### Network

S

#### Steve Bower

Jan 1, 1970
0
Can someone tell me in practical terms, ie. suggested components and
values, how I would match the output of a power op amp to a reactive
load? For this example, we can assume a 100KHz squarewave and a load
impedance of 450 ohms.

Steve Bower

J

#### John Popelish

Jan 1, 1970
0
Steve said:
Can someone tell me in practical terms, ie. suggested components and
values, how I would match the output of a power op amp to a reactive
load? For this example, we can assume a 100KHz squarewave and a load
impedance of 450 ohms.

450 ohms (with no phase angle or other indication of
reactance) indicates pure resistance. Why do you call this

T

#### Tom Bruhns

Jan 1, 1970
0
Can someone tell me in practical terms, ie. suggested components and
values, how I would match the output of a power op amp to a reactive
load? For this example, we can assume a 100KHz squarewave and a load
impedance of 450 ohms.

Steve Bower

impedance versus frequency. If you want to "match" the load to the
amplifier, you need to consider the impedance not only at 100kHz but
at 300, 500, ... for however many harmonics you want to consider.

Also, exactly what does "match" mean? An op amp with negative
feedback applied generally has rather low output impedance, and you
probably do NOT want to present a load to it equal to the complex
conjugate of its output impedance. What are you trying to accomplish
with a "match"? I think you need a clear picture of that before you
can start to worry about how to achieve it.

Cheers,
Tom

P

#### Phil Allison

Jan 1, 1970
0
"John Popelish"
450 ohms (with no phase angle or other indication of reactance) indicates
pure resistance. Why do you call this a reactive load?

** Gotta hunch the OP is really after a "complex conjugate matching network
" to place between his " power op-amp " and the mysterious 450 ohm reactive
load - plus the network needs to be effective from 100kHz to circa
daylight.

Can anyone put him onto suitable CAD software and a procedure to do that ?

Bugger the imaginary need.

Pun intended.

........ Phil

E

#### Eeyore

Jan 1, 1970
0
Steve said:
Can someone tell me in practical terms, ie. suggested components and
values, how I would match the output of a power op amp to a reactive
load? For this example, we can assume a 100KHz squarewave and a load
impedance of 450 ohms.

Well, you certainly can't change the op-amp's characteristics.

What do you mean by 'match' ? You don't 'impedance match' an op-amp to anything.

Graham

S

#### Steve Bower

Jan 1, 1970
0
impedance versus frequency. If you want to "match" the load to the
amplifier, you need to consider the impedance not only at 100kHz but
at 300, 500, ... for however many harmonics you want to consider.

Also, exactly what does "match" mean? An op amp with negative
feedback applied generally has rather low output impedance, and you
probably do NOT want to present a load to it equal to the complex
conjugate of its output impedance. What are you trying to accomplish
with a "match"? I think you need a clear picture of that before you
can start to worry about how to achieve it.

Cheers,
Tom

Maybe an op amp is not the best solution. BTW, I only need to match
one frequency at a time. Not a range, as I understand this is not
practical. The aim is to maximize power transfer.

Disregarding the 450 ohm figure, what is the recommended technique and
signal source for matching a signal to a reactive load? More
specifically, one that has both resistance and capacitance in
parallel.

Steve Bower

E

#### Eeyore

Jan 1, 1970
0
Steve said:

Maybe an op amp is not the best solution. BTW, I only need to match
one frequency at a time. Not a range, as I understand this is not
practical. The aim is to maximize power transfer.

Disregarding the 450 ohm figure, what is the recommended technique and
signal source for matching a signal to a reactive load? More
specifically, one that has both resistance and capacitance in
parallel.

At those frequencies you don't need to match a amplifier to a load !

Just drive the damn load from the output of the op-amp !

What exactly are you worried about ?

Graham

K

#### kell

Jan 1, 1970
0

Maybe an op amp is not the best solution. BTW, I only need to match
one frequency at a time. Not a range, as I understand this is not
practical. The aim is to maximize power transfer.

Disregarding the 450 ohm figure, what is the recommended technique and
signal source for matching a signal to a reactive load? More
specifically, one that has both resistance and capacitance in
parallel.

Steve Bower

Too vague.
What's the app?

J

#### John Popelish

Jan 1, 1970
0
Steve said:
Maybe an op amp is not the best solution. BTW, I only need to match
one frequency at a time. Not a range, as I understand this is not
practical. The aim is to maximize power transfer.

Even if that overloads the power source? By the way, a
square wave includes energy at a lot more than one
frequency, even though it has single number of cycles per
period of time. This is why frequency is measured in hertz,
not cycles per second. I suspect that you want to optimize
power transfer at the fundamental frequency of the square wave.
Disregarding the 450 ohm figure, what is the recommended technique and
signal source for matching a signal to a reactive load?

