Matching source and input impedances in power amplifiers

Maximum power is always transferred when Zo = Zl.

That doesn't mean it's best.. most audio equipment has Zl >> Zo, and speakers handle best when Zl >> Zo; amplifiers tend to go "poof" when Zl ~ Zo!.

RF equipment is (always?) designed to accommodate this matched condition, primarily for two reasons (that I can think of): power efficiency, and mismatched loads cause nasties on your line, namely reflections (manifest as SWR).

BTW, you would want Zin higher than Zs, otherwise your voltage disappears (and most equipment is voltage-sensitive).

As for power delivered to the *load*, more power will be delivered IFF the power amplifier is still linear. If it's saturated, more drive isn't going to do anything.

Tim

--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

 wrote in message 
news:1150852255.387458.116160@i40g2000cwc.googlegroups.com...
> Hi.
>
> I have a question concerning impedance matching in RF power amplifiers.
>
> Given a power amplifier with input impedance Zin and output impedance
> Zo, a source with impedance Zs and a load Zl - does conjugate matching
> Zin and Zs give the highest output power?
>
> I know that matching Zo to Zl will transfer the most power to the load,
> but shouldn\'t Zin be much greater than Zs so the amplifier "sees" the
> whole input signal? (ignoring reflections and noise performance)
>
> Regards
>

Properly "ignoring reflections" is a _sure_ way to invent "0" output power amplifier. If the math has not changed since my learning days the max. power transfer always peaks when Za=Zb. (Z - imaginary number(s) and do the dP/dt calculations)

Have fun.

Stanislaw Slack user from Uladulla.

Yes, if the amplifier is unilateral (S12=0).

In the more general case of a bilateral network (S12 is not 0) changing the output's load will change the impedance looking into the input of the amplifier, and changing the input's load will change the impedance looking into the output of the amplifier.

In terms of reflection coefficients and S-parameters, you end up with a set of two simultaneous equations: conjugate(Gamma_source) = S11 + (S12*S21*Gamma_load)/(1-S22*Gamma_load) conjugate(Gamma_load) = S22 + (S12*S21*Gamma_source)/(1-S11*Gamma_source)

which can be solved to give Gamma_source and Gamma_load required for maximum transducer power gain.

You can't ignore reflections. If you make Zin very high, you will indeed maximise the voltage incident on the Zin terminal, but the magnitude of the reflection coefficient between the transmission line and Zin will be close to unity - i.e. almost all of the power incident on the PA's input will be reflected back to the source. If little of the power available from the source is absorbed by the amplifier, then whatever the power-gain of the amplifier, little will be available at its output for delivery to the load.

--
Rick

So the biggest reason to match is to prevent reflections?

It makes sense why you'd want to match Zl to Zo (assuming everything is unilateral) to transfer the most power FROM the amplifier to the load.

But why would you want to transfer the most power TO the amplifier? Aren't most amplifiers inherently voltage or current sensing?

If the amplifier had say a common gate structure, won't there be more output power (with a matched load) if you maximized the input voltage (vgs) by making Zin >> Zs (given the frequency is low enough to disregard reflections)?

rick H wrote:

At RF frequencies, you can't really speak of voltage or current by its self. At DC you can have impedances high enough that the current can be ignored or low enough that voltage can be ignored. At RF frequencies, you are stuck in the mid band of impedances where neither can be ignored.

The semiconductor devices that do the amplifying, have an input impedance. There is likely some network between them and the input connection of the amplifier that transforms this impedance.

With a common gate amplifier, the impedance is low. You can't ignore the current even at DC.

--
--
kensmith@rahul.net   forging knowledge

Hello Stanislaw,

Learned that the hard way, talking to the amp to hold on just a couple more minutes. POOF. It blew the coax though, not the amp.

Not for the output, usually. Modern broadband stages have an output impedance much lower than what's connected. Pretty much like audio amps do.

--
Regards, Joerg

http://www.analogconsultants.com

Modern amps generally have a lower impedance than the cable and the antenna. What matters is that cable and antenna are matched to each other so nothing is reflected.

As Ken said RF amps are far from ideal. If you drive the gate of a few big FETs or the control grid of a large tube you need very little energy from the driver. But the filters, resonant circuits or whatever might be up front is not going to be without losses. IOW their Q isn't infinite. If you are driving bipolar RF power transistors your drive level will hardly be more than 20dB below the output level for a single stage amp in the MHz range. Above 100MHz that's going to be more like 10dB.

