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Author Topic: LTJT - poynt99 Tests #2  (Read 106250 times)
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Poynt:

Just a quickie.  I am assuming that you are using vanilla 5% tolerance carbon resistors for your current sensing functions.  Are you making actual measurements of the values of those resistors for the math calculations?

Thanks,

MileHigh
   
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PhysicsProf:

I am going to make a few comments about your suggestions in advance of anyone trying them out.  The purpose of commenting ahead is to hopefully shed some more light on the way the Joule Thief works.

This can be expected to have a marginal effect on the operation of the circuit.  Lowering the base input resistor will mean that the transistor switches on slightly faster.  Because the JT does a "snap" when the transistor switches on and off, you are talking about changing the timing by a few microseconds at most.  Also lowering the base resistor will increase the power dissipation slightly. I suspect that lowering the base input resistor will increase the operating frequency of the JT by a very small amount.

[snip]

MileHigh

In fact, by direct observation of a JT circuit in my lab upstairs, lowering the base input resistor (from 1000 ohms to 500 ohms) resulted in a DECREASE of the operating frequency of the JT by a large amount -- 461 kHz to 361 kHz.

I say this not to embarrass you, MH, but rather to show (once again) that one's theoretical understanding of the JT circuit -- or the one I just tested in my home lab, a very simple JT circuit (without the collector LED) -- is in the OPPOSITE direction from what you predicted, and the change is large.  I invite .99 to make the same test and check if this result is more general.  Quick and easy test.

Also, MH, we see again the importance of testing with an actual apparatus -- experiment trumps theory... as previously noted throughout the history of science.
And again I encourage you to set up a JT to play with... not difficult.

   
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.99 "Will you post some input and output shots similar to the ones I posted...with MEAN values instead of RMS?"

Yes of course, when I get time on the Tektronix 3032 (within the next week). I have not been up to the university to use the 3032 since I was there and posted results in this forum last time. 
 I've been using the ATTEN 1062C that I bought and arrived last Thursday.  Lots of fun, and it has the channel 1 * channel 2 math function, which I am using routinely to display the Power = V * I waveform.

However, while this DSO readily calculates the MEAN (and RMS, etc) for channel 1 and channel 2 separately, I have not figured out how to get it to calculate the MEAN for the product V*I ... Not sure it has this capability -- I have written to the tech people at ATTEN.  Also, have not been able (yet) to get the power waveform onto my MAC-Pro laptop where I could further analyze it... and where I could post the waveforms.  I do have waveforms stored on a thumbdrive, if I could just read them-- the disc that came with this system was not helpful.  If anyone knows the "tricks" of this new DSO, I'd appreciate hearing from you.

So for now, to get the MEAN of P = V*I, I have a long trip up to the university and arranging to use that scope.  I am also trying to borrow a more expensive scope that will calculate the MEAN of the V*I waveform directly.  I'm glad you're taking data .99 -- you will no doubt be the first to evaluate P-Mean (and the ratio Pmean-out/Pmean-in) for the Tseung circuit.
   
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In fact, by direct observation of a JT circuit in my lab upstairs, lowering the base input resistor (from 1000 ohms to 500 ohms) resulted in a DECREASE of the operating frequency of the JT by a large amount -- 461 kHz to 361 kHz.

I say this not to embarrass you, MH, but rather to show (once again) that one's theoretical understanding of the JT circuit -- or the one I just test upstairs, a very simply JT circuit (without the collector LED) -- is in the OPPOSITE direction from what you predicted, and the change is large.  I invite .99 to make the same test and check if this result is more general.  Quick and easy test.

Also, MH, we see again the importance of testing with an actual apparatus -- experiment trumps theory... as previously noted throughout the history of science.
And again I encourage you to set up a JT to play with... not difficult.

Actually I am not embarrassed at all.  There are limits to what can be done when you are just commenting and you haven't worked with one on the bench, nor have I seen your actual circuit.  In most of my postings along these lines I make a disclaimer that I am not actually on the bench myself and therefore I can't be 100% certain of what I am stating.  The process of exploring like this is part of the learning experience.

Quote
Also, MH, we see again the importance of testing with an actual apparatus -- experiment trumps theory.

You are missing the point PhysicsProf.  I was wrong with respect to your test.  The empirical evidence on your particular Joule Thief shows large changes in frequency and a decrease in frequency when you lower the base resistance, the opposite of what I said.  There are too many unknowns on both sides to draw any conclusions at all between what I stated and what you tested.

