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Inductive Resistor Open source development of highly efficient inductive resistor circuits.

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  #121  
Old 10-12-2010, 07:29 PM
gadh gadh is offline
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i assume you meant JT = Joule Thief, right ? if so, i do not see why not, although i'm not dealing here with it but with Ainslie circuit, and there are differences (and we're trying to get OU here...)

Harvey, let me add that i built a small farady cage with grounding (to the 555 circuit gnd, hope this is sufficient) and still i notice an immediate rise in the temp. - while the right ambient temp of the water is 25.4 deg., the probe shows me 27.4, but then the temp rise rate is ok, so i'll use it for now, unless you have a better idea.
BTW, the faraday cage is assmbled from a plastic tube wrapped with alum. foil, the gnd connected to the foil. the tube's diameter is about 7mm, so it can block wavelength>1ghz. am i right ?

My other problem is that i cannot get more than 150v peak on the load resistor (with a single 12v battery). while in the same conditions i got before almost 400v peak. what can cause this ? maybe the MOSFET is a bit damaged ?
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  #122  
Old 10-12-2010, 08:26 PM
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Hi Gad,

If you Google Thermocouple RFI you will find a host of information regarding the interactions between RF energy and thermocouples. One book even illustrates how thermocouples are used to detect RF power because the RF heats up the metals.

It is important that your Faraday Cage be designed to absorb the RF energy the resistor is transmitting. The size, shape and material of the cage all play a role in that. Also, note that RF can travel to your meter itself and cause problems too.

Another thing to consider is that the water is a dielectric. This means that you have a very complex multi-node parallel capacitor there between the individual resistor coils and the thermocouple and at certain frequencies that energy can be passed right through the water like a wire even though the DC resistance is high.

You can do a simple test of removing the resistor from the water and testing the system with the resistor in the air about the same distance from the thermocouple as it was under water. If you still get the odd results, then it is most likely RFI. But if the problem goes away, then you should look at AC conduction or even ultrasonic heating. If it is the latter, then it may be that the thermocouple is sensitive to those frequencies and heats up faster than the water. While submersible ultrasonic sensors are available, they are generally narrow band, so using them to detect whether or not your resistor is emanating certain frequencies may be problematic. Designing a wideband submersible ultrasonic detector may be a worthwhile project. This Wideband Sensor doesn't say if it is submersible, but it does have a range of 10kHz to 65kHz.

One of the major problems with this specific circuit is the aperiodic nature - it literally dumps emissions all over the spectrum due to the frequency range and various harmonics. Just have a look at all the screen shots from Glen's many experiments and notice the frequency readings on the scope.

Hopefully the problem is just RFI and the Faraday Cage will help keep that away from the thermocouple.

As far as details go: Thermocouple MFR and model, Wire Lead type and length, Resistor and thermocouple orientation and proximity, Meter Model, Measured Frequency and amplitude, Measured Resistor values like inductance and resistance as well as size, Calorimeter tank size and volume (for sonic resonance evaluation) or any other things that you can think of that may influence the possible RF or UltraSonic effects.

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  #123  
Old 10-12-2010, 08:41 PM
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Originally Posted by gadh View Post
i assume you meant JT = Joule Thief, right ? if so, i do not see why not, although i'm not dealing here with it but with Ainslie circuit, and there are differences (and we're trying to get OU here...)

Harvey, let me add that i built a small farady cage with grounding (to the 555 circuit gnd, hope this is sufficient) and still i notice an immediate rise in the temp. - while the right ambient temp of the water is 25.4 deg., the probe shows me 27.4, but then the temp rise rate is ok, so i'll use it for now, unless you have a better idea.
BTW, the faraday cage is assmbled from a plastic tube wrapped with alum. foil, the gnd connected to the foil. the tube's diameter is about 7mm, so it can block wavelength>1ghz. am i right ?

My other problem is that i cannot get more than 150v peak on the load resistor (with a single 12v battery). while in the same conditions i got before almost 400v peak. what can cause this ? maybe the MOSFET is a bit damaged ?
Hi Gad,

Looks like we cross posted - As regards the voltage, it could be that something is loading the magnetic field so that not all of the energy is able to return on the BEMF spike. In other words, you may be inductively coupled to the Faraday Cage I hate these types of interactive problems where the RFI shielding changes the desired operation of the circuit.

I have to leave the office for about 5 hours, but I will check back after that.

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  #124  
Old 10-13-2010, 03:56 AM
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thanks Harvey.
i'll try first to shield also the whole water tank (in addition to shielding the probe that i'm using now), since i think the other solutions will be more complicated to achieve.
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  #125  
Old 10-13-2010, 05:40 AM
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thanks Harvey.
i'll try first to shield also the whole water tank (in addition to shielding the probe that i'm using now), since i think the other solutions will be more complicated to achieve.
-

If the source of the RF is the Resistor, and if the resistor is in the water, then shielding the tank serves little purpose. The principle of the Faraday Cage is that an electric field cannot exist inside the cage volume when the electric field source is external to the cage - but placing the electric field source inside the volume defeats the purpose of the cage.