That's how it used to be done in tube amplifiers. For the output something similar was done: You would try to find the "sweet spot" for the tube, where it could deliver the maximum of power into the typical

50ohm load without exceeding its plate current limits. This required steep transfer ratios and pretty high-Q inductors. I have spent more than one tube of metal polish to keep that coil shiny so it wouldn't unsolder itself in a big pyrotechnic show.

--
Regards, Joerg

http://www.analogconsultants.com

Been there, done it!

Dust your dictionary and travel to 'efficiency', in the old days it was used in engineering calculations.

Have fun

Stanislaw Slack user from Ulladulla.

That depends a lot on how you define the output impedance. Most good power amplifiers have protection circuits that lower the output voltage if the load impedance is too low. These circuits make the short term output impedance and the long term impedance differ.

If you set out to measure the impedance by connecting a load and very slowly its impedance while watching the output power, you usually get an impedance nearer the intended load.

This is true at low frequencies but at RF frequencies, the gate of a MOSFET can have a significant power requirement.

See

Notice that the power gain is under 20dB at 175MHz.

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kensmith@rahul.net   forging knowledge

I'll confess to not following this complete thread, so if I'm off-base forgive me, but I have to jump in here. If you are speaking of the amplifier output connected to a transmission line, then a mismatched load will cause reflections on the line, but the amplifier output impedance has -nothing- to do with it. The load produces the mismatch, not the source.

More often than not, looking back into the output of an RF PA you will

-not- see an impedance match because the impedance matched condition is not the condition for greatest efficiency.

If you are considering the input of the amplifier as the load for an interstage transmission line, then that might well cause reflections on the line but in narrow-band amplifiers this if often of little consequence.

I have no idea what this means.

But it goes non-linear long before "saturation". Driving beyond the linear region does produce more power, even if it's contained in distortion products. [g]

Wes

Ah yes, good point- the impedance only means something when there's energy coming at it. I've driven 50 ohm cable with a MOSFET before, inarguably a low impedance, and there's no problem whatsoever. Termination at the other end makes a difference, though.

Reflected energy can, of course, be reflected again, when Zo != Zline != Zin.

True, something like a triode amp has Zo (due to internal feedback) about a third of maximum power output load impedance; transistors might present an open circuit...depending on how hard they are pushing.

That should be "if anything, you want Zin > Zo", for instance when connecting audio equipment. Few inputs look like shorts, and few are designed to respond to current rather than voltage as such. Current and voltage are usually closely related for RF so this is more of a low frequency case.

Tim

--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

No. Reflections are just fine; a quarter-wave transformer used to match two different resistances, for example, *relies* on reflections between the transmission line and the source and load to perform the match.

Nor does the OP's question relate to a typical scenario (maximising transducer power gain). Usually one looks at the load-pull data for the amplifier in question, then you match your source and load such that their impedances seen at the input and output ports of the amplifier give you the gain, noise-figure, IP3, efficiency and stability that you want.

Unless you perform some sort of match, which takes into account the length of the transmission line feeding the source to the amplifier, then you've potentially got no idea what the voltage will be at the input terminal of the amplifier. Of course this isn't the case when you've got a nice block diagram with 50 Ohm input/output impedances and

50 Ohm lines everywhere - but this isn't always the case in practice.

If the frequency is "low enough to disregard reflections", then it ain't RF, and you default back to the low-frequency case of the voltage and current looking the same at all points along any given conductor. Under those circumstances, a high input-impedance would indeed cause the input-voltage to look like the unloaded generator voltage. In the RF case, however, this circumstance would just mean that the voltage *incident* on the amplifier's input will looks like the generator voltage, but refelections (which depend on the amplifier's input impedance loading the transmission line) would add to the incident voltage to create the composite terminal voltage. The phase of the reflected wave would be dependent on the capacitance Cgs and could produce a terminal voltage anywhere between zilch (high Cgs providing a low-impedance shunt) to nearly the same voltage as the unloaded generator.

Wes Stewart wrote: ...

In addition, an amplifier is generally designed to produce suitably low distortion products, both in-band (in the modulation bandwidth) and out of band (i.e. harmonics). Proper performance in this regard depends on the amplifier being loaded with an impedance within some tolerance of the one it was designed for. There is NOTHING that says the amplifier's output impedance must be the same as the design load impedance; in fact, with negative feedback, you can cause the output impedance to be just about anything you want, with little effect on the optimal load impedance.

Someone already mentioned that audio amps typically have a high "damping factor"--that is, a very low output impedance relative to the load they are intended to drive. Perhaps a more graphic example of how we use loads which are NOT matched to the source impedance is found in AC power wiring: the source impedance is quite low compared with the load. You would NOT want your 120V line to sag to 60V by putting a matched load on it! (That would blow a fuse/breaker very quickly.)

