load the tuned circuit

[Pat Pending]

High Hfe (or beta if you prefer) often makes very little difference at RF. A BC559C witha typical hfe of over 500 has an Ft of 300MHz. The actual current gain falls to unity at Ft. Fortunately gm (mutual conductance) doesn't fall in the same way. Or have I got that wrong?

I forgot about posting this before I went on holiday!

Jeffrey makes a good point, and in an effort not to be lazy, I dug out my old dynamic gain tester that I built as a lad many decades ago a while ago, from an article in the "Radio Constructor" (May '65 Edition).

With a fresh battery, (and clip!) It still worked, and although it only gives comparative results it showed that the OC170 that had a beta of 85 was only half the dynamic gain of the BC559 with its beta of 600, at 2MHz.

This is all a bit of topic, but may be worth some study and its own thread sometime as many of us use modern general purpose transistors to replace old soldiers in vintage sets now and then.

Andy.

[reset]

A paradox of the impedance matching for power transfer is that half of the total power is dissipated uselessly in the source. The other half usefully in the load.

Maybe I've misunderstood something, because I'm sure transmitters have been made that give better than 50% efficiency from electricty in to power in the aerial.

You haven't missed anything BUT the use of the rule of thumb is misapplied. If you have a power generator that produces 100 mega watts and feeds the power grid it will not consume 100 mega watts at full load. 60 years ago more or less when I was taught Z matching and the Zs = Zl at max power transfer it was to show the wasted power in the source (batteries). It was a bit of common since but seems to have been misused and distorted. At Zs = Zl the voltage across the load is 50% of supply. In other words if I load a 12 volt battery to get max power out I have 50 % of the voltage dropped internally in the battery and only 6 volts at the load.  Any further loading will produce less voltage on the output at higher current. Since power = current X voltage we hit the point of diminishing returns at Zs = Zl. In the '60s this would lead into a rule of thumb that Zs should be a minimum of 1/10th of Zl. Even with the 10 to 1 ratio the source would dissipate 9% which is a lost in a power circuit.
Today you will see recommendation for a 200 to 1 ratio in some manufacturer's data.
Look at any  tube datasheet for plate resistance and plate load. Are they equal? Look at the grid resistor, is it equal to the plate load of the previous stage?
Well I've rambled enough.
Well one other thing. Look up an 811 tube and look at the plate inut power vs the output. It outputs about 5 times what it dissipates. (the only reason I said 811 is Collins used them in equipment I worked on)

[ppppenguin]

I'm well aware that power matching is a nonsense for applications such as mains and batteries. However a transmitter needs to work into a matched load to prevent reflections. If the the load is accurately matched to the feeder then ultimately the tx match to the feeder doesn't matter too much. Perhaps the gap in my knowledge is about the practicalities of tx to feeder to aerial matching.

[reset]

I'm well aware that power matching is a nonsense for applications such as mains and batteries.  However a transmitter needs to work into a matched load to prevent reflections. If the the load is accurately matched to the feeder then ultimately the tx match to the feeder doesn't matter too much.  Perhaps the gap in my knowledge is about the practicalities of tx to feeder to aerial matching.

This is a cut and paste from a discussion on a group that is for broadcaster.

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Here's a bit of technology talk just for information and fun:

We hear many references to "transmitter output impedance" with the "standard" being 50 ohms. It would seem, then, that the best load to present to the transmitter is 50 ohms. But all is not as it appears with output impedance.

The specification that a transmitter has a "50 ohm antenna output" refers to the transmitter being designed to drive this 50 ohm load, but this should not be the output impedance.

The maximum power theorem states that maximum power transfer from a source to a load occurs when the load impedance is the complex conjugate of the source impedance. Complex conjugate means that the reactance in the load is equal and opposite to the reactance in the source and that the source and load resistances are equal. This is essentially what is done when a base loading coil is used with an antenna. The inductive reactance of the coil is equal and opposite to the capacitive reactance of the antenna, in other words it is at resonance.

When the source and load impedances are conjugates the efficiency of power transfer is 50% with half the power being lost as heat in the source. Thus, it is not efficient to have the same source and load resistance so what does this mean? It means that for efficient power transfer the source resistance needs to be much lower than the load resistance (with the reactances equal and opposite). For example, a transmitter designed to drive a 50 ohm load could have an output resistance of typically 5 ohms or less. This would result in 9% of the power being lost in the transmitter for a transfer efficiency of 91%.

The proper load is important for the operation of the transmitter and when it is specified for use with a 50 ohm load this translates into the load for which the final amplifier and filters will work properly and is generally not the actual output impedance.

Neil


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[ppppenguin]

Thanks. That's rather what I suspected. It does mean that the return loss looking back into the tx is poor. But that won't matter too much if not too much power is reflected back from the antenna.