It would be better to shield the meter and the probe instead - so your half way there.

If you still get an odd jump in reading, try lifting the resistor out of the water and see if you still get the same odd jump. If you do, then the shielding is probably not doing it's job. If you don't, then its time to evaluate ultrasonic heating or conductive AC displacement currents in the water.

My first guess is RF, then Ultrasonic, then last AC dielectric pass-through.

Also, I would like to ensure that you are using the correct temperature / voltage chart for your specific thermocouple profile. I know its implied that you have taken care in that regard, but I wouldn't be much help if I overlooked the necessity and sent you off on all those other things without mentioning that sometimes thermocouples don't match the profile being used. I believe you stated that you verified the proper function of the thermocouple in your DC test so I didn't bother mentioning this before - but just in case I misunderstood something I thought I would bring it up.

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  #126  
Old 10-13-2010, 06:42 AM
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regarding the cage - i'll try shielding the probe and meter first.
regarding the probe type - i tried measuring DC heat with another meter and probe (less accurate), i'll now will repeat the DC measurement with the same meter (the better one of course.
hope to get results in a few days
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  #127  
Old 10-13-2010, 06:21 PM
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Hi Gad,

Looks like we cross posted - As regards the voltage, it could be that something is loading the magnetic field so that not all of the energy is able to return on the BEMF spike. In other words, you may be inductively coupled to the Faraday Cage I hate these types of interactive problems where the RFI shielding changes the desired operation of the circuit.

I have to leave the office for about 5 hours, but I will check back after that.

hi Harvey. regarding the voltage - the low voltage spikes occured also in air, and also without faraday cage, so it must be another cause. any new ideas ?

another issue is the energy consumption. when trying to heat the water in a simple DC circuit, i measured that in order to heat 650cc to 5 degrees (celcius) over ambient , it took 72 minutes in a current of 800ma and batt. voltage of about 12v. so the actual battery energy consumed was ~40k joules. the load over the resistor (used for heat) was ~7 watts, so about 30k joules were directly used to heat the water. but the actual energy needed is only 13650 joules by the std. equation ! so where did all the wasted energy go ? and how can i predict exactly how much energy is needed to actually heat the water ?
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  #128  
Old 10-14-2010, 08:39 AM
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Hi Gad,

Is your tank insulated? It sounds as though you are losing energy by conduction to the surrounding environment. Trapped air is a good insulator, so if your tank is not insulated, you can place it on top of some styrene foam inside of another larger tank and trap the air between the two with some styrene foam.

Also, evaporation is a big culprit to stealing energy from a calorimeter so the smaller the water surface area is the better.

These are two of the reasons I have suggested a thermos bottle as a tank for us DIY guys.

Some of the energy can be lost to other energy forms such as sound and RF also. Think of a light bulb, only a portion of the energy is transferred as heat and IR, the rest is transferred as other EM frequencies. The resistor may act similarly for but for RF and ultrasonic frequencies. And when you get the AC going in it, then you may get that audible 'singing' and actually see the moving coils from the magnetic interactions.

As regards the calculations, a 20 Calorie is that energy required to raise 1 gram weight of air-free water from 19.5 C to 20.5 C and is equivalent to 4.182 J (see Calorie - Wikipedia, the free encyclopedia) If you chill your water to 4 C and then carefully measure it into your 650cc tank at that temperature then you can be confident that you have 1g / cc or 650g of water. Then if you raise that to exactly 19.5 C and stabilize it at that temperature, then you are ready for your test. From there, you will activate your heater and time how long it takes to reach 20.5 C. You know it will take 2,718.3 J to do that. Let us suppose it takes 864 seconds to do that (72 / 5 * 60) then we get 3.146 Avg. Watts (2718.3 / 864 = 3.146) If your wondering what the 72 / 5 * 60 is, that is from your 72 minutes, divided by 5 because we only want 1 change, not 5 for 14.4 minutes and since a Joule is a watt-second we multiply by 60 to get the seconds. These calcs are just a general approximation, you will need to do the actual test and verify the results.

Also, we don't go past 20.5 C because the energy curve changes above that and introduces a larger margin of error. And too, the accuracy of your measurements is important to establish your error margins. For example, if the thermocouple is only accurate to +/- 0.25 C then you have a built in error margin of 50% of the total measurement range of 1 C.

You say it is 13.6 kJ in your calcs? Can you share where you got that, perhaps I made a mistake somewhere ?

