Mosfet Heating Circuits

[Harvey]

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!

[FuzzyTomCat]

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



Glen

[gadh]

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

[Harvey]

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!

[Aaron]

555

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.

[gadh]

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

[gadh]

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.

[Harvey]

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: http://www.energeticforum.com/induct...html#post84279

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

[Aaron]

555 measurement

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.

[gadh]

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).

[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.

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...

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...

[gadh]

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 ?

[Harvey]

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,

[Harvey]

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,

[gadh]

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...)