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Author Topic: Circuit sj1. Terse and Technical only.  (Read 167557 times)

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It's not as complicated as it may seem...
Professor, as the circuit stands, the various average power measurements are as follows:

PLED=0.31mW
PRO= 0.76mW
PBAT=-1.82mW

The Battery power wave form is attached.

.99


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"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
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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.
...

.99, great that you have been able to do the non-linearity sim, it would definitely contribute our understanding. thank you.

professor, I wonder if you could also try (with a battery or a cap at the input -- or both, doesn't matter) to measure Pout with that nice scope when a diode and a cap is joined at the output. There are two ways to estimate Pout:

1. Only measure the output energy dumped into the cap
( use math: PoutCap = MEAN[V4*Iout] ).

2. Measure the output energy dumped into the cap and that wasted on the LED/diode
(use math: PoutBoth = MEAN[V3*Iout] ).

Here Iout is the output current through the diode/cap, which can be measured by a CSR as you have done many times before.

In the above measurements, the ground is at the joint of the cap and L1.

The purpose of this experiment is to verify:

PoutCap/PoutLED = (Vc1 + Vc0)/(2*Vdiode)

where

PoutLED = PoutBoth - PoutCap
Vc1: final voltage over the output cap.
Vc0: initial voltage over the output cap.
Vdiode: forward voltage drop over the diode.

If that can be verified, we can greatly simplify our calculations without using a scope later on.

lanenal
   

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It's not as complicated as it may seem...
Professor,

Have you tried changing the polarity of the LED? What differences do you see?

When I reverse the polarity in the sim, the frequency goes back up to 1.68MHz or so, and there is more dissipation and light power in the LED.

.99


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
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Professor,

Have you tried changing the polarity of the LED? What differences do you see?

When I reverse the polarity in the sim, the frequency goes back up to 1.68MHz or so, and there is more dissipation and light power in the LED.

.99

Yes, reversing the LED results in larger Pinput, that is, more dissipation as you say.

Here are the conditions and results for the latest low-power-consumption run:

I used the Cap/stopwatch method.
 P = Ein/time, measuring Ein using a cap, from 2.55 V to 1.5 V, so around 2Volts in.
Best result (to date):  12.7 seconds using a 1000 uF cap for Piin, so

P = 1/2 10mF (2.55**2 - 1.5**2)/12.7s

= 0.17 mW = 170 uW.


That is, "170 microwatts".

Frequency -- power to output leg in spikes, f = 1.44Khz (not MHz) at 2.5 V, and at 1.5V, f = 1.31KHz.

Conditions: Rb = 47 Kohms
Cb = 223 pF (ceramic cap)
MPS 2222 transistor
Ro =  220 ohms
Lo ~ Lb ~ 130 uH
No CSRout or Co or Rr or CSRin.
Red LED lit visibly in lighted room.

With LED reversed, time for 1mF cap to drop from 2.55V to 1.5V is 11.9s, so 180 uW.

   
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My JT could put out 4uW if the math is right.   
7/1000 volts every 10 seconds if I remember correctly.
4700uF 1.3 volts start
Frequency function broke

I also shunt battery through several MOhm resistors in series with LEDs to study input.  Seems like they are close in current.
   
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My JT could put out 4uW if the math is right.   
7/1000 volts every 10 seconds if I remember correctly.
4700uF 1.3 volts start
Frequency function broke

I also shunt battery through several MOhm resistors in series with LEDs to study input.  Seems like they are close in current.

I'm not sure I understand you here, Gibbs.   Are you saying you have a JT that will light up an LED visibly with only 4uW input power?
(Please explain)
   

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It's not as complicated as it may seem...
Professor,

Is there a specific requirement to have a diode or LED in the output?

Could a resistor only be there for determining Pout?

.99


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
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I'm not sure I understand you here, Gibbs.   Are you saying you have a JT that will light up an LED visibly with only 4uW input power?
(Please explain)

hehehe I'm sorry Professor,

I meant to say the input but I think it could be mistaken with "put out".