Cheers,

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  #129  
Old 10-14-2010, 10:17 AM
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Hi Harvey.
i use 2 tanks - the inner is filled with water and the outer contains the inner, and there are 2 inches of air in between. i think that for 5 degrees above ambient its enough insulation for now. (the tanks are both sealed of course ! i'm not stupid...)

my calc. is as follows: 5 degs. X 4.2 joules per degree X 650cc = 13650 joules to heat the water = 13KJ
my total energy drawn from the battery was:
1 - only at load resistor - 7.26 watts X 4320 sec (72min) = 31KJ
(i used a simple DC circuit, in which the total current was ~800ma and voltage about 12v but there were more resistors used in order to stabilize the current so i measured it exactly by multimeter+excel and its 7.26watts)

2 - overall energy - 9.96 watts X 4320 sec (72min) = 43KJ

so i ask - why is the huge difference between 31KJ and 13KJ ?
some friend told me there is more energy needed to heat the load resistor itself and its surface (made of okolon - i think its a type of polystyrene). is he right ?

about the 20.5 degs. limit - i searched on the net and tried to use several online data sources + calculators, but none of them indicated that the heat calc. equation is changed so much in relation to the amb. temp.. only if you use temp > 100 deg. celsius the calc. is different, otherwise its simple as i showed here.
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  #130  
Old 10-14-2010, 08:35 PM
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Originally Posted by Gad
5 degs. X 4.2 joules per degree X 650cc = 13650 joules to heat the water = 13KJ
D'oh! Of course

Okolon?

Okolon is a CSPE Chlorosulfonated Polyethylene compound used as an electrical insulator around wires like this:

http://www.acewireco.com/pdfs/ok42.pdf

I thought you used Teflon

Either way, your friend is right that some energy must be used to heat the solid material and it probably does not heat up at the same rate as the water so you will need to take that into account for the volumes involved. Also, this means you will need to allow the resistor itself to be stable at 19.5 C throughout before starting the test.

One way to avoid all that is to create ratio that buries the differential inside the error margins. If your water has a volume 1000 times that of the resistor, then the energy used to heat the resistor will probably fall three decimal places into you data and if your data is rounded to a single decimal place (Like 7.2W) you can figure that the resistor portion was down in the 0.00x area and of little concern in the final results.

And while we are thinking on it, what does it take to heat the inner tank 1 from 19.5 to 20.5? Does that value impact our data? I guess that depends on the material used and the thickness of that material (volume).

Whatever the problem is here, the indication is that 57% of the energy is going somewhere else in some form or another and by altering the volume ratios you may be able to see what it is. I would be interested to know if that trapped air space is changing in temperature over the duration of the test, just to see if it is not so trapped and internal convection is taking away some energy.

I think now you understand the importance of adhering to the 19.5 C - 20.5 C range. It is only as important as the quality of data you trying to get. There is a large difference in temperature between 4 water and 20 water but the Joule difference to raise either 1 is only 0.022 J. I notice that you rounded your data to 1 decimal place. So that gives you a +/- 72 J

Dissolved gases in the water could play a part in this but nowhere near 57%.

I think it may be time to use a known heat source like a store bought drink heater that you can plug into a watt meter and verify that your tank is working correctly for the tests. If we are unable to establish an acceptable calorimeter function then we will never be able to trust the results during the aperiodic AC testing phases. This process here, could expose a localized problem where the heat is not evenly distributed in the water. In other words, the 57% may be hidden in hot water zones that were never fully distributed to reach the thermometers. It could also tell us if your resistor is specialized in some way. For example, we know that an auto-lamp for cars will convert DC into electromagnetic visible light with nothing else needed to emit that EM radiation but a DC current. Similarly, it is reasonable to conclude that there are configurations of materials and energy that will act this way with RF EM radiation. Infra Red is only one part of the spectrum. (See Thermal radiation - Wikipedia, the free encyclopedia)

Isn't research fun?
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  #131  
Old 10-14-2010, 09:15 PM
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thanks Harvey for all your help in this research...
i think i'll focus on the idea that heating the Okolon uses the unknown energy part we talked about.
the resistor dimensions are:
length - 160mm
diameter - 33mm
its shape - cylinder, full
i'll think more on the drink heater solution, but the complexity of the water tank solution gets me to think on returning back to the simple heat measurement on the resistive wire itself, as ainslie did (using IR themometer).
what do you think ? how can we bring into this heat calc. all the env. params. including the heat loss to the air ? (assuming stable amb. temp. and almost no air flow in a room) is it a better solution ?
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  #132  
Old 10-15-2010, 04:10 PM
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I know it is tempting to take the easy way out and just take a bunch of spot measurements and average them out, but it is impossible to equate them to real energy values. This is especially true when using entirely different driving techniques and frequencies.


The calorimeter (a functioning calorimeter) is the best way to value the dissipated energy.

Likewise, a quality hydrometer is the best way to value the charge in a liquid acid battery (and thus the energy delivered)

When these two are compared to each other, we will get the best data for energy comparison of delivery and dissipation. Any OU will be obvious.

Multiple tests improve the data set giving the best average.

In my opinion, there is currently no other accurate method that can be easily applied to determine these things on this circuit.

A laborious process could be imposed whereby the resistor was profiled for each and every frequency within the operating range in some minute increments like 1Hz and a stationary IR reading taken on the same spot of the resistor in all 500,000 tests (or whatever frequency range was intended to be used) but even then, it could be argued that the pulse shape plays a huge role in the impedance and would need to be adhered to for a baseline. And then a Real Time Analyzer could be used to gather the data of a complete rundown process with no voids in the data. But then we would also need to match precisely in time the IR reading with the frequency in order to apply the baseline power values. It is very problematic.

