalent when it comes to pick up the collapsing magnetic field. With L1, it then becomes easy to loop the device.

[lanenal]

OK, I have my replication running...

I found that when R0=10k, the LED is really dimly lit. So I reduced it to R0=1k, with much brighter LED, and the input is almost the same. My input is about 3mA averaged over time, peak is probably 40mA, duty cycle about 10%. The frequency is measured at 1.37MHz. I have no scope, all measurements done with my DMM. My input V = 5Volts. This seems to be a promising replication, next I think I should try to do a dual DMM with caps test to measure input/output energy ratio.

I will keep it running for a day or so, with a DC converted from the grid.

[poynt99]

What do you have in mind for a Pout measurement method Lane?

.99

[lanenal]

I'd try to charge a capacitor. I've got some 10000uF caps, hopefully will do. Any suggestions?

[PhysicsProf]

OK, I have my replication running...

I found that when R0=10k, the LED is really dimly lit. So I reduced it to R0=1k, with much brighter LED, and the input is almost the same. My input is about 3mA averaged over time, peak is probably 40mA, duty cycle about 10%. The frequency is measured at 1.37MHz. I have no scope, all measurements done with my DMM. My input V = 5Volts. This seems to be a promising replication, next I think I should try to do a dual DMM with caps test to measure input/output energy ratio.

Glad to hear you have a DUT, lanenal.   You asked about what conditions I would recommend; to summarize,

Vin 2.5 V
Rb 2Kohms
MPS2222
C-B 151 pF
D red LED
L-B, L-O bifilar 9turns, ferrite toroid 1"OD,  ~120uH each (recently remeasured the toroid giving best results)
Ro 9.7 Kohms
Rr  4.1 ohms

Testing the input Power using a Cap and a stopwatch, as described earlier in this thread, and independently using the TEK 3230 method, I find Pin ~ 10-17 mW with this DUT   If your Pin is much larger than this, with the conditions given above, then the replication needs to be re-checked for consistency.   With Rb ~ 51Kohms, the LED is seen to glow but very dimly, and Pinput drops (as described above).

Chris S. on the OU.com forum has made several tests and mods and his version 7 is shown below; his "best", IIRC.  I hope Chris does not mind my citing his work; posted here:    http://www.overunity.com/index.php?topic=10773.210

Chris has been very diligent in testing the circuit and making small mods.

[lanenal]

Thanks professor, will try your configuration. My configuration is far from yours. I can hardly charge my 1000uF capacitor in place of R0 to above 2V, after 2 minutes of charging with my input at 5V (after a while the red LED won't lit anymore, and I noticed the circuit stops working after a while, because the frequency drops to zero), this shows that there must be something special in your configuration. Have you tried to replace one of your crucial components at a time (Diode, Transistor, Toroid core) and observe significant changes in your measurements?

[PhysicsProf]

Thanks professor, will try your configuration. My configuration is far from yours. I can hardly charge my 1000uF capacitor in place of R0 to above 2V, after 2 minutes of charging with my input at 5V (after a while the red LED won't lit anymore, and I noticed the circuit stops working after a while, because the frequency drops to zero), this shows that there must be something special in your configuration. Have you tried to replace one of your crucial components at a time (Diode, Transistor, Toroid core) and observe significant changes in your measurements?

Clearly, our DUT's are different.  I don't know why, but I have some further results to share.  I'm working with two capacitors now, 10,000 uF = 10mF each, so a total of 20 mF.  Charged to 2.58V using 2 rechargeable AA's.  I use the caps to run the device until the voltage drops to 1.5V on the caps and note the time t required using a stop-watch.  Thus, I have a reliable measure of Ein and Pin:

E = 1/2 C V**2,  P = E/t,

So Pin = 1/2 20mF (2.58V**2 - 1.5V**2) /t   = 44.06mJ/t

So 44.06 mJ are put into the system, and dividing this Ein by the time, I derive Pin.  The measurements have proven repeatable -- and interesting.
Here are some results this morning:

Basic conditions:
Rb 52Kohms
MPS2222
C-B 151 pF
D red LED
L-B, L-O bifilar 9turns, ferrite toroid 1"OD,  ~120uH each

Ro = 1Kohm
Rr= 0, CSRout = 0 (removed)

Oscilloscope ATTEN used to look at the Pin waveform most of the time in my little lab at home.