I have to go now but I'll provide the schematic when get back.  But like you said, the LED is tuned to just visible condition to achieve this status.  I'm working on increasing lighting condition with little input as possible.  
   
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Alright, back

A few comments about my JT experiment

This design is basically derived from the Professor's JT.  I used a PNP.  It was 3906.  The assorted bag came in 3 types I think, but I never tried the other two types.  I know the NPN types make a different.  The resistor was 1MOhm pot tuned to max.  I tried to add a couple more MOhms but it makes little difference.  The LED is barely visible.  I can see the red LED but I like pink LED better.  There are also 2 positions one can place the LED.  It makes little difference in power consumption if detectable.  I like the emitter-collector position better because it gets a voltage boost from the source.   The Cap and the bifilar can have a good range under similar operation.  I used a .01uF metalized film and forgot what coil I used since I had so many.

WaRnINg,

I've seen the capacitor/resistor combination gives out a pulse once a full moon so I can only give warranty that it does not operates at 1-5Hz .  No refund if you find it operates at 10 or 60Hz. lol 
   
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Professor,

Is there a specific requirement to have a diode or LED in the output?

Could a resistor only be there for determining Pout?

.99

How do you determine Pout for an "erratic" AC output with just a resistor ?  Unless you are using a Thermal Wattmeter or equivalent -- I'm still trying to hunt down one of those.  

In any case, no I would not make a "specific requirement to have a diode or LED in the output" if you're going to measure Pout.  The LED in the output was for the cap/stopwatch test for Pinput, with the LED glowing... something any one can do without a scope.

CORRECTION:  to measure Pout properly, without drastically altering the DUT, the diode in the output leg of the circuit must be present. 
« Last Edit: 2011-06-18, 06:12:08 by PhysicsProf »
   

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It's not as complicated as it may seem...
How do you determine Pout for an "erratic" AC output with just a resistor ?
Quite simply! (I'll get to that shortly)

How does adding a diode make it any more possible?

.99


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
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Alright, back

A few comments about my JT experiment

This design is basically derived from the Professor's JT.  I used a PNP.  It was 3906.  The assorted bag came in 3 types I think, but I never tried the other two types.  I know the NPN types make a different.  The resistor was 1MOhm pot tuned to max.  I tried to add a couple more MOhms but it makes little difference.  The LED is barely visible.  I can see the red LED but I like pink LED better.  There are also 2 positions one can place the LED.  It makes little difference in power consumption if detectable.  I like the emitter-collector position better because it gets a voltage boost from the source.   The Cap and the bifilar can have a good range under similar operation.  I used a .01uF metalized film and forgot what coil I used since I had so many.

WaRnINg,

I've seen the capacitor/resistor combination gives out a pulse once a full moon so I can only give warranty that it does not operates at 1-5Hz .  No refund if you find it operates at 10 or 60Hz. lol 

 Right, 60Hz operation would imply "poaching energy" from the grid, not what we're looking for ...  lol.  

Can you put that circuit or one like it back together, and:
1.  Measure Pinput by the Cap/Time method described above?

Here are the straightforward equations:
Ecap = 1/2 C V**2
so
Pinput = 1/2 C (Vstart**2 - Vstop**2)/time

2.  Measure the frequency?
Yeah, we better check that it's not 60 Hz.

Also, somehow we need to quantify a MINIMUM brightness for the LED to make useful comparisons between circuits...  
I have another idea -- experimenter can replace his LED with a known diode, 1N4148 which is common, and then use the stop-watch/Cap method.

@.99 -- Look, I'm trying to get at ways to compare INPUT power for various circuits -- and I'm recommending a STANDARD using the cap/stop-watch method.

For Pout, adding a diode is not the answer...  a Thermal Wattmeter may be a very good answer, as I've noted before.
   