Battery monitor IC's are available for monitoring DC power consumption, but there is no guarantee that they will function properly in an aperiodic environment.

It would be a very different story if extra energy were pouring out at the claimed values. If we had that type of performance, then the measurement techniques could be greatly relaxed. Extraordinary claims demand extraordinary proof - and that means precision with no room for mistakes. It is my experience that the scientific community will seize any hole and use it as evidence against the claims. For example, if we claim to have a COP > 4 but our data only represents 0.0000024 seconds out of a 1 hour test, they will ask us what was happening the rest of the time and demand continuous data. Glen gave us 5 hours of continuous visual demonstration in a LIVE feed that has been recorded for all to view - but it is difficult to equate the entire period to any energy values. The best we can do is approximate some averages and that is what we have done. But the big gaps in the data are sufficient for the scientific community to discard the tests entirely.

So that is why I feel that we either have to show the circuit running beyond the battery capacity, or we have to do a calorimeter / hydrometer comparison test. I think these are the simplest acceptable methods.

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  #133  
Old 10-15-2010, 04:16 PM
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Quote:
Originally Posted by gadh View Post
thanks Harvey for all your help in this research...
i think i'll focus on the idea that heating the Okolon uses the unknown energy part we talked about.
the resistor dimensions are:
length - 160mm
diameter - 33mm
its shape - cylinder, full
i'll think more on the drink heater solution, but the complexity of the water tank solution gets me to think on returning back to the simple heat measurement on the resistive wire itself, as ainslie did (using IR themometer).
what do you think ? how can we bring into this heat calc. all the env. params. including the heat loss to the air ? (assuming stable amb. temp. and almost no air flow in a room) is it a better solution ?
I think you'll find that Glen used the IR method and the South African group used thermistors or thermocouples.

Where did you get your Okolon? Do we have the thermal specs on that?

EDIT: As regards the IR measurement method the biggest drawback I see is determining the emissivity of the resistor and using a thermometer that takes that into consideration:
Accurate Noncontact Infrared Temperature Measurement - Raytek.com

Infrared Thermometers - Metal Emissivity Table - Raytek.com

Infrared Thermometers - Non-Metal Emissivity Table - Raytek.com
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  #134  
Old 10-16-2010, 09:06 PM
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regarding the measurement method - i think using a thermos + mercury themometer (0.5 deg. accuracy) will eliminate most of the problems I encountered before. i think i'll go for this method now.

about the battery drained energy - i have an accurate digital charger that shows exactly (after recharging the battery) how much energy it used in the last test (in MAH). i think its enough also for now, and of course if i see i'm on the right path - i'll drain the whole battery and test the MOSFET in "ainslie" mode compared to regular AC or DC circuit - these comparisons will be the most proven methods that we have OU. and the okolon/teflon heating is not so important for the proof since this kind of proof is comparative.

an example of an equivalent dc circuit i showed here before (with its energy consumption) - so if i see i got 5 deg. rise in 72 minutes, and i get the same time (roughly) for the same temperature rise for a DC circuit using 800ma for example - i have a real basis for comparison. (using roughly the same curve its better of course)

i do not think we have to be so accurate since we aim to get COP > 6 at least, as spoken before in this forum (I think Glen got it) so our margins of error are much bigger than we need.

i can tell you also that in my last test i think i got COP~6 since the battery voltage almost did not drop during the test and the battery recharging data confirmed that i was right in this calculation (while in a DC circuit i used much more energy to get the 5 deg. rise as i said). but i certainly have to do more testing since i cannot rely on the temp. probe because of the possible RF like you said.
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  #135  
Old 10-16-2010, 09:52 PM
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Using your battery recharge method (interesting approach BTW) along with the calorimeter is all that is needed if we can rely on the recharger. I know I have cell phones that say they are charged when they are not and only last a few minutes when used - so I don't trust those chips so much. But if you have confidence in your device then there is no need to do a DC 'baseline' because the recharge and calorimeter give you precisely what you are comparing, the battery supply to the thermal output in joules.

However, if you are wishing to provide an 'Efficiency' curve, where you pit a DC system against the Aperiodic system - then that is fine for the comparison. In that case you are saying the DC is some value and the Aperiodic is either more or less efficient than the DC system. This has nothing to do with a coefficient (mathematical number) of performance.

The Coefficient of Performance relates to two thermal reservoirs (or energy reservoirs if you prefer) where some value is applied to move energy from one to the other.

From:Coefficient of performance - Wikipedia, the free encyclopedia

So, let's suppose your battery shows a 1W recharge, while your thermal output shows 4W dissipated regardless of the method used to do the heating. This would indicate that you have released 3W energy from one reservoir and dissipated it into another while adding 1W of energy from your battery. Your delta Q is 4 (heat reservoir increases by 4W) and your delta W is 1 (the energy applied to do the work). Therefore, the COP = 4/1.