1.  With the above conditions, the caps discharge from 2.58V to 1.5V in t = 37.8 seconds, so Pin = 44.06mJ/37.8s = 1.17 mW.
LED dim but clearly visible throughout the run.

2.  Next, I removed the Ro/LED from the system, so current flows back to ground through L1 loop only, t = 37.0 s, Pin = 1.19mW.
 A bit of a surprise, repeatedly, with the Ro/LED out of the circuit, without that load, the input power drain INCREASES.
 A bit hard to explain without OU perhaps, but not sure.  In any case, the power drain through the LED and 1Kohm resistor is small (if not negative )...

3.  So I take the output of L1 and connect it to point 6 (instead of to point 4), so that the return path is through the 1Kohm Rout.  Rout/LED back in the circuit.
Now the LED is extremely dim, but visible in a darkened room.  t= 52s, Pin = 0.85mW.  Pin went down, as might be expected since the current through L1 is impeded by the 1Kohm Ro.

4.  Next, change Ro from 1Kohm to 220ohms.  Same test as in 3, now t = 62.2 sec, Pin = 0.71mW.  Now this is surprising to see Pin go down with Ro reduced, since  the current through L1 is impeded LESS by 220ohms than by 1Kohm Ro -- we are approaching the situation in 1 where the L1 output goes directly to ground.

5.  Back to condition as in 1, but with Ro=220ohms, t= 38.2 s, Pin = 1.15mW (about the same as with Ro=1Kohm).

6.   Next, I removed the Ro/LED from the system, so current flows back to ground through L1 loop only, t = 37.1 s, Pin = 1.19mW.
 A bit of a surprise, repeatedly, with the Ro/LED out of the circuit, without that load, the input power drain INCREASES.  As before.


Next I went to my own "replication", with the conditions as in 4 above, and found t=54 seconds (0.82mW), rather than 62.2 s with the 1st DUT.  So small variations make a difference (not too surprising) -- in particular, the toroid/windings differ in my own replication.  But the power draw (Pin) is much less than your Pin, Laneal...

Now, this gives us an idea of how the circuit behaves, and a measure of the low power consumption.

[PhysicsProf]

Now with my replication, more results:

1.   I take the output of L1 and connect it to point 6 (instead of to point 4), so that the return path is through the 220ohm Rout.  Rout/LED back in the circuit.
Now the LED is dim, but clearly visible.   t= 54s, Pin = 0.82mW.  (repeat from above)

2.  Connect L1 direct back to point 4.   t= 50.8s, Pin = 0.87mW.   I observed that with this condition, the LED is growing dimmer as the voltage from the caps drops -- until about 1.2Vin, then the LED suddenly gets brighter -- unexpected.  I repeated this experiment and observation.

3.  Place a 220-ohm R in series with L1, then to point 4.   t= 59.8s, Pin = 0.74mW.

[PhysicsProf]

  I can play with the circuit until Pin is quite small, and the LED lights up for longer and longer on the two caps...but I need to get  Pout measured in the best way possible, without an oscilloscope preferably.  I want to check the scope results and derive a method I can do in my home lab.  The goal is maximum Pout/Pin.   Working on that.

Meanwhile, this weekend I have been puzzling over a relativity puzzle.  Also a lot of fun, and a mental challenge.  At a conference a week ago, a fellow posed a question and I've modified it so that now it looks like a real puzzle, in that momentum conservation appears to be violated...  which "cannot happen".
OK, so show me what's wrong.  I like to pose puzzles like this to other physicists, and if any of you have answers, pls let me know.  I admit haven't found an answer yet... and somehow, it MIGHT relate to what we're doing here.  Of course, at present its a thought-experiment, not done physically yet...

Consider two loops of wire facing each other, A and B, 3 cm apart as shown in the attached.  We're going to use the fact that magnetic fields  propagate at the speed of light so that it takes time for the field generated in A to reach B, and vice versa,
t = 3cm/3X10**10cm/s = 0.1 nanosecond (ns) for a field generated at A by a current pulse to reach B. 

Sure, edge effects, etc. -- I'm not worrying about those, yet.

