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Old 5th May 2010, 07:02 PM   #121
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Quote:
Originally Posted by geraldfryjr View Post
"Each panel now sees 3.01db less drive, but the panel area is doubled."
Does this mean that the efficiency gain is equal (0db gain) or 3db more?
Starting with one panel and a transformer step-up ratio of 100:1.
If you double the number of panels, and reduce the step-up ratio by sqrt(2)=0.7071

Then:
You gain 6dB from doubling the panel area: 20*LOG(2) = +6dB
You lose 3dB from reducing the step-up ratio by sqrt(2): 20*log(0.7071) = -3dB

So, with both changes, you gain +3dB in efficiency.


Be aware though, that doubling the number of panels doubles the panel capacitance in series with the leakage inductance which defines the upper bandwidth limit. If you want to keep the same high frequency extension, you will need to reduce the leakage inductance of your transformer along with reducing the step-up ratio. No free lunch...

Last edited by bolserst; 5th May 2010 at 07:24 PM.
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Old 5th May 2010, 09:22 PM   #122
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Thanks,Steve,for verifying that.
That is what I thought also.
So in order to reduce my transformation ratio from 1:200 to 1:100 I would have to double my area to have the same efficiency.
And if I quadruple the area I would net an extra +6db gain in efficiency.
And if I kept the same transformation ratio of 1:200 I would gain +12db in efficiency.
If this is True, It seems to go along with my recent findings.
The rest of the stuff I understand.
This helps me more in my understanding of the design criteria.
As I had mentioned before, the reason I had built the two different sizes in that one of them is aproximately 4 times (actualy 5.26 times) the area as the other ,was too investigate this princpal.

My other question now is, does raising the capacitance of the panel by using an insulating material with a higher dielectric constant (k) raise the efficiency as well? jer
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Old 5th May 2010, 10:34 PM   #123
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This is a SPICE model I created of a transformer and ESL load. The items in the box represent the transformer with leakage inductance, stray capacitance, and wire resistance. To the left is the source and damping resistor. To the right is a 1000pf ESL.

Click the image to open in full size.

The next screenshot is the voltage across the load without the damping resistor. Note the high Q resonance at about 31.6khz.

Click the image to open in full size.

This screenshot shows the impedance without damping. It is about 1.12 ohms at 20khz and about .175 ohms at 31.6khz

Click the image to open in full size.

We can calculate the unloaded Q at about 6.73

We will design for a Q of about .707. In order to accomplish this we will add a resistor in series with the primary to make the total Q .707.

Here are the calculations...

Fo=31600 hz
Xc = 1/(2*pi*31600*C)
C=1700/10^12
Xc=2962
Rsec=190 ohms
Rpr=.1 ohms
Q=2962/(.1*50^2+190)=6.73

Desired Q= .707

Rtot=2962/.707=4189 ohms

Rds=4189-(.1*50^2)-190=3749

Rdp=3749/50^2=1.4996 ohms

Therefore we need a resistor in the primary of about 1.5 ohms.

Here is the result of that addition:
Click the image to open in full size.

The impedance plot follows:
Click the image to open in full size.

Note that at 20khz the impedance is about 2 ohms. The absolute minimum impedance is 1.67 ohms at about 31.6khz.

The frequency response shows about .655 db down at 20khz

I did this quickly, so I do not take any responsibility for my errors

Later,

Jim
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Old 5th May 2010, 11:24 PM   #124
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Very cool!
I have not worked with spice yet, but I have been working with circuitmaker 2000 which uses spice.
I'm wondering if it could be implemented in that program as well?
I think it can, but there are some control functions that I have a hard time using or havn't figured out how to use.
Even though it was a great program in the day, it has and still does serve it's purpose well, except for the poor support and no support of today.
You can't beat the fact that you can draw the schematic ,simulate,make changes and simulate until the circuitworks ,dump the netlist to traxmaker manauly or automaticily route and print the board trace to a laser printer,iron,etch,drill, populate,solder and your done with nothing left but the smoke(ha ha ha).It's great!
Anway I'll give it a try and see if it works. jer
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Old 6th May 2010, 03:47 AM   #125
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Thanks for the cool info. I'm going to have to go back a few posts with paper and pencil to really get my head around this.
What a way cool program.
Paul
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Old 6th May 2010, 10:02 AM   #126
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Jelanier,

cool !

I have some remarks on the values:

1. Primary inductance of 288mH appears too high. This would request a very big core and a lot of windings. I think about 100mH is the limit

2. Your leakage indutance L5 and L6 is too low. Its ratio is 0,000042 (L5/L2). Even you make a lot of intersection windings you will hardly target 0,0005, which corresponds to 90mH leakage.

Would be intersting to see, how those values will modify performance

Capaciti
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Old 6th May 2010, 10:13 AM   #127
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Those figures are accurate and represent the actual values of a commercially available product. The secondary inductance is 720H with leakage inductance of 15mh. 720/50^2=.288H for the primary

Jim
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Old 6th May 2010, 01:03 PM   #128
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Jelanier,

Wow, if thats true, this needs to be an outperforming stepup. Would be interesting to know, how this stepup is designed in order to reduce leakage down to 0,000042, means 0,042 promille !!

Capaciti
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Old 6th May 2010, 01:25 PM   #129
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Quote:
Originally Posted by Capaciti View Post
Jelanier,

Wow, if thats true, this needs to be an outperforming stepup. Would be interesting to know, how this stepup is designed in order to reduce leakage down to 0,000042, means 0,042 promille !!

Capaciti
I could be wrong, but the leakage inductance, winding capacitance, and primary & secondary winding resistances match the Plitron 50:1 toroidal step-up transformer.

Jelanier, if you are reflecting the ~720H secondary inductance to the primary side, L1=288mH, what do L2 & L3 represent in your model?
Attached Images
File Type: gif plitron_4133.gif (151.0 KB, 78 views)

Last edited by bolserst; 6th May 2010 at 01:37 PM.
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Old 6th May 2010, 02:01 PM   #130
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Quote:
Originally Posted by bolserst View Post
I could be wrong, but the leakage inductance, winding capacitance, and primary & secondary winding resistances match the Plitron 50:1 toroidal step-up transformer.

Jelanier, if you are reflecting the ~720H secondary inductance to the primary side, L1=288mH, what do L2 & L3 represent in your model?
Inductance varies as to the square of turns. L2 and L3 share the same field.
L2 represents half the number of turns. Twice the number of turns would yield 4 times the inductance. The load is connected at 2X.
Since L2 is half the number of turns it represents a step up ratio of 25:1. So .288H x 25^2=180H



If you model this as a single ended design (a single winding of 720 with 15mh leakage) the result is the same.

That is also is a hint as to how low leakage transformers are created. I dabble with winding audio transformers myself and I can tell you that it is as much an art as a science. It requires mathematical gymnastics to achieve your goals. I would recommend Terman's engineering handbook if you wish to gain more knowledge on winding transformers.

Jim

Last edited by jelanier; 6th May 2010 at 02:23 PM.
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