So you see that the COP has nothing to do with DC to Aperiodic comparison. It has to do with your source and your output.

In Glen's tests we did not use a calorimeter. The ONLY reason we did a DC baseline (which really bit us in the data) was to get some arbitrary referential between the power used and the heat produced. The process failed us for many reasons and we need a calorimeter test to help solidify the existing data. IOW, if the calorimeter agrees with the DC baseline we used, then we can rely on the data - otherwise it will need to be discarded and redone. And of course we would need to impose on Glen to do the calorimeter test, because he has the actual parts used in the previous data.

But in your case Gad, you have the advantage of doing your tests direct with the calorimeter so the data is accurate to begin with. And you have the added advantage of showing the differentials between DC, AC and Aperiodic methods of producing heat with the same resistor and same energy input.
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Old 10-16-2010, 10:53 PM
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Hi Gad,

It occurred to me also, that you could do a simple test to determine the specific heat of your resistor similar to these:

http://class.phys.psu.edu/250labs/Calorimetry.pdf

Cheers!
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Old 10-21-2010, 03:59 AM
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Hi everyone,

There was a very good posting at another web site on a mosfet High Frequency Induction Heating circuit by Richie Burnett . His published circuit diagrams are all missing the "mosfet" switching circuitry but that technology is somewhat similar to the 555 Mosfet Heating circuit diagram used in this thread. The combination of diagrams and possibly other design modifications to the metal object being heated enclosing it inside some type of borosilicate glass tubing or ceramic fixture may make a nifty steam generator or .....

High Frequency Induction Heating



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  #138  
Old 11-01-2010, 12:04 PM
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energy calculations in Ainslie circuit - conclusions

Hi all.
in my last 3 thorough experiments i measured:
1. the total *energy* consumed by heating 650cc of distilled water by my load resistor (Harvey - sorry, my mistake - its really made of Teflon...), insterted in an insulated plastic thermos. i used mercury thermometer (accurate by 0.2 degrees - celsius) since the digital thermometer was not stable because of the RF emmited by the resistor (i assume).

2. the total *energy* drained from the 12v batteries (i used 1 12v for the 555 circuit, 1 12v for the load resistor circuit), measured by a digital charger that shows the total MAH used for recharging the battery to 14v from its current state.

the waveforms were pretty good (see attached) in all these experiments.
the results were BAD though (no energy gain):
1. water heated to 10 degs. above ambient (24-34 celsius), with counting the energy needed to heat the load resistor too (C teflon=1.09, C water=4.186) and with calc. the energy needed to preserve the heat during the test (1 degs per hour) = overall 30 kilojoules (KJ) energy.

2. battery: 1200 MAH charged, 12.5 volts avg. during the test = 54KJ. even if we drop the part of the MOSFET heating , we reach 38KJ (although it breaks the logic behind the energy consumption which suppose to incl. total waist, not partially)
so my conclusion for now is that i do not see any gain here !
BTW, the 555 battery gives about 10% of the energy to the system ! (measured by charger also) - there is absolutely energy pass from the 555 to the load, though relatively small, but significat.

but i'm still not giving up, and i think of 2 possible ways to overcome this (for now):
1. use 24v for the load - maybe we have to pass some voltage threshold in order to get the effect. currently i get peaks of 170-200v with 1 battery, but with 2 i could get 400-700v.
2. maybe the waveform was not good enough, although it seems i cannot get better than i already got. i tried many configurations (adjusting the 3 potentiometers)

note that i emphasized *energy* and not power (watts), since energy is the right way to measure the possible gain in the system.

i would like your help in this process, so suggetions please...
Gad
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  #139  
Old 11-01-2010, 05:15 PM
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24V Good Idea

Hi Gad,

Thank you very much for posting your results. I was hoping for gain too

Your method of comparing energy is the correct method. You have confidence in your battery energy measurements, personally I would like to see that part supported with a hydrometer, but sometimes I can get a bit pedantic on stuff and your method may even be more accurate for all I know.

I am concerned that the thermos was plastic rather than glass. The plastic variety are known to leak heat more and I don't know if they have a good RF reflection in their coating like the glass ones do.

I think 24V is important in this particular configuration. Remember that the on resistance of the FET is about 2 Ohms. When we add that to the 10 ohms load you get 12 ohms and so you are limited to how much energy you can push into the inductor and have it push back. In my 3 part analysis earlier in this thread I showed that the bulk of the dissipated energy occurred from BEMF collapse because of the increased dv/dt in that waveform. Therefore, you want that large BEMF spike and the FET can handle up to 1kV there. So the increased voltage plays a big part there. At one point I think I posted that a 52V rail would optimize that aspect of the circuit for low frequency tests, but Glen proved that he could get 1kV spikes with just the 24V supply. Above 1kV the avalanche diode conducts and you lose energy in the system. In the very early thread there was some discussion as to whether or not that avalanche feature played a part in the aperiodic operation of the circuit, but it was later proven to not be a factor.

