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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, 04:52 AM   #1
dch53 is offline dch53  Australia
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Default LTSpice model for a commercial toroidal output transformer

Hi. I've created a model for a commercial toroidal output transformer with cathode feedback. It's not behaving as expected in amp simulations or simple test simulations.

Here are the relevant specs:

Ultralinear tap 33%
CFB Windings 10% Ra
Secondary Impedance 4 and 8 Ω
Primary Impedance 4 kΩ
Turns Ratio (Np:Ns) 31.62:1 (4Ω) , 22.36:1 (8Ω)
Primary Inductance Lp 580 H
Primary Leakage Inductance Lsp 4.31 mH
Total Primary DC Resistance 65Ω
Effective Primary Capacitance 1.8 nF

For 8Ω Ls = Lp/n**2 = 580/22.36 = 1.16H.
A UL tap of 33% means primary windings of 193H, 97H, 97H and 193H.
CFB: not really sure about "Ra" but anyway10% CFB means CFB windings of 580/2 * 0.1 = 2.9H.

I've distributed series resistance proportionally.

I've assumed that the Effective Primary Capacitance is distributed amongst the parallel capacitance parameters assuming the primary winding capacitances are in parallel.

Here is the transformer model in a test simulation:

Transformer test model.png

In this simulation, if I try to calculate Rp as Vin RMS/Iin RMS I don't get 4K as I should.

So, what have I done wrong here?
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Old 6th July 2018, 05:56 AM   #2
PRR is offline PRR  United States
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Quote:
Originally Posted by dch53 View Post
....In this simulation, if I try to calculate Rp as Vin RMS/Iin RMS I don't get 4K as I should.
So what DO you get?

Counting on thumbs, I get 3,887r, which is "exactly 4K" for practical purpose.

L1 L2 on coupled core sum as the square of the roots. So 193H+97H makes 563H. 563H/1.16H makes 486:1 impedance ratio. 486*8r makes 3,887r. *For one side!*

In _my_ simulator (which I did not try), I know the floating windings ends will make it complain directly. Of course it would work in practice, which shows that simulators are awful dumb. Are you sure your sim isn't throwing an error but not displaying it to you?
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Old 6th July 2018, 06:10 AM   #3
PRR is offline PRR  United States
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Ah, I see what you've done.

The primary values you propose do not account for coupling. I re-figured to get proper end-to-end values, and UL taps, and K taps though I did it too fast and do not trust me.
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File Type: gif dch53-2.gif (9.8 KB, 268 views)

Last edited by PRR; 6th July 2018 at 06:18 AM.
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Old 6th July 2018, 08:30 AM   #4
dch53 is offline dch53  Australia
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Quote:
Originally Posted by PRR View Post
Ah, I see what you've done.
The primary values you propose do not account for coupling. I re-figured to get proper end-to-end values, and UL taps, and K taps though I did it too fast and do not trust me.
Thanks very much for that. The amp simulation is much better now. A few questions:

1. How did you calculate those values? I've done a lot of searching but couldn't find anything helpful. The links I've used for modeling transformers don't mention anything about converting actual inductances for modeling purposes.

2. If I stick your values in my test model:

Transformer test model.png

Vp over Vs is 20.76 - close to the specified 22.36. However, if I measure and calculate Rp from Vin RMS over the RMS current in L1, I still don't get 4K. Am I missing something fundamental here?

3. Do I have to tell LTSpice that my transformer has a air core?

4. How do I distribute the 1.8n effective primary capacitance across the 4 primary windings?

5. No secondary resistance is specified. My original Tx model used 0.7 for the secondary effective resistance. Use that?
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Old 6th July 2018, 08:52 AM   #5
Koonw is offline Koonw
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Quote:
1. How did you calculate those values?
You verify the values like this: insert a series a resistor value=Ra in between the generator and primary coil and verify that all the voltages are 1/2 as if no resistor is in series. There is more in Intactaduio: :: View topic - Drop down transformer models for LTSpice
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Old 6th July 2018, 11:12 AM   #6
tikiroo is offline tikiroo  New Zealand
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Inductance from centre tap to UL tap is:

Ls = Lpp*(UL fraction)^2/4 = 580 * 0.33^2/4 = 15.8H

Inductance from UL tap to plate is:

Lp = Lpp*(1-UL fraction)^2/4 = 580 * (1-0.33)^2/4 = 65.1H
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Old 6th July 2018, 11:30 AM   #7
45 is offline 45  Italy
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Quote:
Originally Posted by dch53 View Post
CFB: not really sure about "Ra" but anyway10% CFB means CFB windings of 580/2 * 0.1 = 2.9H.
10% CFb is referred to Za-a so its 400R cathode-to-cathode which means a step-down ratio of 7.071.

Because the plate windings have 22.36:1 step down ratio then when CFb is on the total step-down ratio is 29.431:1 or in other words the CFb windings cathode-to-cathode have about 24% of the total turns. So the total inductance of the cathode windings is 580*(0.24/0.76)^2=57.84H

Last edited by 45; 6th July 2018 at 11:52 AM.
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Old 6th July 2018, 11:33 AM   #8
dch53 is offline dch53  Australia
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Quote:
Originally Posted by dch53 View Post
Vp over Vs is 20.76 - close to the specified 22.36. However, if I measure and calculate Rp from Vin RMS over the RMS current in L1, I still don't get 4K. Am I missing something fundamental here?
The end-to-end rms voltage divided by the rms current should equal ~4k, you seem to be only looking at centre tap to end voltage.

Quote:
Originally Posted by tikiroo View Post
Inductance from centre tap to UL tap is:

Ls = Lpp*(UL fraction)^2/4 = 580 * 0.33^2/4 = 15.8H

Inductance from UL tap to plate is:

Lp = Lpp*(1-UL fraction)^2/4 = 580 * (1-0.33)^2/4 = 65.1H
Thanks. I understand how this works now.

Quote:
Originally Posted by 45 View Post
10% CFb is referred to Za-a so its 400R cathode-to-cathode which means a step-down ratio of 7.071.

Because the plate windings have 22.36:1 step down ratio then when CFb is on the total step-down ratio is 29.431:1 or in other words the CFb windings cathode-to-cathode have about 24% of the total turns. So the total inductance of the cathode windings is 580*(0.24/0.76)^2=57.84H
Thanks. So each cathode winding is 28.92H.

Quote:
Originally Posted by tikiroo View Post
The end-to-end rms voltage divided by the rms current should equal ~4k, you seem to be only looking at centre tap to end voltage.
So, my test circuit looks like this:

Transformer test model.png

When I run it I get Vin RMS = 705.55mV and I L1 = 3.72mA RMS giving a Rp of 189.66Ω.

So I don't have something right.

I tried putting a 4K resistor between the input and L1 and this didn't halve the voltage at the top of L1.
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Old 6th July 2018, 02:44 PM   #9
dch53 is offline dch53  Australia
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So, my output stage now looks like this (inductor series resistances shown in blue):

output stage.png

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.

Quote:
I tried putting a 4K resistor between the input and L1 and this didn't halve the voltage at the top of L1.
Should be 2K resistor, Ra=2k, Raa=4k, no?
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File Type: png P-P xformer test-1.png (66.5 KB, 137 views)
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Old 6th July 2018, 03:11 PM   #10
PRR is offline PRR  United States
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Quote:
Originally Posted by 45 View Post
10% CFb is referred to Za-a so its 400R cathode-to-cathode which means a step-down ratio of 7.071....
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.
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