Adding either parallel or series reactance of the opposite
type to resonate with the reactance at a given frequency, is
very common. Either way, you can hide the energy transfers
through the reactive part. The difference is what happens
at other frequencies, like those harmonics in your square
wave. You might also need a transformer or a more
complicated form of resonator if you need to shift the
resistive component to a different effective value (change
the ratio of voltage to current) to get the desired energy
transfer from source to this load.
More
specifically, one that has both resistance and capacitance in
parallel.

I would probably first try a series inductance that has an
equal magnitude impedance as the capacitive part of the load
at the square wave fundamental frequency (series tuned).
This has the advantage of looking inductive (rising
impedance as frequency rises) at the higher frequency
harmonics of the square wave, so that the opamp doesn't work
very hard to drive them. Don't be surprised to find that
the inductor has more voltage across it than the opamp puts out.

R

#### Rich Grise

Jan 1, 1970
0
What exactly are you worried about ?

His grade on the exam? ;-)

Cheers!
Rich

S

#### Simon S Aysdie

Jan 1, 1970
0

Maybe an op amp is not the best solution. BTW, I only need to match
one frequency at a time. Not a range, as I understand this is not
practical. The aim is to maximize power transfer.

The point with power amplifiers is not to "maximize power transfer."
The point is to maximize efficiency. They are not the same thing.

"Maximizing power transfer" is a small signal concept, and has to do
with internal impedances of the devices.

Maximizing efficiency has to do with extracting as much power from the
power supply as possible, and dissipating as little as possible in the
device. It is a large signal concept.

efficiency := P_out/P_dc

efficiency_pae := (P_out - P_in)/P_dc

Note that in neither is there any discussion whatsoever of "maximizing
power transfer."

Disregarding the 450 ohm figure, what is the recommended technique and
signal source for matching a signal to a reactive load? More
specifically, one that has both resistance and capacitance in
parallel.

You transform to the desired "characteristic impedance" (really a
resistance) and conjugate the reactance.

There is an infinite set of solutions for a given problem. For
matching at a single frequency, a minimum of two elements are
typically needed. Distributed line transformation can do it too.

Can you use a Smith Chart? A Smith Chart makes it bonehead simple,
you know.

T

#### Tom Bruhns

Jan 1, 1970
0

Maybe an op amp is not the best solution. BTW, I only need to match
one frequency at a time. Not a range, as I understand this is not
practical. The aim is to maximize power transfer.

Disregarding the 450 ohm figure, what is the recommended technique and
signal source for matching a signal to a reactive load? More
specifically, one that has both resistance and capacitance in
parallel.

Steve Bower

It would help, as others have suggested, if you could state your
problem in terms like "I want to deliver xxx watts of power to a load
that looks like yyy ohms in parallel with zzz pF. The waveform from
the source is a square wave at 100kHz, and I (do/do not) care that the
voltage across the load looks pretty much like a square wave. As a
source, I have a power op amp that's rated to deliver up to qqq volts
peak, at up to rrr amps." If you can put the problem in terms like
those, I suspect we'll be able to give a lot more specific answers.

Another couple examples of common situations where you do NOT want to
maximize "power transfer":
--When you plug a load into a standard wall outlet, the source
resistance of the power supplied through that outlet is generally
under an ohm. You'd theoretically get maximum "power transfer" if you
put a load on it equal to the source resistance. But you'd drop the
voltage to half the open-circuit voltage for an instant before the
circuit breaker tripped.
--An audio power amplifier is typically designed to deliver its power
into loads around 4 to 8 ohms, even though its output impedance is
almost always a small fraction of an ohm. If the amplifier didn't
have built-in current overload protection, you could get a lot more
power out of it if you put a much lower load resistance on the output--
but only for a short time till the output stage went into melt-down.

You may save yourself a lot of trouble if you put your circuit into a
program like the free LTSpice. You'll be able to check current and
voltage waveforms in an instant, and try all sorts of different
matching networks till you get something you're happy with. You can
even check power dissipations--in the load, in the op amp.

Cheers,
Tom

T

#### Tony Williams

Jan 1, 1970
0
Tom Bruhns said:
It would help, as others have suggested, if you could state your
problem in terms like "I want to deliver xxx watts of power to a
load that looks like yyy ohms in parallel with zzz pF. The
waveform from the source is a square wave at 100kHz, and I (do/do
not) care that the voltage across the load looks pretty much like
a square wave. As a source, I have a power op amp that's rated
to deliver up to qqq volts peak, at up to rrr amps." If you can
put the problem in terms like those, I suspect we'll be able to
give a lot more specific answers.

W

#### Winfield Hill

Jan 1, 1970
0

Steve is thinking hard.

T

#### Tony Williams

Jan 1, 1970
0
[/QUOTE]
Steve is thinking hard.

Let's have a WAG then.........

The quoted 458 ohms load impedance might be something
like a 1k in parallel with 3nF (at 100KHz sine).