I am very interested in knowing more about that 555 battery and the 10% energy transfer. Are you saying that energy is being transferred to the load resistor or are you saying that 10% of the total energy was consumed in the 555 circuit?

What frequencies does your equipment show the circuit running at?

Are there any other places that heat can be lost in the circuit or voltage being dropped due to capacitive or inductive reactance?

Keep up the good work! As far as I can tell, you are the only person presently experimenting with this and publishing the results. I think Glen is hoping to get some means to record continuous data and fill in the gaps in the data. But I do think from a low cost perspective that the calorimeter is the best approach to quantifying the energy ratio.

Cheers!
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  #140  
Old 11-01-2010, 06:53 PM
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555

Quote:
Originally Posted by gadh View Post
BTW, the 555 battery gives about 10% of the energy to the system ! (measured by charger also) - there is absolutely energy pass from the 555 to the load, though relatively small, but significat.
10% sounds WAY too high even if your load side draw is low. You can also
drop the 555 draw to the bare minimum that you actually need. If you're
replicating the quantum article, you can use a separate 555 battery, but
if you're just wanting to experiment with the concepts, you can power
the 555 from the same load battery and measure the load of both
simultaneously.
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Old 11-01-2010, 08:09 PM
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Hi Harvey and many thanks for your kind support !

see my comments below:

I am concerned that the thermos was plastic rather than glass. The plastic variety are known to leak heat more and I don't know if they have a good RF reflection in their coating like the glass ones do.

GAD: i measured the heat loss to be as i wrote. the measurement was to see how many degrees i lose per hour after the test ends. this energy is added to the KJ energy calc.
what do i need RF reflection for ? i do not use digital thermometer anymore.

I think 24V is important in this particular configuration. Remember that the on resistance of the FET is about 2 Ohms. When we add that to the 10 ohms load you get 12 ohms and so you are limited to how much energy you can push into the inductor and have it push back. In my 3 part analysis earlier in this thread I showed that the bulk of the dissipated energy occurred from BEMF collapse because of the increased dv/dt in that waveform. Therefore, you want that large BEMF spike and the FET can handle up to 1kV there. So the increased voltage plays a big part there. At one point I think I posted that a 52V rail would optimize that aspect of the circuit for low frequency tests, but Glen proved that he could get 1kV spikes with just the 24V supply. Above 1kV the avalanche diode conducts and you lose energy in the system. In the very early thread there was some discussion as to whether or not that avalanche feature played a part in the aperiodic operation of the circuit, but it was later proven to not be a factor.

GAD: i'll try that

I am very interested in knowing more about that 555 battery and the 10% energy transfer. Are you saying that energy is being transferred to the load resistor or are you saying that 10% of the total energy was consumed in the 555 circuit?

GAD: i cannot tell that the energy is being transferred to the load from the 555, since i do not have such knowledge to prove that . i can tell the latter only, by charging both batteries and see the charge, for example: 120 mah in the 555 battery, 1200 mah in the load battery. by doing some simple calc., we can tell that the avg. power used by the 555 was 1.01 watts (during 91 minutes) which means avg. current of 79ma, at avg. voltage of 12.8volts. in my opinion the current that should be enough for the 555 can be as low as 10-20 ma . 78 seems a bit high to me, which can lead to the conclusion that this current somehow leaks into the load circuit thru the MOSFET gate connection
the battery used was the same as the load circuit battery - 4.5AH, 12V sealed lead-acid PB battery.

What frequencies does your equipment show the circuit running at?
GAD: 150-300khz (each experiment i had diff. freq.)

Are there any other places that heat can be lost in the circuit or voltage being dropped due to capacitive or inductive reactance?

GAD: other than the MOSFET + load, i do not think so. i replicated the wiring of Glen's also, as i reported few months ago in my early experiments.


Q from GAD: do you think the gate. pot resistance should be lower than 2 ohms ? i could not get lower than that (1.8ohms) , and i noticed that the lower it gets, the "nicer" the waveforms get...

Gad
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  #142  
Old 11-01-2010, 08:11 PM
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Quote:
Originally Posted by Aaron View Post
10% sounds WAY too high even if your load side draw is low. You can also
drop the 555 draw to the bare minimum that you actually need. If you're
replicating the quantum article, you can use a separate 555 battery, but
if you're just wanting to experiment with the concepts, you can power
the 555 from the same load battery and measure the load of both
simultaneously.
how can i "drop the 555 draw to the bare minimum that you actually need" ? please elaborate since i did not get the point. i do not think using the same battery will do that, its just for convenience.
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Old 11-01-2010, 11:08 PM
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Hi Gad,

One of the places that I identified very early as being a potential power hog and poor engineering was in the way that the current path exists in the original 555 circuit. Even with Glen's modifications, the problem is still there.

See this drawing: Mosfet Heating Circuits

Now, follow the 12V+ path through D1 and R5. Now imagine that R4 is set to zero ohms. When the 555 switches, Pin 7 becomes ground. If D1 has a 0.6V drop, this still places 11.4V across R5 which equates to 103 mA of current just for that single path not counting all the other current paths through the chip.

