I'm looking at a RF transformer filter design that appears to be using the primary-side transformer inductance as part of a second-order band-pass filter (first schematic). This isn't the only filter in the design, but I believe I have isolated it properly. The center frequency is approximately 14MHz. My understanding is that the transformer is providing the inductance necessary for the filter as well as presenting the load impedance in parallel.


simulate this circuit – Schematic created using CircuitLab

Now I'd like to use a generic "pulse" transformer, like those used in Ethernet or ADSL modems, but those have much higher winding inductances that make it practically impossible to design as part of a HF resonant circuit (e.g. 100µH or more). To compensate for that, I would need to add a separate inductor to get the effective parallel inductance to be correct (as shown in the second schematic). This eliminates the economy of the transformer acting as the inductor, but what else do I lose?


simulate this circuit

As requested, I went ahead and simulated the circuit using LTSpice, assuming a perfect magnetic coupling. The -6dB is due to the source impedance of the source. As modeled, the simulation results are identical.

enter image description here

  • $\begingroup$ As drawn, schematic #2 has L2 and L3 wired in parallel, providing a very different parallel resonant frequency with C2. I'm curious what simulation shows the effect to be. I would not predict the results you are looking for. $\endgroup$ – user2338215 Aug 3 '14 at 14:55
  • $\begingroup$ @user2338215 But inductors in parallel act like resistors in parallel. The effective inductance on the primary is the same in both circuits. I ran the simulation and I have edited the question to include the simulation results. The simulation does in fact show that they behave the same. $\endgroup$ – W5VO Aug 3 '14 at 15:12
  • $\begingroup$ My bad. Just eyeballing it, the equivalence wasn't apparent to me. One consideration though: perfect magnetic coupling may not be a good assumption here. L2 and L3 will be sharing the current (like parallel resistors) and it seems that the coupling to L4 will be effectively reduced, given it comes strictly from L3. $\endgroup$ – user2338215 Aug 3 '14 at 15:36
  • $\begingroup$ @user2338215 When I talk about "coupling", I mean how the simulator treats the two inductances. With an "ideal" coupling of 1, that only means that L3 and L4 are magnetically linked to form an ideal transformer. This is a fairly reasonable first-order approximation since they share a magnetic core. In the case of the second schematic, the transformer is only presenting the 50Ω load from the secondary back to the primary and no real power flows into L2. $\endgroup$ – W5VO Aug 3 '14 at 17:52
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    $\begingroup$ Thanks for addressing my questions. My own inexperience with tuned transformer theory was confusing me. I found in engr.uky.edu/~gedney/courses/ee521/notes/… a design similar to what you are proposing in your second schematic. As jcoppens suggests, the reference includes an adjustable C2 to cancel the shunt inductance and compensate for other uncertainties introduced by the transformer. Thanks again. $\endgroup$ – user2338215 Aug 3 '14 at 20:31

It seems an excellent idea. The only possible problem I could imagine, is maybe the internal capacitance of the 1:1 pulse transformer -I've seen numbers in the order of 30 - 50 pF. This isn't a problem in non-resonant (ethernet) apps, but you might have to adjust your 220 pF to compensate.

Just found another datasheet with only 15pF inter-winding capacitance.

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