Do you know if your transistor was fully on or in partial conduction mode when you used a 1K base resistor?  What about when you switched to a 500-ohm base resistor?  My statement was based on the presumption that for both the 1K and 500-ohm case that the transistor would be functioning as a switch.  Can you explain why the frequency dropped when you changed the base resistor?  It would be fun to work together one day, time permitting, to figure out why the frequency changed.

There is a logical reason for the decrease in the operating frequency of your Joule Thief that is backed up by theory.  So the challenge for those that are interested in pursuing the research is to explain your observations with sound theory.  That's where the fun comes in as part of the intellectual journey.

Let's go back to the Wikipedia formula for the operating frequency of the Joule Thief:

F = ((V_batt x R_batt) / L_mutual).  Note R_batt is the output impedance of the battery.  This formula presumes that the transistor is acting like a switch.

On the the other hand, supposing the transistor is in partial conduction mode when the base input resistor is 1K ohm.  That would effectively make R_batt become (R_batt + R_transistor) which is higher, which according to the formula would increase the JT operating frequency.

Supposing the transistor is in full conduction mode and acting like a switch when the base input resistor is 500 ohms.  That would effectively make R_batt lower, which according to the formula would decrease the JT operating frequency.

So, based on the limited amount of information that I have on hand, it would appear to me that when you have a 1K base input resistor the transistor is operating in partial conduction mode.  That would explain the decrease in frequency when you switch the base input resistor to 500 ohms.

As far as I understand, the desire is for the Joule Thief circuit to operate the transistor as a switching device because this then limits the power dissipated across the collector-emitter junction to a minimum.

This suggests an interesting experiment:  As you lower the base input resistance, you should see the operating frequency of the Joule Thief decrease.  Eventually you should get to a point where lowering the base input resistance even further does not change the operating frequency (i.e.; the basis for my original comments).  That point where the frequency stops changing would be your ideal base input resistance.  I am assuming that a standard Joule Thief circuit assumes that the transistor should be operating as a switching device.

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It's not as complicated as it may seem...
Poynt:

Just a quickie.  I am assuming that you are using vanilla 5% tolerance carbon resistors for your current sensing functions.  Are you making actual measurements of the values of those resistors for the math calculations?

Thanks,

MileHigh

MH,

No I am not at the moment using the actual value of the "1 Ohm" CSR resistor, as I see no need until or unless the "n" begins to approach 100%. The resistor is 5% carbon I believe, and a value of 0.98 Ohms iirc.

.99


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

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It's not as complicated as it may seem...
An open question for anyone who may wish to reply:

With reference to the scope shot below, what factor(s) determines the current reached before the transistor switches OFF?

In the shot below (TEK00012.PNG), the current (blue trace) rises to about 85mA.



.99


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  I'm finding you are raising some very good points, and questions.  I certainly don't know all the answers.   \
MH:  "As far as I understand, the desire is for the Joule Thief circuit is to operate the transistor as a switching device because this then limits the power dissipated across the collector-emitter junction to a minimum."  Agreed... how best to achieve this is an important question.

  The  circuit frequency is sensitive to just about any change I make -- resistances, LED's in or out, toroid windings, etc.  It does seem to reach a natural resonance condition quickly in almost all cases.
   
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An open question for anyone who may wish to reply:

With reference to the scope shot below, what factor(s) determines the current reached before the transistor switches OFF?

In the shot below (TEK00012.PNG), the current (blue trace) rises to about 85mA.


.99

This is the reason why I joked with you earlier to get a good battery.  I noticed that the 6kHz on test 1 and the 50kHz on test 2 have the same amount of current.   The slope of the current looks too linear so I suspected that current rise in L1 and L2 coil is controlled by the switching.  I think the reason is that the battery has too high impedance.  
   

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It's not as complicated as it may seem...
This is the reason why I joked with you earlier to get a good battery.  I noticed that the 6kHz on test 1 and the 50kHz on test 2 have the same amount of current.   The slope of the current looks too linear so I suspected that current rise in L1 and L2 coil is controlled by the switching.  I think the reason is that the battery has too high impedance.  


The supply voltage does have an effect on the oscillation frequency, true. However, the battery impedance does not have that much effect on things at the moment, and you even pointed out yourself that the current reached the same amount, even when the Fo was about a decade higher.

So, what are the factors that determine the maximum rise of the current? Why is the current slope too linear, what do you mean?

.99


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"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
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An open question for anyone who may wish to reply:

With reference to the scope shot below, what factor(s) determines the current reached before the transistor switches OFF?

In the shot below (TEK00012.PNG), the current (blue trace) rises to about 85mA.