Now see my suggested modification:Harvey's Suggested Modification

Here we find that the discharge pin still provides the same action it was intended to in the original design, i.e. clamp charge source to allow the cap (c3) to fully discharge, but we no longer have that high current path. And we have eliminated one part, the 110 Ohm resistor.

Granted, Glen has reduced the resistor value of R6 from 1K to 330 Ohms, but with my suggested modification we now have 11.4V across 330 Ohms instead of 110 Ohms. So in this way we reduce the current to 34.5 mA in that leg.

=====================

Regarding the Gate pot:
This was an area of problem for everyone who has worked on this circuit including me. The wire wound pots simply don't have a clean enough resolution in the lower ohm ranges to get things dialed in and they seem to change value when they get warm. Because of this I did make a suggestion at one point to make a wire slider out of a section of nichrome to act as a fine tuning mechanism. By sliding a shorting jumper of sufficient gauge along its length you have an infinitely variable low ohm resistor capable of carrying high currents.

The current in this line should be limited to the instantaneous NE555 output current of 200 mA. This means if you have a 12V rail you do not want to charge the Gate with more than 200 mA of current. If your Ciss is 2800 pF and you are running at 300 kHz, the Xc is 5.3 milliohms so the Gate looks like a short circuit at that frequency. I say If here because the 2800 pF Ciss specification for that FET is based on a Vdd of 25V and a frequency of
1Mhz - "so your results may vary" . This tells us that our gate pot 'shouldn't' be lower than 60 ohms to protect the NE555. The interesting thing here is that we have pushed the current up here and the NE555 has endured in Glen's tests for hours upon hours of continuous operation. In fact, it seems to be required in order to precipitate the aperiodic operation.

Look at the following diagram:



Everyone who understands bipolar transistors knows that there is a diode junction from the base to the collector. Now look at Q20 connected to Q23 (it is called Q because they literally use a transistor as a diode there). Follow the path from Q20 emitter back to R16 and note that this will activate Q14 with a positive current flow. If the output pin 3 is back fed with sufficient voltage, the Base-Collector junction of Q20 can avalanche like a zener diode and precipitate an early discharge cycle. Conversely, if a sufficient negative voltage exists there, it could cause Q25 to reset the chip through the same path. Because of the astable way the external circuit is designed, when the discharge is activated, the chip is re-triggered and this is what causes the aperiodic operation. Evidently, this feedback action is precipitated by excessive output currents.

Additionally, we cannot overlook the fact that the gate pot is inductive. Therefore, some interaction exists between that inductance and the Ciss of the FET.

If you have a way to properly measure the instantaneous output current from the NE555, then you could adjust your pot up to the rated maximum. Beyond that and you are in uncharted territory test piloting the real field limits of the chip

In my spice simulations, the avalanche breakdown seemed to occur around six ohms and was sensitive to changes as small as 1/10 of an ohm causing the simulation to crash. Part of that was traced back to the mathematical treatment of repeating decimals in my spice program. When I chose values that did not result in converging repeating decimals, the simulation ran and did provide aperiodic operation.



EDIT To Add: Regarding the RF, we know for a certainty that this circuit dissipates energy as RF from the Inductive Resistor. Therefore, if the Calorimeter can trap that RF energy in the vessel, it should eventually be absorbed by the water, glass, Teflon etc. and converted to heat. The same can be said for IR energy in the water. I think even the plastic has IR reflective material though
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Last edited by Harvey; 11-10-2010 at 03:33 PM.
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  #144  
Old 11-02-2010, 12:00 AM
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555 measurement

Quote:
Originally Posted by gadh View Post
how can i "drop the 555 draw to the bare minimum that you actually need" ? please elaborate since i did not get the point. i do not think using the same battery will do that, its just for convenience.
Gadh,

You can put a 10 ohm resistor for example in line with a 1k or higher
variable pot and increase resistance on the negative line from the 555
battery until it doesn't work cleanly anymore. Measure what resistance
then put fixed resistors or leave the variables. You'd be surprised at
how low you can power these cleanly without any problems.

I don't recall off hand but there are some 555 models that are intended
to be lower power devices, you could start with those and they're just
as inexpensive. Not that power draw is a major issue for the 555 but
you might as well use as little as necessary, which from my experience
is way less than the specs. And it won't be under-driving them as some
people claim.

Mentioning drawing both from the same battery was just for simplicity of
using one current sensing resistor for example off the negative terminal
and you will have no question what the draw of both are at the same
time from one single probe/ground from a scope.

Most scopes, you can measure the current sensing resistor off the load
battery and the 555 timer battery and they will cross talk and you will
not get an accurate reading unless you measure one separately, figure
it out and then remove then connect to the other and measure it
separately and add that to the other.