We send a pulsed current through A and (with time delay as shown) through B.  A "positive current pulse" is such that a North pole points to the right, a negative current pulse generates a North pole pointing left.  When the first + pulse goes through A, B is off.  But as the field from A arrives, B switches on with a negative pulse and thus is REPELLED, pushed to the right.  We can end the thought experiment there, with A turned off now and so feels no effects, while loop B (free to move) travels to the right.  There is motion to the right only, which does not conserve momentum... Can you find an error?

But we can go on and get loop A to move also.  The field from B travels back towards A and when it arrives 0.1 ns later, A has a negative pulse and so is pulled to the right and begins to move right.  THAT field from A propagates at the speed of light towards B... which is pushed to the right when it arrives, and so on.
Thus, the whole "system" moves to the right (one can connect the loops on a platform, or a space-ship , with an apparent violation of conservation of momentum...

I haven't found the error if there is one...  let me know why this won't work.  Thanks.   

[PhysicsProf]

"We send a pulsed current through A and (with time delay as shown) through B.  A "positive current pulse" is such that a North pole points to the right, a negative current pulse generates a North pole pointing left.  When the first + pulse goes through A, B is off.  But as the field from A arrives, B switches on with a negative pulse and thus is REPELLED, pushed to the right.  We can end the thought experiment there, with A turned off now and so feels no effects, while loop B (free to move) travels to the right.  There is motion to the right only, which does not conserve momentum... Can you find an error?

But we can go on and get loop A to move also.  The field from B travels back towards A and when it arrives 0.1 ns later, A has a negative pulse and so is pulled to the right and begins to move right.  THAT field from A propagates at the speed of light towards B... which is pushed to the right when it arrives, and so on.
Thus, the whole "system" moves to the right (one can connect the loops on a platform, or a space-ship , with an apparent violation of conservation of momentum...
I haven't found the error if there is one...  let me know why this won't work.  Thanks.   

I can't believe this would work ...
But if it did... following some experiments certainly...

Then, I would arrange the loops (small, short-wire coils probably) into a wheel, and let the push-pull described above generate circular motion -- There's your motor, to drive a generator or a car...  you see where this could lead.. 

[lanenal]

I can't believe this would work ...
But if it did... following some experiments certainly...

Then, I would arrange the loops (small, short-wire coils probably) into a wheel, and let the push-pull described above generate circular motion -- There's your motor, to drive a generator or a car...  you see where this could lead..  

Very thought provoking...it seems that Newton's third law does not hold here (that's is, force and its anti-force are not symmetric).

I can certainly lower my input to your level (my LED is brighter than yours and my input power is about 15mW). Would be very interesting to see your alternative measurements of output. Charging a cap via a diode is probably a good idea. The power wasted on the diode can also be easily calculated: because there is a nearly constant voltage drop over the diode when there is current through it (thus the total energy spent on the diode is: C * deltaV * Vdiode, where C is the capacitance, deltaV is the voltage change over the capacitor, and Vdiode is the constant forward voltage drop on the diode).

lanenal

[PhysicsProf]

Thanks, laneal -- yes, violating Newton's third law is essentially equivalent to non-conservation of momentum.

More studies with the circuit, trying to achieve low Pin (< 1mW), then look at Pout --

Conditions:
Ein by 10,000uF cap = 10mF
Rb 52Kohms
MPS2222
C-B 372 pF (note increase)
D red LED
L-B, L-O bifilar 9turns, ferrite toroid 1"OD,  ~120uH each
L1 connected to point 6 (between LED and Ro)
Ro = 220 ohms

Rr= 0, CSRout = 0 (removed)

Then, using Cap + stopwatch method, Pin is 0.41mW with LED glowing dim but easily seen in lit room

Next, to look at Pout -- crudely.  I simply replaced Ro with a 10mF capacitor


Ro = > replaced by 10,000 uF cap, same as the input-power cap (yes, I need to measure the actual capacitances -- using my colleague's meter).

Using cap and stop-watch method for BOTH caps, get following results.