I'm pretty sure that it's the leveling off in the current rise that causes the transistor to switch off, not the absolute amount of current flow.  It's the change in the rate of change of the current that determines when the transistor starts to switch off.  So that's the second derivative (the "accelerated acceleration") of the current with respect to time that is the determining factor.

As the current climbs linearly as the core get's energized the positive EMF on the L1 coil keeps the transistor switched on with a constant voltage.  Then the current starts to level off as the output impedance of the battery (plus the resistance of the transistor junction assuming it is in partial conduction mode) starts to become a factor.

This leveling off of rate of the current rise causes a slight drop in the positive EMF from L1.  This triggers a positive feedback loop, (a 'cascade' effect) where the dropping EMF from L1 starts to switch off the transistor, and that in turn causes more leveling off of the rate of increasing current (eventually becoming decreasing current), which in turn reduces the EMF from L1 (eventually going negative), which turns the transistor off even more, etc.

This positive feedback loop causes the transistor to "snap" of very very quickly, much faster than a typical Bedini motor shuts off the transistor.  You note that it's so fast that you can't even see it in your scope captures.  They look like normal linear ramp functions but I have to assume that the at the tail end of the ramp the slope changes just slightly for a fraction of a microsecond and then all hell breaks loose.

A great analogy for a positive feedback loop that is somewhat applicable comes from trying to adjust the position of your car seat while driving.  If your seat is free to move forwards and backwards and you reflexively touch the brake while you are doing 100 km/hour, then you will floor the breaks and will be unable to stop this from happening.  The slightest touch on the brake pedal will slow the car down resulting in the seat sliding forwards which results in you pushing harder on the brake pedal, etc.  It happens in a flash and you have no control over it.

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It's not as complicated as it may seem...
OK, the feedback does sound plausible MH.

Are there any other factors involved with this Imax? What factors determine the Fo?

.99


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

Does your scope have low-pass filtering?  It might be possible to see the slope change at the end of the ramp if you filtered out the noise.

As far as the Fo goes, we can see that for the charge cycle, it's related to R/L, and then I assume the discharge cycle is related to how fast you discharge the energy in the core.  So I assume that part is load dependent, and depends on what is connected to the collector terminal and/or L3.

I just realized too that you can't forget the "backwards" thinking that you have to do for inductors.  I slip there sometimes and I made some incorrect statements about the discharge of L3 recently.

So with the "inductor backwards" thinking cap on, the lower the load resistance the slower the core will discharge, and the higher the load resistance the faster the core will discharge.

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It's not as complicated as it may seem...
Poynt:

Does your scope have low-pass filtering?  It might be possible to see the slope change at the end of the ramp if you filtered out the noise.

As far as the Fo goes, we can see that for the charge cycle, it's related to R/L, and then I assume the discharge cycle is related to how fast you discharge the energy in the core.  So I assume that part is load dependent, and depends on what is connected to the collector terminal and/or L3.

I just realized too that you can't forget the "backwards" thinking that you have to do for inductors.  I slip there sometimes and I made some incorrect statements about the discharge of L3 recently.

So with the "inductor backwards" thinking cap on, the lower the load resistance the slower the core will discharge, and the higher the load resistance the faster the core will discharge.

MileHigh

Yes, the scope has an acquisition "averaging" function, and when utilized, it cleans up the traces quite a bit.

The battery voltage is one factor that affects Fo. One other factor that affects Fo is the time constant of the base drive, L/R, which is the reason changing the 1k to 500 Ohms decreases the frequency, i.e. smaller R yields longer tau. Incidentally, this also affects the Imax value. A smaller Rb and you have a larger Imax ("Imax" being the value at the top of the ramp).

So: decR, decFo, incTau, incImax,

What other parameter change might have a similar effect on Imax and Fo?

Why did Imax remain the same after I cut the core to increase Fo?

.99


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

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...
Can you explain why the frequency dropped when you changed the base resistor?  It would be fun to work together one day, time permitting, to figure out why the frequency changed.

There is a logical reason for the decrease in the operating frequency of your Joule Thief that is backed up by theory.  So the challenge for those that are interested in pursuing the research is to explain your observations with sound theory.  That's where the fun comes in as part of the intellectual journey.
...

MileHigh


Some variants of the Blocking Oscillator will show an "OFF
time" which is dependent upon an RC time constant.

This variant seems to be governed by the L/R time constant.

Was the slope and time of the inductor current increase relatively
constant in both cases?


I'm pretty sure that it's the leveling off in the current rise that causes the transistor to switch off, not the absolute amount of current flow.
...
MileHigh

Yes, that is the "trigger" which initiates the cutoff
cycle;  inductor current flow ceases to increase.


