16
\$\begingroup\$

According to the maximum power transfer theorem, when a fixed source impedance is given, the load impedance must be chosen to match to the source impedance to achieve maximum power transfer.

On the other hand, if the source impedance is not out of reach from the designers, instead of matching the load to the source impedance, the source impedance can simply be minimized to achieve maximum efficiency and power transfer, it's a common practice in power supplies and audio-frequency amplifiers.

However, in RF circuits, to avoid signal-integrity issues, reflection loss, and damages to the high-power RF amplifier due to reflection, impedance matching must be used to match all the source impedance, load impedance, and also the characteristic impedance of the transmission line, and finally the antenna.

impedance matching

If my understanding is correct, a matched source and load (for example, a RF amplifier output and an antenna), forms a voltage divider, each receives half of the voltage. Given a fixed total impedance, it means there is always 50% of power wasted in burning and heating the RF transmitter itself.

So, is it correct to say impedance matching implies the efficiency of any practical RF transmitter cannot be greater than 50%? And any practical RF transmitter must waste at least 50% of energy?

\$\endgroup\$
  • \$\begingroup\$ cas.ee.ic.ac.uk/people/dario/files/E418/ch2.pdf \$\endgroup\$ – Sorenp Jul 23 at 9:10
  • 1
    \$\begingroup\$ quick summary: although in a 50 ohm environment, there need be no 50 ohm resistances to cause losses. (thanks to Andy for this) \$\endgroup\$ – analogsystemsrf Jul 23 at 11:22
  • 1
    \$\begingroup\$ Thing about an audio amplifier, to exclude complications caused by RF. A typical good quality amp is intended to drive loudspeakers with a 4 or 8 ohm impedance, but the output impedance of the amp is around 0.01 ohms. Almost all the output power is dissipated in the speakers, not in the amp's output stage. The output power is limited by the maximum voltage swing at the output and the external load impedance, not by the impedance of the amp itself. \$\endgroup\$ – alephzero Jul 23 at 20:42
19
\$\begingroup\$

If your power source is a zero ohm output voltage source, followed by a 50 ohm resistor, then yes, what you think is correct.

However, practical RF amplifiers (at least ones designed to be efficient) are never built like that. They tend to have a low impedance common emitter or source stage followed by reactive impedance matching, all designed to operate into 50 ohms.

Interestingly, if you buy a general purpose signal generator, the output is usually built as a voltage source followed with a real 50 ohm resistor, as efficiency is not an issue, and having an accurately defined output impedance over a very wide frequency range is the main design goal.

\$\endgroup\$
  • 2
    \$\begingroup\$ So what is the actual efficiency, in practical terms? (You've already implied that it is greater than 50 percent). \$\endgroup\$ – Robert Harvey Jul 23 at 18:10
  • 3
    \$\begingroup\$ @RobertHarvey It's whatever it is. Get a specific circuit, model or measure it, and find out. In fact, one of the signal generators we produced, we needed more efficiency than a 50 ohm resistor would give us (heat dissipation in a small space), so at the expense of output resistance tolerance, we used a series 22 ohm resistor, and servoed the output of the amplifier to synthesise an effective 50 ohms at the output. \$\endgroup\$ – Neil_UK Jul 23 at 20:55
9
\$\begingroup\$

RF amplifiers do NOT in general have an output impedance remotely close to 50R..... They are however designed to drive a 50R load!

Much like audio amplifiers the source impedance is generally far from the design load impedance, because you DONT want maximum power transfer, you want something closer to maximum efficiency!

Depending on the topology, the things approximate either voltage sources (Low output impedance) or current sources (High output impedance).

If you think about a for example, HF, Push pull output stage, the devices are operating at some designed voltage and current, hence some 'impedance' (Usually quite low), which is then transformed to an industry standard 50R.

This impedance is set by the designer to result in some voltage across that 50R load that will deliver whatever the designed power level is. Note that those output devices could be in deep class C or even class F and be operating essentially as switches dissipating near zero power, but I as a designer still need to decide what voltage and what current to chose as an operating point and hence what transformation I need to get to the target power at the output.

Now clearly if you try to run such an amplifier into a load far removed from 50R then the voltages and currents seen by the power devices will be other then the designer intended, and if you go to far the smoke comes out.

A further complication is the output filters and (at UHF and up) the possibility of a terminated circulator at the output which actually makes the thing look like 50R looking back into the input.