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I use the LED to detect if my circuit is on.  I also use it as a rule of thumb for energy output since the amount of light is proportional to energy (just a rule of thump for nonscopers).  I also use it to charge up a capacitor to quantify somewhat of the output. 

Professor,

I think my last results holds.  .007V/10 seconds.  I also see a person name xee on OU that shares the same circuit in NPN configuration.  His stated power was 10uW so I think we do have something in common.  My frequency function broke.

I tried to charge up an identical 4700uF cap and the input:output is 3:1 ratio (every 3 unit volt input I recover 1 unit volt ouput).  I believe the rate is constant though.  Also the LED stop when I charge the recover cap to about 2.4V from outside source.  This indicate my peak voltage is below 4-5V.  The diode also leaks. 

I have put all this behind and moved on.  Now I have roughly the same input with 2 LEDs lit.  LOL  that means my peak voltage is double?  LOL how could this be? Of course there is a trick to this. hehehe
   

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It's not as complicated as it may seem...
Also, somehow we need to quantify a MINIMUM brightness for the LED to make useful comparisons between circuits...
Yes, that's why I made that post at OU.

Quote
I have another idea -- experimenter can replace his LED with a known diode, 1N4148 which is common, and then use the stop-watch/Cap method.
This does not solve the problem of having a baseline for comparison. What criteria would there now be for the circuit operating?

Quote
@.99 -- Look, I'm trying to get at ways to compare INPUT power for various circuits -- and I'm recommending a STANDARD using the cap/stop-watch method.
I'm trying to help you in both regards; Pin and Pout. No need to get aggravated.

Quote
For Pout, adding a diode is not the answer...  
You misread me. I was not trying to suggest that. I was trying to suggest the contrary actually.

.99


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
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I use the LED to detect if my circuit is on.  I also use it as a rule of thumb for energy output since the amount of light is proportional to energy (just a rule of thump for nonscopers).  I also use it to charge up a capacitor to quantify somewhat of the output.  

Professor,

I think my last results holds.  .007V/10 seconds.  I also see a person name xee on OU that shares the same circuit in NPN configuration.  His stated power was 10uW so I think we do have something in common.  My frequency function broke.

I tried to charge up an identical 4700uF cap and the input:output is 3:1 ratio (every 3 unit volt input I recover 1 unit volt ouput).  I believe the rate is constant though.  Also the LED stop when I charge the recover cap to about 2.4V from outside source.  This indicate my peak voltage is below 4-5V.  The diode also leaks.  

I have put all this behind and moved on.  Now I have roughly the same input with 2 LEDs lit.  LOL  that means my peak voltage is double?  LOL how could this be? Of course there is a trick to this. hehehe

Hmm...  I'm asking both you and Xee2 to use this STANDARD cap/time method if you would.
1.  Measure Pinput by the Cap/Time method described above?

Here are the straightforward equations:
Ecap = 1/2 C V**2
so
Pinput = 1/2 C (Vstart**2 - Vstop**2)/time


But I can see that a minimum frequency might also be needed for the test... see below.

Some of you may not realize this discussion is going on in parallel over at OU.com -- and to understand the dialog, you may want to read there also:
http://www.overunity.com/index.php?topic=10773.330

My last reply:
Quote
This does not solve the problem of having a baseline for comparison. What criteria would there now be for the circuit operating?

With your current criteria, I am certain I can beat everyone. I can build a circuit that will drop from 2.5V to 2.4V in 10 seconds.....do I win?

.99

Required is a JT-type circuit, with at least one transistor and one bifilar-wound toroid and it must light an LED to observable brightness in a lighted room (all as I said before IIRC) before the final test with a 1N4148 replacing the LED.

And the winner will check the FREQUENCY also...  
I'm going to add a provisional stipulation that the frequency of input-power pulses be above 200 Hz, to prevent pulsing "once in a blue moon" and also to avoid 60-Hz pick-up.  If you can persuade me this is not a "fair" stipulation, I'm listening.

