Matching source and input impedances in power amplifiers

[[email protected]]

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

 

[Tim Williams]

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

 

[Stanislaw Flatto]

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.

 

[rick H]

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?

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.

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)

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.

 

[[email protected]]

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)?

 

[Ken Smith]

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?

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.

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)?

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

 

[Joerg]

Hello Stanislaw,

Properly "ignoring reflections" is a _sure_ way to invent "0" output
power amplifier.

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.

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)

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

 

[Joerg]

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.

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.

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

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.

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)?

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.

 

[Stanislaw Flatto]

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.

Been there, done it!

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

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

Have fun

Stanislaw
Slack user from Ulladulla.

 

[Ken Smith]

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.

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.

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.

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

See
http://www.semiconductors.philips.com/acrobat_download/datasheets/BLF242_3.pdf

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

 

[Wes Stewart]

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).

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.

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

I have no idea what this means.

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.

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

 

[Tim Williams]

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.

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.

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.

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.

I have no idea what this means.

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

 

[rick H]

So the biggest reason to match is to prevent reflections?

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.

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?

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 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)?

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.

 

[Tom Bruhns]

Wes Stewart wrote:
....

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.

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 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.

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.

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

Yes, indeed.

Cheers,
Tom

 

[Joerg]

Hello Stanislaw,

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

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).

 

[Joerg]

Hello Ken,

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.

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.

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

See
http://www.semiconductors.philips.com/acrobat_download/datasheets/BLF242_3.pdf

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

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.

 

[Stanislaw Flatto]

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).

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.

 

[Tom Bruhns]

Joerg wrote:
....

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.

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

 

[Ken Smith]

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.

That has nothing to do with the amp's output impedance itself. It is
just a protection against over-current[/QUOTE]

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.

because they won't select
transistors with too much excess capability here for cost reasons.

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

[....]

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.

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.

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.

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.

 

[Tom Bruhns]

....

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

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