A power opamp was mentioned so the voltage could be
in excess of 20V pk-pk across that 458 ohms.

Sine or square? A 100KHz square wave was mentioned.

So the problem might be the best way to drive a
1k//3nF load with at least 20V pk-pk and with (say)
better than 1uS rise and fall times.

T

#### Tom Bruhns

Jan 1, 1970
0
Steve is thinking hard.

Let's have a WAG then.........

The quoted 458 ohms load impedance might be something
like a 1k in parallel with 3nF (at 100KHz sine).

A power opamp was mentioned so the voltage could be
in excess of 20V pk-pk across that 458 ohms.

Sine or square? A 100KHz square wave was mentioned.

So the problem might be the best way to drive a
1k//3nF load with at least 20V pk-pk and with (say)
better than 1uS rise and fall times.
[/QUOTE]

Put about 10.3uH between a 100kHz square wave generator and that
load...my simple brain is fascinated by the resulting waveform.

Cheers,
Tom

J

#### joseph2k

Jan 1, 1970
0
John said:
Even if that overloads the power source? By the way, a
square wave includes energy at a lot more than one
frequency, even though it has single number of cycles per
period of time. This is why frequency is measured in hertz,
not cycles per second.

Oh no, John, say it isn't so. I "grew up" with cycles per second, surely
you are not quite so young to not remember changing to Hertz in honor of
Heinrich Hertz. Oh that's right it is a Metric unit. See:
http://www.diracdelta.co.uk/science/source/h/e/hertz/source.html
I suspect that you want to optimize
power transfer at the fundamental frequency of the square wave.

Adding either parallel or series reactance of the opposite
type to resonate with the reactance at a given frequency, is
very common. Either way, you can hide the energy transfers
through the reactive part. The difference is what happens
at other frequencies, like those harmonics in your square
wave. You might also need a transformer or a more
complicated form of resonator if you need to shift the
resistive component to a different effective value (change
the ratio of voltage to current) to get the desired energy
transfer from source to this load.

I would probably first try a series inductance that has an
equal magnitude impedance as the capacitive part of the load
at the square wave fundamental frequency (series tuned).
This has the advantage of looking inductive (rising
impedance as frequency rises) at the higher frequency
harmonics of the square wave, so that the opamp doesn't work
very hard to drive them. Don't be surprised to find that
the inductor has more voltage across it than the opamp puts out.

Resonance is full of surprises for the uninitiated.

J

#### joseph2k

Jan 1, 1970
0
Simon said:
The point with power amplifiers is not to "maximize power transfer."
The point is to maximize efficiency. They are not the same thing.

"Maximizing power transfer" is a small signal concept, and has to do
with internal impedances of the devices.

What in the pluperfect hell do you think you are saying? Are you advocating
serious mismatches in the kilowatt, megawatt and above ranges?
Maximizing efficiency has to do with extracting as much power from the
power supply as possible, and dissipating as little as possible in the
device. It is a large signal concept.

efficiency := P_out/P_dc

efficiency_pae := (P_out - P_in)/P_dc

Note that in neither is there any discussion whatsoever of "maximizing
power transfer."

What part of P_out must not be transferred efficiently to the load?
You transform to the desired "characteristic impedance" (really a
resistance) and conjugate the reactance.

There is an infinite set of solutions for a given problem. For
matching at a single frequency, a minimum of two elements are
typically needed. Distributed line transformation can do it too.

Can you use a Smith Chart? A Smith Chart makes it bonehead simple,
you know.
Smith charts are not all that useful or relevant at 100 kHz. Nor for
wideband signals for that matter.

S

#### Simon S Aysdie

Jan 1, 1970
0
Simon S Aysdie wrote:

What part of P_out must not be transferred efficiently to the load?

No part. That's why efficiency is the concern and not "maximum power
transfer." Again, MPT says nothing of converting power supply energy
into signal energy delivered to a load. It is a small signal model.
That's why MPT is not a good concept for PAs.
Smith charts are not all that useful or relevant at 100 kHz.

What is it about a smith chart that makes it not work at 100kHz (or
1Hz for that matter)?
Nor for wideband signals for that matter.

You missed the "single frequency" comment. There is no DC to daylight
solution anyway.

T

#### Tom Bruhns

Jan 1, 1970
0
Simon S Aysdie wrote: ....

Smith charts are not all that useful or relevant at 100 kHz. Nor for
wideband signals for that matter.

Why not? They certainly can be for me, for both 100kHz and for
wideband. Hint: they are at this point primarily a visualization
tool, not a calculation tool. So long as I'm dealing with networks of
series and shunt reactive components and transmission lines (ladder
networks), a Smith chart is a valuable tool. That's as true at 1Hz as
at 10GHz, though it's less likely that I'll be dealing with such a
ladder network at audio frequencies than at RF.

Cheers,
Tom

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