Mosfet Heating Circuits

[gadh]

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 ?

[Harvey]

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.

[Harvey]

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.

[gadh]

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.

[Harvey]

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.

[gadh]

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

[gadh]

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 ?

[Harvey]

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,

[gadh]

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.

[Harvey]

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?

[gadh]

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 ?

[Harvey]

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.

[Harvey]

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

[gadh]

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.

[Harvey]

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.