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Author Topic: partzmans board ATL  (Read 6613 times)
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Thank you Gyula.
 
The bias current for both L5 and series connected L2 comes a power supply with a 50 ohm resistor in series.  The voltage is simply adjusted to arrive at the current required in a static state.  The circuit is then single cycled at a rate of 10 cycles per second.  During the interval between cycles, the constant current settles to the desired value.

At the beginning of each cycle, switch S6 closes thus clamping the non-dotted end of L5 to ground.  This removes the current from the power supply from having any influence on the circuit operation.  IOW, only L5 and the input primary are sources for energy for the complete cycle.

No, you're not missing anything and this is a good question.  If one was to run this device without any bias current in the secondary, you would see an increase in the secondary current from zero and if all the energies were accounted for, it would be conservative.  If a bias current is now added to the secondary, there will an increase in the output energy simply because the delta I is between two larger current magnitudes.  This in itself produces a gain.  The method chosen here to produce this secondary bias current is a constant current inductor which is basically not "seen" by the primary because there is very little delta I in the secondary during the ramp up of the L1 primary.  Thus little to no Lenz reflected back to the primary.

Regards,
Pm

Good day PM

Can the 'bias' current  be considered a DC source?  If it is, could it be pushing/helping the L2 core unto the brink of saturation (similar to bias current of a MagAmp)? Does the device give COP > 1 w/o the bias current?
Or am I missing something fundamental here also?

take care, peace
lost_bro
   
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Good day PM

Can the 'bias' current  be considered a DC source?

Yes, the bias supply can be a constant current source from some dc supply and depending on the self capacitance of the windings, it may be less or more efficient than a current inductor.

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If it is, could it be pushing/helping the L2 core unto the brink of saturation (similar to bias current of a MagAmp)?

It will promote saturation in L2 in the same manner as the constant current inductor.  L2 is non-linear but really never goes into a hard saturation like a mag amp.  The main difference is that a solid state current source will have no current loss like the inductor but it will instead create a large power dissipation in the power supply feeding it.  This is due to the fact that the supply voltage for the solid state current source must be at least equal to the peak ac voltage at the hot end of L2. 

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Does the device give COP > 1 w/o the bias current?

No it will not.  Neither will it work without a constant current source supplying L2 during the time L1 is charging.

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Or am I missing something fundamental here also?

Sounds to me like you've got a pretty good handle on it!

Regards,
Pm

Quote
take care, peace
lost_bro

   
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Hey partzman,

I have been a little confused on the location of the L5 inductor.  Is this a separate winding that is wound along the same bobbin and cores as L2? 

Thanks,

Dave
   
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Hey partzman,

I have been a little confused on the location of the L5 inductor.  Is this a separate winding that is wound along the same bobbin and cores as L2? 

Thanks,

Dave

Dave,

L5 is the constant current inductor that is wound on a separate core and is totally independent from the L1/L2 cores.

Regards,
Pm
   
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Dave,

L5 is the constant current inductor that is wound on a separate core and is totally independent from the L1/L2 cores.

Regards,
Pm

Good day PM

OK, so regarding the config of the L5 inductor:
It is magnetically independent from both L1 & L2,  and has a galvanic series connection to L2.
So the output of the device is across L2 and L5?
Or is the output taken from only across L2?

Sorry for so many questions, just trying to visualize what is happening.

take care, peace
lost_bro



   
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Good day PM

OK, so regarding the config of the L5 inductor:
It is magnetically independent from both L1 & L2,  and has a galvanic series connection to L2.
So the output of the device is across L2 and L5?
Or is the output taken from only across L2?

With the polarities shown in the schematic for L1 and L2, when the field is collapsing in L1, the current in L2 will increase in a positive direction.  This will occur if there is any bias current in L2 or not.  In this case however, there is a bias current from L5 so the current in L2 will increase from that bias current level to some higher positive level.  The output is considered to be at the point when the current in L2 reaches a peak level and basically the cycle stops at that point.  The output is then taken across L2 so to speak by taking the difference between the starting and ending energies in L2.   This is a key point in the energy gain mechanism.

L5 on the other hand will have a loss in energy depending on the voltage across L2 due to di=E*dt/L .  So the average voltage across L2 affects the loss in L5.

The constant current in L5 imposed on the secondary L2 is basically not seen by the primary L1 during it's charging phase as explained in my attached paper below.  This too is a key point in the energy gain mechanism.

Regards,
Pm


Quote
Sorry for so many questions, just trying to visualize what is happening.

take care, peace
lost_bro
   
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With the polarities shown in the schematic for L1 and L2, when the field is collapsing in L1, the current in L2 will increase in a positive direction.  This will occur if there is any bias current in L2 or not.  In this case however, there is a bias current from L5 so the current in L2 will increase from that bias current level to some higher positive level.  The output is considered to be at the point when the current in L2 reaches a peak level and basically the cycle stops at that point.  The output is then taken across L2 so to speak by taking the difference between the starting and ending energies in L2.   This is a key point in the energy gain mechanism.

L5 on the other hand will have a loss in energy depending on the voltage across L2 due to di=E*dt/L .  So the average voltage across L2 affects the loss in L5.

The constant current in L5 imposed on the secondary L2 is basically not seen by the primary L1 during it's charging phase as explained in my attached paper below.  This too is a key point in the energy gain mechanism.