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1) current rise is linear with slope di/dt=V/L
2) Max current x number of turns produces core saturation. Double the turns and max current is cut in half.
The rest operates like MH pointed out.

PS.  The horizontal segment before the current ramp is mainly governed by an RC time constant.  C is the BE transistor junction capacitance.  So, very some of these parameters and you vary Fo.
« Last Edit: 2011-02-09, 07:31:47 by EMdevices »
   
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PS.  The horizontal segment before the current ramp is mainly governed by an RC time constant.  C is the BE transistor junction capacitance.  So, very some of these parameters and you very Fo.

EMdevices,

Thanks for your comments but I will respond to the point above.  Note that the transistor will also "snap" back on with another positive feedback loop.  This will happen when the rate of decreasing flux in the core starts to level out when the core is running "out of gas."  So the horizontal segment before the current ramp is associated with the toroidal core inductor discharging its stored energy through L3.  So it's another L/R type of time constant at play.

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MH, I'm assuming a sharp discharge cycle, and the graph seems to show the quick discharge at less than 1% of the time duration of the level current, if I'm not mistaken.  During kickback, L/R will govern, true, but this is really small (occurs fast) like I said. The longer horizontal is really charge buildup in the transistor. 
« Last Edit: 2011-02-09, 07:52:47 by EMdevices »
   

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MH, I'm assuming a sharp discharge cycle, and the graph seems to show the quick discharge at less than 1% of the time duration of the level current, if I'm not mistaken.  During kickback, L/R will govern, true, but this is really small (occurs fast) like I said. The longer horizontal is really charge buildup in the transistor. 


By "scoping" the signal at the transistor
from base to emitter it is possible to
arrive at a clearer understanding of the
nature of the "cutoff" portion of the base-
emitter signal.

It should show a significant reverse bias
during that time;  the result of some
'discharge' current.


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It's not as complicated as it may seem...
All good comments, thanks. However, may I reiterate these questions:

1) What other parameter change might have a similar effect on Imax and Fo?

2) Why did Imax remain the same after I cut the core to increase Fo?

and I'll add one more since this has been mentioned a few times...

3) Is core saturation necessary for oscillation, and is it present with this LTJT?

Thanks,
.99


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

I'll take a stab at the questions at risk of repeating some points because I am not going to go back to reread both threads.

Quote
1) What other parameter change might have a similar effect on Imax and Fo?

During the core charging cycle I'll go back and say it's the battery output impedance plus the transistor impedance that dominates.  It's this resistance that determines when the slope of the core charging current starts to level off.  So if you lower the effective resistance here you increase the Imax, and you decrease the Fo.

Also, if the source battery voltage is higher, then the initial slope of the current increase is higher.  That means you will reach the the leveling off point in the slops faster, and this will result in increasing Fo.  Note that the battery voltage will not have any affect on Imax.

If you increase the number of turns in L1 and L2 you increase the inductance with a risk of saturating the core.  If we assume that the core does not get saturated then this will lengthen the charging and discharging time and decrease Fo.  It should not affect Imax.

I am assuming that I might be missing something here and not really answering your question.

Quote
2) Why did Imax remain the same after I cut the core to increase Fo?

Same answer as above, it's the resistance that determines Imax.

I am actually a little puzzled here about the frequency increase.  If you have a core where the number of turns in L1 and L2 are constant, and then you make a cut in the core, you are reducing it's cross-sectional area.  Won't that make the the core saturate sooner but not affect the value of the inductance associated with L1 and L2?  But it is apparent that cutting the core is decreasing L and therefore increasing Fo.  I am having some trouble visualizing this because the current waveforms show a linear rise (see below), indicating core saturation is not a factor here.

Quote
3) Is core saturation necessary for oscillation, and is it present with this LTJT?

No, and in fact the linear rise in the current waveform is telling you that the core is not saturated.  If the core got saturated you would see the current slope increase.  One more time, it's the series resistance causing the slope to level off that triggers the discharge cycle.

That's my best crack and answering the questions!

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1) What other parameter change might have a similar effect on Imax and Fo?

Imax is changed with changes in:    number of turns, different core properties, radius of toroid
F0 is changed with changes above plus, change in output resistor value, higher R values increase F0.

2) Why did Imax remain the same after I cut the core to increase Fo?

given the same number of turns, same toroid radius, core saturation occurs at a certain H intensity, and that is proportional to N*I,   or the number of turns times the current, and this quantity remained unchanged.  

3) Is core saturation necessary for oscillation, and is it present with this LTJT?