Ein = 1/2 C (Vstart**2 - Vfinal**2)   -- discharging to provide Pin
Eout = 1/2 C (Vofinal**2 - Vostart**2)  -- charging to estimate Pout.  Note that the LED is glowing, and that energy is NOT captured, (various losses also not counted) so this provides a conservative estimate of Pout.

n = Eout/Ein -- since Cin ~ Cout,

n ~ (Vofinal**2 - Vostart**2) / (Vstart**2 - Vfinal**2)  -- conservative estimate, made using matched capacitors for input and output energy.

Typical result:
n ~ (1.58**2 - 1.30**2) / (2.54**2 - 2.25**2)  = 0.58 = 58% (conservative)

Note, Pin = Ein/time ~ 0.25 mW  (Pin is less with this system, with the cap replacing the 220 ohm Ro)

Best result tonight:
n ~ (1.366**2 - 1.258**2) / (2.0**2 - 1.89**2)  = 0.7 = 70% (conservative)

Now, this is with matched caps for input energy and output energy, with the energy in the LED "thrown away", as it was lit visibly.  Hope you're following what I'm doing here -- achieve a low Pin while LED is still lit visibly (quite bright), then get a FIRST CONSERVATIVE ESTIMATE for Pout (or Eout in this case).

Next -- a simple test, 10mF cap at 2.53V, so Ein = 32 mJ,  connect matching cap in parallel --> both caps now at 1.22 volts (volts drops by half as the charge is shared, conservation of charge).  But
E = 1/2 10mF V**2

So Ein = 32 mJ, Eout = 15 mJ (measured using voltages before and after joining caps in parallel) -- so HALF THE ENERGY IS LOST JUST CHARGING ONE CAP WITH A CHARGED-CAP (= capacitances).  n = 50% is the best one can do under these circumstances.  Not certain if that applies in this circuit...  a sim might tell that...  

Here, I get n ~ 58% typically and n~ 70% best run.

Conclusion:  interesting, not definitively OU in this configuration -- still would like a better way to measure Eout or Poutput routinely so I can tune caps and resistors -- and the wound-toroid -- to maximize Pout/Pin = Eout/Ein.

[lanenal]

Thanks, laneal -- yes, violating Newton's third law is essentially equivalent to non-conservation of momentum.

More studies with the circuit, trying to achieve low Pin (< 1mW), then look at Pout --

Conditions:
Ein by 10,000uF cap = 10mF
Rb 52Kohms
MPS2222
C-B 372 pF (note increase)
D red LED
L-B, L-O bifilar 9turns, ferrite toroid 1"OD,  ~120uH each
L1 connected to point 6 (between LED and Ro)
Ro = 220 ohms

Rr= 0, CSRout = 0 (removed)

Then, using Cap + stopwatch method, Pin is 0.41mW with LED glowing dim but easily seen in lit room

Next, to look at Pout -- crudely.  I simply replaced Ro with a 10mF capacitor


Ro = > replaced by 10,000 uF cap, same as the input-power cap (yes, I need to measure the actual capacitances -- using my colleague's meter).

Using cap and stop-watch method for BOTH caps, get following results.

Ein = 1/2 C (Vstart**2 - Vfinal**2)   -- discharging to provide Pin
Eout = 1/2 C (Vofinal**2 - Vostart**2)  -- charging to estimate Pout.  Note that the LED is glowing, and that energy is NOT captured, (various losses also not counted) so this provides a conservative estimate of Pout.

n = Eout/Ein -- since Cin ~ Cout,

n ~ (Vofinal**2 - Vostart**2) / (Vstart**2 - Vfinal**2)  -- conservative estimate, made using matched capacitors for input and output energy.

Typical result:
n ~ (1.58**2 - 1.30**2) / (2.54**2 - 2.25**2)  = 0.58 = 58% (conservative)

Note, Pin = Ein/time ~ 0.25 mW  (Pin is less with this system, with the cap replacing the 220 ohm Ro)

Best result tonight:
n ~ (1.366**2 - 1.258**2) / (2.0**2 - 1.89**2)  = 0.7 = 70% (conservative)

Now, this is with matched caps for input energy and output energy, with the energy in the LED "thrown away", as it was lit visibly.  Hope you're following what I'm doing here -- achieve a low Pin while LED is still lit visibly (quite bright), then get a FIRST CONSERVATIVE ESTIMATE for Pout (or Eout in this case).