The winning device cannot be poaching from the grid.  

How's that?  and I do appreciate your checking the criteria, .99.  Rigor without rancor.

@.99 -- do you have something new for measuring Pout?  I'm very interested if you've come with something better than the Thermal Wattmeter...

   
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I don't realized that there is a contest going on here.  I'm not entering the contest but my interest is to compares input and how much light one can get. 

1.3V - 1.293V on 4700uF in 10 seconds

Pin = 1/2 * .0047 *(1.3*1.3 - 1.293*1.293)/ 10sec

Pin = .000004267 W
Pin = 4.26 uW

I'll check on the frequency later.

There are also an old video I see that related.

http://www.youtube.com/user/jonnydavro#p/u/14/tQecrP5ZPYQ

@2:15 the voltage reading is 14.03
@2:20 the voltage reading is 13.87
2200uF
joule ringer
frequency-unknown

Pin=1/2 * .0022(14.03*14.03 - 13.87-13.87)/5 sec
Pin = .000982 W
Pin = 982 uW

@4:20 6.56V
@4:32 6.27V

Pin = 1/2 *.0022(6.56*6.56 - 6.27*6.27)/12 sec
Pin = .000341 W
Pin = 341uW
   
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... my interest is to compares input and how much light one can get. 

1.3V - 1.293V on 4700uF in 10 seconds

Pin = 1/2 * .0047 *(1.3*1.3 - 1.293*1.293)/ 10sec

Pin = .000004267 W
Pin = 4.26 uW

I'll check on the frequency later.

Remarkable, Gibbs!  is there a video of this?  I agree with your goal of comparing Pin and light output.

  I don't suppose you could dig up (or rebuild) the device and get similar results with a 1N4148 diode replacing the LED, and taking the measurement at 2 volts?  so we can have a standard for comparisons with other circuits ...

   
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Remarkable, Gibbs!  is there a video of this?  I agree with your goal of comparing Pin and light output.

  I don't suppose you could dig up (or rebuild) the device and get similar results with a 1N4148 diode replacing the LED, and taking the measurement at 2 volts?  so we can have a standard for comparisons with other circuits ...



No video but sure I'll take a note of the 1N4148 and 2 volts.

There is something unusual about the video I posted.  The meter next to the voltage read .42 mA @ 13.87V and .57mA @ 6.27V
13.87 * .42 = .0058 = 5.8mW    Cap method is = .982mW
6.27 * .57 =  .00357 = 3.57mW  Cap method is = .341mW

Ah... I have to think about this.
   
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No video but sure I'll take a note of the 1N4148 and 2 volts.
Wow, did you just say 1N4148? That was exactly the diode I used in my simulation. See reply #125 on page 6 of this thread.
   
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Thanks, Gibbs -- pls get back to us on this.

Wow, did you just say 1N4148? That was exactly the diode I used in my simulation. See reply #125 on page 6 of this thread.

Yes, I see that.  I take it your frequency there was approx 1/1.8us ~ 550KHz, is that right?   Do you have a measure of the Pinput from this simulation?
   
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  The discussion on the contest is continuing with some enthusiasm at OU.com, so I will focus on that, at this time.
e.g., from nul-pts:  http://www.overunity.com/index.php?topic=10773.msg291497#msg291497
Quote
i've been running my SJ1 variant tests from a single  (depleted) AAA NiMH, and the voltage has been around the 1.12-1.15V range

just for fun i constructed another unit to add to the comparative results you're getting back from folks

the circuit is a minor mod to one of my earlier tests with a variant of your SJ1 circuit (see schematic below for this test circuit)

C1 0.022uF
D1 1N5187 (schottky)
Q1 2N3906
T1 approx 50:50:100; 0.45mm wire; tri-toroid (ferrite)
C2 1000uF (nominal)
L1 approx 0.5mH
LED1 6mm(?) HiBrite (visible, but not bright)

i'm using a tertiary winding to decouple the AC o/p from the DC operating conditions of the oscillator


this circuit takes 433 seconds (7min 13sec) to discharge a nominal 1000uF cap from 2.24V to 1.5V

C2 1000uF (nominal)
2.24V => 2.509mJ
1.50V => 1.125mJ
         -------
    Ein: 1.384mJ

Pav: 1.384/433 = 3.2uW


this is all just ballbark at the moment, obviously
- to be more accurate, the 1000uF cap would need measuring...  " etc.