Yes, and Yes,  (this is the fundametnal quality of BOs, as opposed to regular oscillators.)


Some Discussion

This circuit is very simple and has been around for decades in one form or another.   Core saturation defines the end of the ON time of the BJT and the length of the OFF state can be controled with RC time constants or power ouput impedance.

MH,  you are absolutely right,  I looked at the charts again, and I can see the OUTPUT waveform labled by p99.   Disregard my RC and charge buildup in the transistor comments,  not applicable here.

the figure I include shows the ON and OFF states.


During the ON state the current ramp is linear and dictated by di/dt = V/L,  like I previously mentioned.

The end of the ON state occurs when saturation is reached and the coupling between the base coil and the collector coil begins to drop off sharply under positive feedback.    A carefull analysis of these switching events involves magnetics and I can expand on this if needed.

During the OFF state the current is routed through the output resistors and the energy stored in the magnetic field is dissipated by the output resistors and the LED or diode and it follows a roughly L/R timeconstant decay.   This off period/output phase seems to last about 3 or 4 time constants, and the exponential decay is clearly seen.

When the energy is pretty much disipated and the voltage has droped to a certain threshold, where the base circuit can now begin to function again, it will repeat the cycle.


Most Important parameters;

1)   By far, the Core properties, number of turns and operating voltage dictate the duration of the ON time.
2)   the Off time is mainly dictated by output impedance (in additon to the core properties, energy stored, etc...)
3)   there are other parameters of course, but we don't need to get into them here since they don't have a big impact.

P.S.   I should expand on core saturation,    the complete formula is   H = N*I / (2*pi*r),  and H and B are governed by the material properties (see a typical B-H curve)   So inspecting the equation,  another thing that can change the performance is obviously the radius of the toroid, but not it's thickness (in the z-direction)
« Last Edit: 2011-02-09, 21:37:29 by EMdevices »
   
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EMdevices:

It's fun to "look into the guts."  Of course there is only so far I myself can go without being on the bench and poking around with a scope probe.  I don't have a bench and I don't have a scope!  I also downloaded pSpice and it's sitting somewhere in my download directory.

Honestly at this point I am content to see what the results of the real analysis are for the new LTJT sample due to arrive very soon, i.e.; the average output power vs. the average input power.

In the discussions so far for anyone that has been following, it should be pretty clear that the Joule Thief simply charges a toroidal magnetic core with energy supplied by the battery and then discharges that energy through some sort of a load, then repeats the cycle.  This is an "astable multivibrator" pulse circuit and it "resonates" but it is not directly related to "resonance" in the way that we talk about an LC tank circuit.  It's a kind of distant cousin to an LC resonant tank circuit.

Looking forward to the numbers!

MileHigh

P.S:  Great composite graphic!
   
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Absolutely MH,

I've worked on these things for years and they are realy simple and neat circuits, I love them.    they can be used to do DC-DC conversion, trickle charge batteries, etc...,  

But one thing is clear, they are not Over Unity devices, barring any external influences.

I'm going to share a measurement tip.   If these devices are operating at almost 99% efficiency,  it's very easy to make measuremnts errors and conclude they are over unity.   I've made those mistakes in the past and I have a feeling I'm not the only one.  

Here it is:   if the scope probes are not properly calibrated, pulses voltage will overshoot more then reality and that will get you so excited when you calculate Pin vs. Pout by looking at the voltage spikes on the scope.   I jumped up and down with joy many times, then I remembered the probe business and fell flat on my face.   So, check those darn probes and use the same one for Pin and Pout calculations to minimize variability.
   

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It's not as complicated as it may seem...
The FedEx package arrived today at about 13:00, so I already have the device at hand.

Hopefully I will be able to "sneak" some time away tonight to do a quick test of Pin and Pout.

.99

PS. I have a different view of those 3 questions (I fully agree with MH on #3 though), and I'll perhaps elucidate on those a little later on.


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My answers in a non-technical manner, I would say are as follows;

1a) Add another secondary in parallel, or,

1b) let's say the secondary was 20 turns, then remove the secondary and wind four 5 turn secondaries connected in parallel.

2) I would say because the core transfer is not required to move more then a few degrees since the windings are overlapping. If the core had two 180 degree wound coils non-overlapping, then the core is required to really move the flux 180 degrees. If each of the winds is going all around the core, then the flux only needs to just nudge and the transfer is complete. Making a cut in the core will make not much difference under the later.

3) Here is where I differ. I doubt if a 1.5 volt driven core will achieve core saturation of more then 40% given the amount of voltage/amperage that core could really handle. So my answer is no and no.

wattsup



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