\$\endgroup\$
5
\$\begingroup\$

So, is it correct to say impedance matching implies the efficiency of any practical RF transmitter cannot be greater than 50%? And any practical RF transmitter must waste at least 50% of energy?

No, that's wrong. The diagram in your post lacks the essential building block in this discussion: the amplifier itself.

All amplifiers can be described according to their PAE (Power Added Efficiency).

$$ PAE = \frac{P_{out}-P_{in}}{P_{supply}} = \frac{P_{out}-\frac{P_{out}}{G}}{P_{supply}} = \frac{P_{out}}{P_{supply}}\left({1-\frac{1}{G}}\right) = \eta\left({1-\frac{1}{G}}\right) $$

PAE is the key parameter here, because the amplifier gain is likely to be very high. The power transferred TO THE AMPLIFIER by the generator, when impedances are matched, will be only 50% of the maximum generator power indeed. But if gain is high enough then the power wasted in the internal impedance of the generator will be very low compared with the power delivered BY THE AMPLIFIER to the load. Thus, the impact in total efficiency is likely to be low.

What matters here is (mainly) the output stage of the amplifier having a high raw efficiency \$\eta = P_{out}/P_{supply}\$, which depends on the amplification class (A, B, AB, C, D, F, etc.) and the operating point of the amplifier.

\$\endgroup\$
4
\$\begingroup\$

So, is it correct to say impedance matching implies the efficiency of any practical RF transmitter cannot be greater than 50%? And any practical RF transmitter must waste at least 50% of energy?

No it's not correct to say that.

What you must ensure when connecting amplifier to antenna via a cable (normally coax) is that there are no significant reflections of power from the load (antenna) that can damage the amplifier or make it less effective.

If the antenna impedance matches the characteristic impedance of the coax then the amplifier can drive the feed end of the coax without requiring any series source resistance. The impedance seen at the driven end will be the antenna impedance because it matches the characteristic impedance of the cable.

\$\endgroup\$
  • \$\begingroup\$ Let's say I have an RF amplifier, inside, it has a RF power MOSFET with very low output resistance, and the output is connected to a 50 Ohm coax connector, which is then wired to a coax cable to the antenna. In this case, do you mean that the only actual series resistance between the low-impedance MOSFET output and the coax connector is the characteristic 50 ohm impedance of the PCB trace, a RF connector and the coax itself, and it fact there is no actual 50 Ohm resistance anyway in the transmitter, as a result it only dissipates little power? Is this the correct understanding? \$\endgroup\$ – 比尔盖子 Jul 23 at 8:37
  • 1
    \$\begingroup\$ @比尔盖子 That is how I would see it but there may be some reason to have a series resistance in the amplifier output but this need not be for matching purposes. For instance a series resistance may be used to prevent any filters used from having too-high of a Q factor. Other uses are for current limit protection circuits. \$\endgroup\$ – Andy aka Jul 23 at 9:02
  • \$\begingroup\$ And another downvote from the phantom. \$\endgroup\$ – Andy aka Jul 23 at 19:34
2
\$\begingroup\$

Impedance consists of both real (resistive) and imaginary (reactive) parts. Only the real (resistive) part dissipates power. Theoretically one could have a purely reactive impedance with a magnitude of 50 ohms and not dissipate any power in it.

The units of impedance are Volts per Amp. When talking about the impedance of a transmission line we are actually talking about how much current would need to be fed into the line to cause a voltage of a certain magnitude to propagate along the line. Meaning the ratio of the voltage and current.

For example CAT-5 cable has a propagation velocity of about 0.64 * C. It also has a capacitance of about 15pF per foot (48pF per meter). Its impedance is primarily determined by the capacitance between the twisted pairs (there are of course some small inductive and resistive components).

If we put a 1V signal onto one end of the line, the signal will propagate at 192,000,000 m/s, for every 1 meter that the signal travels it will need to charge 48pF to 1V (so 48pC).

1V * 48pF/m / (180M m/s) = 9.44mA.

1V / 9.44mA = 105.9 ohms (which is very close to the nominal 100 ohm impedance).

\$\endgroup\$
-1
\$\begingroup\$

This is correct. A "practical" Amplifier will have to match the output consisting of connectors, cables, antenna. For eventual maximum power delivery to the antenna, >= 50% will get wasted elsewhere.

\$\endgroup\$
  • \$\begingroup\$ Yes, for maximum possible power delivery. However a practical RF transmitter may not be designed to produce the maximum power that it possibly could. \$\endgroup\$ – Bruce Abbott Jul 24 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.