LTSpice model for a commercial toroidal output transformer

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.
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?
 

PRR

Member
Joined 2003
Paid Member
....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?
 

PRR

Member
Joined 2003
Paid Member
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.
 

Attachments

  • dch53-2.gif
    dch53-2.gif
    9.8 KB · Views: 594
Last edited:
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?
 
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:
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.

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.

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.

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.
 
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.

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?
 

Attachments

  • P-P xformer test-1.png
    P-P xformer test-1.png
    66.5 KB · Views: 422
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.

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.
 
So, my output stage now looks like this (inductor series resistances shown in blue):

View 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).
 
Disabled Account
Joined 2013
Google Groups

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.
 
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:
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...0.1² gives induction value of the the CFB winding wrt total primary inductance..
0.1² 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; 3,3² x 1,45 = 15,8.
Simple math :rolleyes:
Unless the specifications of the transformer are incorrect :eek:
 
Last edited:
Unless the specifications of the transformer are incorrect :eek:[/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....
 
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:
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:
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.