How do I calculate a RF transformer ?

[nenea dani]

Hello colleagues,
   I want to couple a 10,7Mhz ceramic filter with Zo = 330 ohm to AD8307 with Zi = 1,1Kohm differential, 1,4pF each to ground through a ferrite ring transformer Al = 40microH / 100sp.   For each turn in the primary of the transformer I have to have 1.8 turns in the secondary. How do I calculate the number of turns at one of the windings for low losses in 180Khz? I attach the scheme,  C1 = C2 = 100nF .

[David Hess]

Why is 180kHz important for a 10.7MHz bandpass filter?

[Yansi]

The transformer needs to have a impedance transformation ratio of 1:3.33,  that is a 1:1.83 turns ratio.  So you can use for example a 6 to 11 turns ratio (close enough I think).

[nenea dani]

  I know transformers that adapt two Z,  limit  somehow the  BW.  I do not want to go too far below that of the ceramic filter that is 180KHz. It's a typical layout in the AD8307 data sheet at the end (page 21) with this filter. There the adjustment is made with a resistance R4= 464 Ohm. I think that resistance steals too much of the signal strength instead of being used by AD8307. At the same time, I understand that any reactance can disrupt the filter transfer feature.

[Yansi]

Yes, it will consume a few dB. Does it matter though? The overall range of the circuit is still 120dB (at least written there).

Also, the resistor may achieve a tad better match, than using a transformer. Transformers aren't ideal, they have also a magnetizing and leakage inductance. That also will have some effect on the overall impedance, as seen by the filter. I think the more resistive termination is the more predictable one.

[David Hess]

I know transformers that adapt two Z,  limit  somehow the  BW.  I do not want to go too far below that of the ceramic filter that is 180KHz.

A 180kHz difference in frequency at 10.7MHz is insignificant unless the transformer is tuned and even then, it will not matter because otherwise you could just use the transformer as the filter.

It's a typical layout in the AD8307 data sheet at the end (page 21) with this filter. There the adjustment is made with a resistance R4= 464 Ohm. I think that resistance steals too much of the signal strength instead of being used by AD8307. At the same time, I understand that any reactance can disrupt the filter transfer feature.

The only requirement is that the transformer impedance be significantly higher than the source impedance.  When this condition is met, then the load impedance is reflected through the transformer's turns ratio and becomes a majority of the load on the source.

So for instance if the source impedance is 330 ohms, the transformer impedance should be say a minimum of 3.3k (x10) at 10.7MHz which is 50uH.  If I understand your AL specification correctly, that will be about 12 turns.

[nenea dani]

I used a computer on the net, and for T-37-2 /  Al= 40microH /100turns, 50microH give me 112 turns. I probably got something wrong. In principle, I understand that in the primary I have 12 turns and in secondary 21.8 turns? I mean 1.1KOhm from AD8307 is reflected as 330 Ohm in front of the filter and vice versa?   For the time being, I tested several  filters with a 270 Ohm series resistance to a ready-made module by SV1AFN with 50 Ohm at input to AD8307. Only the Murata and TDK filters have almost symmetrical characteristics, centered on 10.7 MHz and low spurious. I attach 3 pictures  for fairness.

[Yansi]

T37-2 core has 4nH magnetic conductivity per turn squared. It is also a low permeability iron powder core (mix no. 2, permeability of about 10), unsuitable for this application (it would be good to make a resonant transformer though).
For 50uH, you would need sqrt(50e-6 / 4e-9) = 112 turns, you have calculated that correct.

To make a wideband un-tuned transformer, you need a high permeability ferrite core instead.

[ogden]

I want to couple a 10,7Mhz ceramic filter with Zo = 330 ohm to AD8307 with Zi = 1,1Kohm differential, 1,4pF each to ground through a ferrite ring transformer Al = 40microH / 100sp.   For each turn in the primary of the transformer I have to have 1.8 turns in the secondary. How do I calculate the number of turns at one of the windings for low losses in 180Khz? I attach the scheme,  C1 = C2 = 100nF .

You anyway do not use transformer as a balun, there is no isolation as well - then why bother? Use just parallel 470 Ohm resistor which together with 1.1K input impedance will appear as around 330 Ohms to the filter output. Further info: check datasheet figure 32, chapter "basic connections".

[nenea dani]

Although it's simple, this 470 Ohm resistor appears as an unwanted consumer.

[ogden]

Although it's simple, this 470 Ohm resistor appears as an unwanted consumer.

It depends. Sometimes you plan to lose some signal power in minimum loss (resistive) matching pad. What is planned signal level range you want to detect?

[edit] As an example check AD607 datasheet, table "Filter Termination Resistor Values for Common IFs" - where both input and output of filter is matched with resistor.

[David Hess]

If you have excess signal to noise ratio, then resistive impedance matching has some advantages.  It is very simple.  The match is broadband so there is none of this nonsense about using a triplexor to terminate the filter into a proper load at other frequencies.

[nenea dani]

[/quote]

"It depends. Sometimes you plan to lose some signal power in minimum loss (resistive) matching pad. What is planned signal level range you want to detect?"

          I have no plan regarding the level of the input signal. I want to build an SA and start from the end to the beginning. I understand from the net that the conception and construction of the SA is a kind of black art. It seems that all sensitivity is in this logarithmic detector and that's why I want it to have no losses.   It just remains to compensate in the IF amp. for losses from mixers and filters.  I just want to adjust the filter between AD603 and AD8307. It seems that the solution with the resistance of 470 Ohm is the most suitable.