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LTSpice model for a commercial toroidal output transformer
LTSpice model for a commercial toroidal output transformer
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Old 6th July 2018, 09:23 PM   #11
45 is offline 45  Italy
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Quote:
Originally Posted by PRR View Post
I believe it means 10% Voltage ratio, not impedance ratio.

So would be a 0.1*0.1 or 0.01 impedance ratio, 40 Ohms.
No PRR. I know this transformer. It's made by Toroidy if Poland. CFb is 10% of Za. Clearly stated by the manufacturer.

TTG-CFB4000PP - Tube output CFB transformer [4kOhm] Cathode Feedback Push-pull - Shop Toroidy.pl

I very much prefer turns % because it is more straightforward to use in practice.

Quote:
Originally Posted by dch53 View Post
Thanks. So each cathode winding is 28.92H.

As an individual winding the inductance will 1/4 of the two windings in series because inductance depends on (turns)^2. So if you have 1/2 turns alone then when you square it it becomes 1/4 of total L.
I have no idea of how the simulator works. I don't use any. Sorry.
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Old 6th July 2018, 11:38 PM   #12
pieter t is offline pieter t  Netherlands
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Quote:
Originally Posted by dch53 View Post
So, my output stage now looks like this (inductor series resistances shown in blue):

Attachment 690455

I've set the series resistance of the CFB windings to 10R for convenience.

I'll run some tests tomorrow and see how much cathode feedback I'm getting and what the THD looks like.
For proper simulation the winding ratios must be translated to inductance ratios.
The output transformer specifications are:
- 580 H total primary inductance;
- 33% UL taps;
- 10% CFB windings.
UL taps and CFB ratios are normally specified being winding ratios.
L1,L2,L3,L4 and L5 are correct.
The CFB windings L6 and L7 however should be 1,45 H each (5,8 H when in series).
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Old 7th July 2018, 12:31 AM   #13
45 is offline 45  Italy
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Quote:
Originally Posted by pieter t View Post
The CFB windings L6 and L7 however should be 1,45 H each (5,8 H when in series).
That's 1%. 10% of 580 is 58 and the single alone is 14.5.

I did some calculations to show the relationship with turns ratios with minimal approx. and getting 57.84H....
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Old 7th July 2018, 07:01 AM   #14
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Please read this above different interpretation of cathode feedback % Ra. It appeared that amount of CFB is about 5% voltage ratio if 16 ohms tap is used as CFB winding, for 10% CFB L6, L7 in all your calculations based of inductance ratio is too large (except 1.45 as previous post), 1.5H is closer.
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Old 7th July 2018, 09:46 AM   #15
45 is offline 45  Italy
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To be sure I have an e-mail to the manufacturer. We should know soon.

Inside the link you provide there is no interpration to do. It only says % CBf which is clearly voltage ratio or they talk about 16 ohm winding which can't cause interpretations. What is Ra? The only thing that comes to my mind is the primary load...that's my interpretation. Anyway this doens't stop simulations both transformers can be simulated....

Last edited by 45; 7th July 2018 at 10:12 AM.
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Old 7th July 2018, 10:09 AM   #16
pieter t is offline pieter t  Netherlands
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Quote:
Originally Posted by 45 View Post
That's 1%. 10% of 580 is 58 and the single alone is 14.5.

I did some calculations to show the relationship with turns ratios with minimal approx. and getting 57.84H....
No it is 1%.
CFB winding is 10% of total primary.
Total primary is 580 H.
10% is 0.1; this squared...0.1 gives induction value of the the CFB winding wrt total primary inductance..
0.1 is 1%, so CFB winding inductance is 5,8 H; half winding is 1,45 H.
Also check with the correct values of L3 and L4 in post #13 (15,8 H) for the 33% UL taps.:
33 / 10 = 3,3; 3,3 x 1,45 = 15,8.
Simple math
Unless the specifications of the transformer are incorrect

Last edited by pieter t; 7th July 2018 at 10:16 AM.
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Old 7th July 2018, 10:17 AM   #17
45 is offline 45  Italy
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Unless the specifications of the transformer are incorrect [/QUOTE]

Elementary math, poor irony! It's a matter of definition given by the manufacter.
That is only true if it is 10% CFB. The manufactures says it's 10%Ra. So what is Ra? To me it is 1/2 of plate-to-plate load and referred to the individual winding. That results in a different ratio when referred to voltage or turns. We' ll know soon....
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Old 7th July 2018, 10:25 AM   #18
pieter t is offline pieter t  Netherlands
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Quote:
Originally Posted by 45 View Post
The manufactures says it's 10%Ra. So what is Ra? To me it is 1/2 of plate-to-plate load and referred to the individual winding. That results in a different ratio when referred to voltage or turns. We' ll know soon....
Yes, the manufacturer will mean 10% to Ra being the half primary; it remains 10% for the whole primary.
The manufacturer says:
UL taps 33%;
CFB 10%.
Normally the percentage of UL taps, in this case 33%, is a winding ratio (or voltage ratio if you prefer).
Then the 10% CFB also is a winding ratio.
Ra is not interesting now to get the right inductance values into the simulation; that's the next step (the R values in the sim).

Last edited by pieter t; 7th July 2018 at 10:34 AM.
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Old 7th July 2018, 10:33 AM   #19
45 is offline 45  Italy
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Cathode feedback referred to impedance is NOT wrong. It's an alternative way. ALL Japanese transformers with cathode windings are defined this way in my knowledge. For example, the Tamura F2021 has a specific 16R CFB winding.

Last edited by 45; 7th July 2018 at 10:36 AM.
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Old 7th July 2018, 10:40 AM   #20
pieter t is offline pieter t  Netherlands
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Nobody says it is wrong; yes an alternative way.
But for getting the right inductance values into the simulation, it does not make a difference wether you calculate in impedance or voltage ratio; the outcome will be the same.
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