If the amplifier takes significant power to drive, it would be a good idea for the amplifier to present a reasonable load to the driving stage, so the driving stage doesn't have to be over-rated for the job its doing. And in some cases, reflections are bad. You'd want to avoid them in a video system, for example, because they'd cause a "ghost" image. Given that the driving stage output impedance may not match the line used to connect to the amplifier, you'd do well to have a reasonable match at the amplifier end, at least. Also, if the amplifier is linear, the output power will be proportional to the input power, up to the point where the amplifier becomes nonlinear. So if the driving stage can produce only enough clean power to drive the amplifier, you'd want a match to get that power to the amplifier.

In summary, I don't think there's one simple "one size fits all" answer for the O.P. There are many nuances to consider. Simple complex-conjugate matching per EE/physics circuit theory is the least of your worries, and likely isn't even the best place to begin thinking about the issue. I'd start with something more like, "what system will give me the power I want, with low enough distortion products, and high enough efficiency?" The matching will fall out of the answers to that.

:-) Yes, indeed.

Cheers, Tom

Hello Stanislaw,

Done that back at the university. Told the professor that if he was right about Z_out = Z_antenna the German Radio transmitter about 20 miles from there would have melted down. Transmitters do have efficiencies around 80-90% and that's not possible unless Z of the final stage is lower.

Take a look at a modern transistorized short wave power amp. Load it with 50ohms. Disable the SWR bridge, add 100ohms or even another 50ohm in parallel and see what happens (provided the power supply can stomach it and you don't do it for long).

--
Regards, Joerg

http://www.analogconsultants.com

Hello Ken,

That has nothing to do with the amp's output impedance itself. It is just a protection against over-current because they won't select transistors with too much excess capability here for cost reasons. Pretty much like what you find in high-end audio amps that could drive

2ohm speakers but where the mfg prevents users from doing that.

I have tried it but that was because I had to drive a 30ohm transducer and needed the power. The 50ohm amp did that no sweat after disabling it's protective gear. Basically I stripped it to be a pure amp, no current monitoring and no SWR monitor. However, I had to stop after a few minutes because the insulation of the wire on the output toroid began to smell.

FETs aren't the ideal part for 175MHz. I did it once with a VMOS device but never again. Too fickle and bipolars were cheaper.

--
Regards, Joerg

http://www.analogconsultants.com

Tested or calculated?

150mW into 6m Andrew hi-performance dish gives reliable link over 100Km.!! "Stealth"(hic!) planes can be detected reflecting mobile phones "noise". Math, what is it? Forgive me that I mentioned "efficiency".

Have fun

Stanislaw Slack user from Ulladulla.

Joerg wrote: ...

I hope he/she told you to go think more deeply about it. A model which assumes a source built internally like a Thevenin equivalent is not necessarily an accurate model. I can, using feedback, adjust the output impedance of an amplifier over a large range, with essentially no change in efficiency or operating point for a given load. Would an amplifier with a negative output impedance have efficiency greater than

100%? Of course not, but the Thevenin equivalent might predict so.

If you're still not convinced, consider a class-C RF amplifier stage which is optimally loaded with 5000 ohms at the plates of the output tubes/valves, and which presents a source impedance, as you say, somewhat lower at the plates, let's say 1000 ohms. It can be matched to a 50 ohm load (so the 50 ohm load presents a 5000 ohm resistive load to the plates) with a network of capacitors and inductors which, depending on the design, will transform that 1000 ohms into something very different from 10+j0 ohms, likely rather reactive. In fact it could also be coupled to the 50 ohm load with a quarter wave of 500 ohm transmission line, which would present the desired 5000 ohms to the plates, but would transform the 1000 ohms to 250 ohms, making the transmitter output impedance much HIGHER than the optimal load impedance.

Cheers, Tom

In article , Joerg wrote: [...]

It depends on how you define "output impedance". If you define it as the variation in output voltage vs load current, the impedance is as I stated.

The protection reduces the output voltage when the load current increases therfor, it is part of the output impedance of the circuit.

[....]

You changed the output impedance by those modifications. If you hadn't made them, the output voltage would have been less at that load current.

Maybe not, but the question was about the power into the gate of a MOSFET when doing RF. I picked a case, I knew, to use as a counter example.

--
--
kensmith@rahul.net   forging knowledge

...

That's also true of an RF power tube/valve, even when there is no DC grid current. The power gain at DC may be very high indeed, but not so at high frequency.

Cheers, Tom