This is close to the number you mentioned, Gibbs... y'all may wish to follow the discussion there at OU.
   
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Yes, I see that.  I take it your frequency there was approx 1/1.8us ~ 550KHz, is that right?   Do you have a measure of the Pinput from this simulation?

I looked into the way of doing it in LTSpice, and it can do RMS and MEAN calculations over many cycles. I found that input is 955uW and output is 213uW. Not surprising here. These numbers shows us that a lot of power is spent at the transistor/coil/core/diode. Therefore when we get a n=50% when only considering the power stored in a cap at the output, that is a strong indicator of ou. As I have shown in the other posts, including the power dissipation on the diode might be enough to show ou.
   
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  The discussion on the contest is continuing with some enthusiasm at OU.com, so I will focus on that, at this time.
e.g., from nul-pts:  http://www.overunity.com/index.php?topic=10773.msg291497#msg291497
This is close to the number you mentioned, Gibbs... y'all may wish to follow the discussion there at OU.


Yes, I just read that.  I am very impressed.  He beat me but not just by a little bit, a whole lot!  I have to put my hands around my LED to see it and the worst case, turned off the light lol.  I'm also glad we coming to a standard agreement.  Just shake it around until everyone feel generous to give in. 

Wow, did you just say 1N4148? That was exactly the diode I used in my simulation. See reply #125 on page 6 of this thread.

Lane,

I was very impressed with your looping technique.  I had a lot of fun trying to looped the JT past weeks with your technique.  This is why I know COP>1 is real, but it won't be in the form of charge.  One can see if the circuit reached COP>1 easily with both cap/stopwatch and dual DMM method! The cap/stopwatch method is always be less than or equal to Dual DMM method.  The energy consuming from the capacitor is a real component, there is no if, and, or but about it.  The dual DMM method is not totally invalid.  We can use it to determine the phase angle and calculate the ratio of heat&radiation: energy dissipate from cap.  It is now important to get this thing in a chamber where not only heat, but radiation can be efficiently captured. 
   
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I think we should reconsider the condition for this contest.  Here is my argument.  

What if a guy with a circuit COP 5 that has an input of 5 watts vs a circuit with COP .5 with 1 watt input?  The only way we can tell is to compare input to output and it's not an easy task up to now.  I think the quality of a circuit should depends on its COP and  not input or output. But how?

I propose that we should want the quality of the circuit in relation to how much it is out of phase.  As I mention earlier about the cap/watch method and the dual DMM method.  The cap/watch method is a sure way to know how much we put in.  But what does the dual DMM method means?  Here is what I see:

Consider the attached image below.  With the dual DMM method when voltage and current are in phase, it should multiply the Average of the voltage and the average of the current.  This is the power consumption and should agree with the cap/watch method.  Now how about an out of phase scenario?  The average of the voltage and current are exactly the same as the in phase condition and thus the dual DMM method gives the same measurement.  But we know this is incorrect if there is an out of phase condition, thus by comparing the real input from the cap to the dual DMM method, we have a mean to know how much the phase angle is off.  The quality of the circuit can be express as the ratio of the two.  
   

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It's not as complicated as it may seem...
Gibbs,

If your input source is DC, the phase angle between the input voltage and current is always 0ยบ.

Aside from the very low net input currents involved, and the potential difficulty in getting a nice clean average reading of it, the INPUT power measurement with a non-inductive CSR and single (or dual) DMM is a snap.

.99


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
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