Next -- a simple test, 10mF cap at 2.53V, so Ein = 32 mJ,  connect matching cap in parallel --> both caps now at 1.22 volts (volts drops by half as the charge is shared, conservation of charge).  But
E = 1/2 10mF V**2

So Ein = 32 mJ, Eout = 15 mJ (measured using voltages before and after joining caps in parallel) -- so HALF THE ENERGY IS LOST JUST CHARGING ONE CAP WITH A CHARGED-CAP (= capacitances).  n = 50% is the best one can do under these circumstances.  Not certain if that applies in this circuit...  a sim might tell that...  

Here, I get n ~ 58% typically and n~ 70% best run.

Conclusion:  interesting, not definitively OU in this configuration -- still would like a better way to measure Eout or Poutput routinely so I can tune caps and resistors -- and the wound-toroid -- to maximize Pout/Pin = Eout/Ein.

Thanks professor for sharing those measurements.

So with a 10mF capacitor in place of R0, the voltage rises to 1.58V from 1.30V.
For a red LED, the forward voltage drop is about 1.67V (I measured mine with a DMM, please replace it with your own measurement).
Therefore, the power spent on the LED is: C * deltaV * Vdiode = 10mF * (1.58 - 1.30) * 1.67 =  4.676mJ.
The total energy stored in the cap is  (1.58**2 - 1.30**2) * 10mF /2 = 4.03
So, total output is: 4.676+4.03 = 8.706mJ.

Total input: (2.54**2 - 2.25**2) * 10mF/2 = 6.9455

Therefore n = 8.706 / 6.9455 = 1.253

Hey, that's more like an OU . But clearly it needs an accurate measurement of your Vdiode.

P.S.: my calculation shows that as long as the Vdiode > 1.04125V, we will have n>1.

[lanenal]

Best result tonight:
n ~ (1.366**2 - 1.258**2) / (2.0**2 - 1.89**2)  = 0.7 = 70% (conservative)

For this best case, if Vdiode > (1.366V+1.258V)/2 = 1.312V, then n > 140%.
Vdiode is the voltage drop over the diode at the moment when the capacitor is being charged.

For this best case, if Vdiode > 0.667, n>1.
Of course, this is still a conservative computation of n, as we have not calculated power wastes in the transistor and toroid.

I wrote this Python function to compute n and vpar. see below:


def computen(vi0, vi1, vo0, vo1, vd, ci, co=None):
   '''compute the n = Eout/Ein, where
vi0 = initial voltage of input capacitor,
vi1 = final voltage of input capacitor,
vo0 = initial voltage of output capacitor,
vo1 = final voltage of output capacitor,
vd  = forward voltage drop over the diode.
ci = capacitance of input capacitor,
co = capacitance of output capacitor,
   it defaults to None, meaning co = ci.

Ein = (vi0**2 - vi1**2)*ci/2,
Eout= (vo1**2 - vo0**2)*co/2 + vd*(vo1-vo0)*co

This function also computes the critical vpar
such that when vd = vpar, n = 1.
'''
   if co is None: co = ci #default
   Ein = (vi0**2 - vi1**2)*ci/2.
   Eoc = (vo1**2 - vo0**2)*co/2.
   Eout =  Eoc + vd*(vo1-vo0)*co
   vpar = (Ein - Eoc)/(vo1-vo0)/co
   return Eout/Ein, vpar

[lanenal]

I further modified my simulation, now the voltage power turns off (drops to zero volts) after 2 ms. See below the simulation of the first 10 ms. It clearly runs on the charged capacitor for the last 8 ms. V(vtop) - V(vbot) is the voltage difference over the voltage power source. I(L1) and I(C1) are the currents in L1 and C1.

[PhysicsProf]

  Good points, laneal.    Here I attach a photo of the set-up, using one cap for Ein and the second cap to capture some of the Eout.  Two DMM's read the voltage, one on each cap.
 
   Also a screen shot of the voltage across the LED, while powered by the Ein cap alone, with the second cap charging (from approx 0 volts, starting voltage).

The measurements read:
Vmean 40 mV, which is approx what I read with the DMM across the LED...  but --
Vpp  4.84V
Vrms 680mV
Vtop 3.12V -- this is the voltage in the forward direction, the direction of the current flow allowed by the LED

So you tell me, you asked for the voltage across the LED -- but which voltage does one use??

