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switch mode RF PA current drain vs. load Z-matching / resonance

Once upon a time, one could manually tune up the Pi-network of a class- C tube amp by minimizing plate current. I never really understood the math of this behavior (in fact, I never SAW it explained anywhere) and did not make intuitive sense to me either.

Why should plate current rise if e.g. the load Z was higher than the device output Z?

OTOH I can imagine that moving away from a parallel LC-resonance in the load might cause the load to absorb exponentialy more current on either the capacitive or the inductive node, reflected by increased plate current. This may mean that a well-matched Pi-network indeed looked like a parallel-resonant resistive load.

I now wonder if switched-mode RF power amplifiers behave the same way.

CON: SMPAs are at heart near-zero Z sources. Within I and V limits of switching devices, their output is determined by how much current a given-Z load will take at a given power supply. The lower the transformed Z seen by the amplifier, the higher the current. Changing the output transformer winding ratio changes the power. On the other hand a C-class amp did have an impedance defined by the amplifying device, which never reached near-zero resistance.

PRO: At a given output transformer winding ratio (or 1:1 in the case of E-class), a matching LC network may still somehow look like a parallel resonant load. The only time I played with a SMPA (push-pull into a balun uptransformer, followed by a low-pass T-network) I noticed a sharp and marked power maximum into a dummy load at a given capacity setting in the T - not just a rise as capacity decreased. I didn't check the input current, alas...

Enlightenment available, anyone? Pointers to the maths?

Reply to
spamhog
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It did not.

When the plate circuit is off-tune, it is a low impedance seen by the RF, so the plate current dips and the plate RF voltage peaks at resonance. For the perfectionsts: not exactly if the tank is a low-Q pi network, but quite.

The tuning procedure is:

Close the load (output) capacitor. This minimizes the coupling to the load. Set the tuning (input) capacitor to approximate resonance position. Apply RF input and tune the input capacitor for a plate current minimum, called a dip. Open the output capacitor slightly and re-tune input for a dip, which is now shallower than at start. Continue the procedure until the dip is a the rated plate current.

The method applies to class AB and B amplifiers in the same way as for class C.

For grounded-grid amplifiers the initial tuning has to be performed with limited excitation to prevent excess grid current.

--

The explanations and math were at least still a
few years ago in the ARRL Handbook. I'm not sure
for the newest editions, though.
Reply to
Tauno Voipio

Thank you Tauno!

Beginning to understand.

- IF impedance rises, THEN plate current has to drop.

- IF the output circuit goes from tuned to off tune impedance and voltage drop and current rises.

I already knew how to go through the motions, but now I understand a bit more. Indeed, there is a resonance, "Tune" ("Plate") and "Load" control do what they say.

I would say that also in the case of a switched circuit detuning would raise the current and peaking should reduce it. In principle it might be possible to do a tuneup by looking at the current variations. Maybe dealing with extremely low impedance devices will keep loaded Q low and make adjustment more mushy, gradual, and potentially confusing than with tubes, but it should work.

Does it make sense?

I'll soon have a sturdy and well balanced (low 2nd harmonic) ~40W push- pull switching amp, currently filtered for 10MHz. I'll do some tests at low power and report here. I want to try a simplified 7-pole "thing" where you first turn 4-section variable to tune a constant-K constant-Z double Pi-section LPF into a dummy load, and then switch in a standard Pi to match the antenna load. With some coil switching it should allow clean output and matching over a very broad frequency range.

Reply to
spamhog

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