Anyway, I got to the point where I could take a bit of the recovery from
the inductive resistor, charge a cap and feed it back in an isolated way to
the 555 battery so that it was practically running for free. I took a bit
of the recovery by placing a small coil in the vicinity of the inductive
resistor and that charged a cap that powered my 555 in some experiments.
That was just for fun to see that it could be done.
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  #145  
Old 11-02-2010, 06:19 AM
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thanks Harvey for your thorough explanation, but since i do not understand all that you wrote, i cannot make of it a practical solution...

just tell me - "then you could adjust your pot up to the rated maximum" - why not the minimum ? since in my observations the max. gate pot. resistance did not give any resonance freq, only the opposite - the close to the minimum. and certainly if you close the pot. (aspire to zero ohms) you get flat line on the scope (maybe its what you mean as "avalanche" ? - short circuit ?)

Nevertheless, i'll try your suggested circuit changes and i'll report my results, but i'll give this change a lower priority than using 24v on the load (my/your former suggestion).
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Old 11-02-2010, 06:31 AM
gadh gadh is offline
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Quote:
Originally Posted by Aaron View Post
You can put a 10 ohm resistor for example in line with a 1k or higher
variable pot and increase resistance on the negative line from the 555
battery until it doesn't work cleanly anymore. Measure what resistance
then put fixed resistors or leave the variables. You'd be surprised at
how low you can power these cleanly without any problems.
but if the resistance on the gate pot. will be so high then i wont notice any resonance freq. , wont i ? if you'll attach some scehmatics i think it would be easier to understand for me. sorry for some lack of knowledge in this area...

Quote:
Originally Posted by Aaron View Post
Anyway, I got to the point where I could take a bit of the recovery from
the inductive resistor, charge a cap and feed it back in an isolated way to
the 555 battery so that it was practically running for free. I took a bit
of the recovery by placing a small coil in the vicinity of the inductive
resistor and that charged a cap that powered my 555 in some experiments.
That was just for fun to see that it could be done.
wow ! i think this direction can lead us to using the extra energy of the load to feed back the main battery instead of wasting to heat ?! i think i read in this thread also about using a strong flyback diode to do so. or to charge another battery ? this way it would be a lot easier for us to measure the extra energy by comparing the charge on all the batteries after usage. maybe you can continue testing this thesis ? (i think you should open a new thread for that) - i sense it is promising...
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Old 11-02-2010, 06:39 AM
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Quote:
Originally Posted by Harvey View Post
Granted, Glen has reduced the resistor value of R6 from 1K to 330 Ohms, but with my suggested modification we now have 11.4V across 330 Ohms instead of 110 Ohms. So in this way we reduce the current to 34.5 mA in that leg.
Please note that i already use Glen's modified circuit of the 555 from 26/11/2009. can you clarify what are your sugested modifications over his circuit ? do you mean to use Aaron's circuit from Aug 9 ,2009 as you pointed me to in your link ?
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Old 11-02-2010, 07:34 AM
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Quote:
Originally Posted by gadh View Post
thanks Harvey for your thorough explanation, but since i do not understand all that you wrote, i cannot make of it a practical solution...

just tell me - "then you could adjust your pot up to the rated maximum" - why not the minimum ? since in my observations the max. gate pot. resistance did not give any resonance freq, only the opposite - the close to the minimum. and certainly if you close the pot. (aspire to zero ohms) you get flat line on the scope (maybe its what you mean as "avalanche" ? - short circuit ?)

Nevertheless, i'll try your suggested circuit changes and i'll report my results, but i'll give this change a lower priority than using 24v on the load (my/your former suggestion).
I should have been more precise in my wording, my apologies for that. I clearly see how it could have been read as resistance rather than amperage.

A better way to write it would have been:
"then you could adjust your pot up to the maximum 200 mA supported by the output stage of the NE555".

Cheers,

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Old 11-02-2010, 07:41 AM
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Quote:
Originally Posted by gadh View Post
Please note that i already use Glen's modified circuit of the 555 from 26/11/2009. can you clarify what are your sugested modifications over his circuit ? do you mean to use Aaron's circuit from Aug 9 ,2009 as you pointed me to in your link ?
Hi Gad,

If you look carefully at that link, you will see that was a suggested modification very early in our trials to figure out why nobody was able to get the circuit to work properly. Some of the replicators were complaining about the chip getting hot and I took a look to see what could be causing that problem.

The red lines are mine on Aaron's previous schematic. The principle still holds for Glen's. If you move the connection point as shown, you can replace the 110 ohm resistor with a wire which gives a wider range of adjustment on the frequency pot and still reduces the current in that leg which was causing he heating during the discharge cycle.

Cheers,

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Old 11-02-2010, 08:40 AM
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Harvey , please let me see if i understood your modifications correctly:
1. replace R5 with 1.5kohms resistor or with another pot. ?
2. R5 connected to RST leg. R1 (pot.) connected to OUT leg. do you mean that changing R5 will affect R1 sensitivity ?
3. replace all components (R/C) according to your "red" circuit modifications or just R5 ?

(you need to bring your explanations "down to the ground" for me...)
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