Regards,
Pm

Thank you for this explanation.  I believe I understand what you are doing, although it did require the document as I wasn't sure the 'whys' of your results til now.  I am wondering how this could be cascaded to multiple transformers into a function of growth and utilize the COP>1.  hmm....  O0  Thanks for sharing!

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I mentioned in a previous post that a 'U' type core with a shunt could be use instead of a small/large core combo.  An equivalent to this would be a modified 'E' core with equal core area on all legs.  The attached simulation includes the model of such a transformer of a bench version that uses modified Magnetics P42515 'E' cores.  The sim and bench results match within ~5% or so after carefully modelling the non-linearity of each leg of the core.

The attached sim shows the results.  The magnetic and electric models of the transformer utilize capacitor/gyrators and the arbitrary behavioral voltage sources B1, B3, and B4 model the non-linear core characteristics.  P1 and P3 represent the low level permeance (reciprocal of reluctance) of the outside legs while P2 is the center leg with gap.

Also shown on the schematic is a table of the what I call "recovery inductance values" which are determined from the actual discharge energy of L2 from the various peak currents shown.  Using these inductance values will guarantee 100% accurate recovery energies in the final calculations even though the inductances are non-linear.  The actual inductance levels of L2 are higher than stated in the table because the losses are taken into account.  For example, let's say that the power recovered from an inductor back to a 30vdc supply over 66us from a 200ma peak was 2.7w or 178.2uJ.  To convert this to an inductance value that will yield this same energy level we use L = 2*UL/I^2 = 2*(178.2e-6)/.2^2 = 8.91mh.  We now have an inductance value that when used in the recovery calculations of the secondary L2 will give accurate results.

L2 is the 267mH constant current inductor and switches S2, S4, and S8 form a pseudo bridge that drives the primary.  During the charging phase of the primary, L2 is connected to the secondary thus supplying a constant current of 200ma to said secondary.  During the second phase for the primary it's energy is returned to the supply Vs.  At the same time, L2 is disconnected the secondary and is clamped to freeze it's current while the hot end of the secondary is grounded to allow an increase in it's bias current.  The cycle is complete at the end of a 10us period and the energies are calculated.

The mean input energy consumed from the supply Vs is shown in the plot math as V(Vs)*I(V1) = 1.715uJ.

Cursor #1 shows the ending secondary current I(V8) = 231.92ma peak for an energy gain of (.23192^2-.2^2)*.0088/2  = 60.66uJ.  The 8.8mH inductance for the secondary was interpolated from the table.

Cursor #2 shows the ending current in L2 that is I(L2) = 199.54ma for an energy loss of (.2^2-.19954^2)*.267/2 = 24.54uJ.

The net COP = 60.66/(1.715+24.54) = 2.31.  This is ~5% higher than the actual bench device tested.

Regards,
Pm
   
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I mentioned in a previous post that a 'U' type core with a shunt could be use instead of a small/large core combo.  An equivalent to this would be a modified 'E' core with equal core area on all legs.  The attached simulation includes the model of such a transformer of a bench version that uses modified Magnetics P42515 'E' cores.  The sim and bench results match within ~5% or so after carefully modelling the non-linearity of each leg of the core.

The attached sim shows the results.  The magnetic and electric models of the transformer utilize capacitor/gyrators and the arbitrary behavioral voltage sources B1, B3, and B4 model the non-linear core characteristics.  P1 and P3 represent the low level permeance (reciprocal of reluctance) of the outside legs while P2 is the center leg with gap.


Can you provide the spice sim file? Thank you.
Very interesting!
   
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Can you provide the spice sim file? Thank you.
Very interesting!

I'm sorry but the .asc file is not being made available at this time.

Pm
   
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The method I used for the recovered energy level from L2 results in inaccurately overstated energy levels in most all of the previous tests.

Regards,
Pm
   
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The method I used for the recovered energy level from L2 results in inaccurately overstated energy levels in most all of the previous tests.

Regards,
Pm

Good day Pm

Been following along everyday........
Sorry to hear that, but 'in most all' of previous tests would signify that at least one configuration gave anomalous results?
Even one configuration showing anomalous results is enough to progress forward.

take care, peace
lost_bro
   
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Good day Pm

Been following along everyday........
Sorry to hear that, but 'in most all' of previous tests would signify that at least one configuration gave anomalous results?
Even one configuration showing anomalous results is enough to progress forward.

take care, peace
lost_bro

Yes, there are still gains in the different configurations but in the range of 1.2 to 1.5 at present.

Regards,
Pm
   
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This is an explanation of the logical error made in my recovery inductance analysis of the sim in post #182.  The first plot below shows the general charge and discharge profile of the secondary and the cursors are placed in the area of interest that is, the area between the maximum and minimum currents encountered in the secondary.  From this area, the corrected recovery inductance will now be calculated whereas before, the entire discharge curve area was used which overstated the actual recovery inductance.

The second plot shows the expanded area between the cursors.  The supply voltage is 50v dc so we can calculate the inductance from L=E*dt/di = 50*3.606e-6/.0307 = 5.87mH.  This compares to the previously interpolated value of 8.8mH.  To prove this value, we see the plot math indicates an energy level of -38.507uJ and we compare that to the calculated energy gain of ((.2307^2)-(.2^2))*5.87e-3/2 = 38.81uJ which is reasonably accurate.

So, using this new recovery inductance with it's attendant energy gain in our previous COP calculations, we have a new COP = 38.81/((1.715+24.54) = 1.48. 

All my previous tests will have a similar correction both on the bench and in simulation.

Regards,
Pm
   
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