Simulate coaxial ferrite transformer with QUCS?

Thread Starter

BRUNET Joss

Joined Nov 1, 2018
26
Hello everyone,

I am using coaxial ferrite transformer for two years in my "wideband" power amplifier (MPN SD2931-10 / approx. 70-170 MHz / P1dB 200W).
I was only following application note but now, I am looking to understand further.

To do that, I plane to use QUCS to simulate. First, I need to write some specs:
Hybrid combiner at working frequency 1.8-50 MHz with power approx 1000W.
Equivalent commercial product can be founded on ethernet with key word "1.8 to 60MHz 2.5KW Hybrid Splitter/Combiner Set".
Material ferrite will be 31 or/and 43.
I am planing to realise the hybrid transformer of the figure 8 or 11 of the attach application note (Motorola AN-749).

To begin, I want to get a good model of a transformer that I can confirm with VNA measure.
I have bought some Fair-Rite 5943003801 and I plane to use 50R semi-rigid coaxial around it.
So, below, the first simulation

1643210697303.png

For TR1, threes parameters need to be define: Lpri, Lsec, K and L1.
The problem is how can I know this 4 parameters?
Have you already done similar simulation?
 

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UweX

Joined Sep 2, 2020
16
Somehow the pages are unsorted, it took me a while to figure it out. So don't be confused !
As a rule of thumb I would use ten times the impedance involved, so 10*50 Ohm = 500 Ohm at the lowest operating frequency. Depending on the winding ration you get the secondary inductance. The ideal coupling factor is 1.0, start with that.
In the power level you are in it maybe a good advice to also take the wire resistance of the rigid cable into account. Besides the magnetic losses that maybe the second largest loss.
 
Hi Joss.
This get complicated and the traditional transformer model does not work too well. I suggest a model that looks like this as a starting point.
Transmision line transformer model.png
Line1 and Line2 together represent the length of coax you have used. L1 and L2 are the same value and can be found by multiplying the Al value of the core by the square of the number of turns. Use a k value of 1 for Tr1. R1 and R2 should be of equal value and represent the core losses. This part is very rough approximation as the permittivity of ferrites like the type 43 vary a lot over frequency. Inductors wound on these ferrites tend to be more resistive than inductive at VHF and UHF. You can get a clue to the value from graphs for the material.
2022-02-01 11_37_47-Start.png
This shows the crossover between inductive and resistive at 60MHz for type 43.
For PA applications the core will also be non-linear if it is going into saturation. This in not covered in this model.

Chris.
 

Thread Starter

BRUNET Joss

Joined Nov 1, 2018
26
Hi UweX and Chris,

Thank you for your both answer.

For the inductance value, I have an Al coefficient of 1075, so 1075*12^2=0.155 mH.
I'll try the model of Chris.
For the u', I have understood the difficulty to keep the right value for all the frequency band.

I begin to get some results in practice but not in Qucs.
I'll let you inform.

Joss
 
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