[lanenal]

 Good points, laneal.    Here I attach a photo of the set-up, using one cap for Ein and the second cap to capture some of the Eout.  Two DMM's read the voltage, one on each cap.
 
  Also a screen shot of the voltage across the LED, while powered by the Ein cap alone, with the second cap charging (from approx 0 volts, starting voltage).

The measurements read:
Vmean 40 mV, which is approx what I read with the DMM across the LED...  but --
Vpp  4.84V
Vrms 680mV
Vtop 3.12V -- this is the voltage in the forward direction, the direction of the current flow allowed by the LED

So you tell me, you asked for the voltage across the LED -- but which voltage does one use??

Professor, thanks for the diagrams. Conceptually it should be V4 - V3 as labeled in your schematic 2Cap13Jun2011.jpg. However, there is a simple way to get a good estimation of it, see below (quoted from http://www.repairfaq.org/sam/semitest.htm#sttdj):

On a (digital) DMM, there will usually be a diode test mode. Using this, a silicon diode should read between .5 to .8 V in the forward direction and open in reverse. For a germanium diode, it will be lower, perhaps .2 to .4 V or so in the forward direction. Using the normal resistance ranges - any of them - will usually show open for any semiconductor junction since the meter does not apply enough voltage to reach the value of the forward drop. Note, however, that a defective diode may indeed indicate a resistance lower than infinity especially on the highest ohms range. So, any reading of this sort would be an indication of a bad device but the opposite is not guaranteed.


Therefore, you may also replace the LED with a germanium diode and probably get better n (without calculating the energy wasted on the LED).

Another way to get better n is this: let the output capacitor having a higher initial voltage relative to the diode's forward voltage.

lanenal

[poynt99]

I'd try to charge a capacitor. I've got some 10000uF caps, hopefully will do. Any suggestions?

For a possible Pout measurement, we can try using a single diode or FWBR output to charge a capacitor. Then connect a potentiometer across this output cap and monitor the DC output with a DMM. Slowly decrease the resistance of the potentiometer until the DC output voltage begins to drop steadily. Back it off until the output voltage holds steadily. Now use the output voltage and resistance reading on the pot to compute power. I would suggest somewhere between a 10k to 50k pot.

.99

[poynt99]

Professor, Lane,

I have tried my own sim based on the values shown in schema01.png. I used a modeled non-linear core of 1" diameter and 9 turns each coil.

Fo is approximately 1.68MHz. Correction, the Fo is about 174kHz. It was about 1.68MHz before I added the LED and RO to the output.

Is the wave form shown, taken off the emitter, similar to what you are seeing?

I've not done any power measurements yet, as I want to know if I'm close or not first.

Regards,
.99

Edit: Corrected Fo frequency.

[PhysicsProf]

I like this idea, .99:

"For a possible Pout measurement, we can try using a single diode or FWBR output to charge a capacitor. Then connect a potentiometer across this output cap and monitor the DC output with a DMM. Slowly decrease the resistance of the potentiometer until the DC output voltage begins to drop steadily. Back it off until the output voltage holds steadily. Now use the output voltage and resistance reading on the pot to compute power. I would suggest somewhere between a 10k to 50k pot.

We can know the Pinput from the cap/stop-watch tests and can then run off a battery at a steady Vin, and with your cap/poten. measurement for Poutput, we can keep conditions steady.  It concerned me that with variable voltages on Vin and on the output cap, one could not really "tune" the circuit.  This way is much better.  Thank you!

Professor, Lane,

I have tried my own sim based on the values shown in schema01.png. I used a modeled non-linear core of 1" diameter and 9 turns each coil.

Fo is approximately 1.68MHz. Correction, the Fo is about 174kHz. It was about 1.68MHz before I added the LED and RO to the output.

Is the wave form shown, taken off the emitter, similar to what you are seeing?

I've not done any power measurements yet, as I want to know if I'm close or not first.

Regards,
.99

Edit: Corrected Fo frequency.

Yes, the output waveform is similar to what I'm seeing, so pls carry on with power measurements.

I would like especially to see:

1.  Is the average input power small, as I'm observing with the actual DUT -- that is around 1 mW?

2.  What is the shape of the input power waveform?

These will provide (IMO) a good test of the SIM.