Matching reactive load

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

Reply to
Steve Bower
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450 ohms (with no phase angle or other indication of reactance) indicates pure resistance. Why do you call this a reactive load?
Reply to
John Popelish

In addition to what John P. already posted, I have to ask about the 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

Reply to
Tom Bruhns

** 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

Reply to
Phil Allison

I am already worried ;-)

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

Reply to
Steve Bower

Too vague. What's the app?

Reply to
kell

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.

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.

Reply to
John Popelish

His grade on the exam? ;-)

Cheers! Rich

Reply to
Rich Grise

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

or for power added efficiency

efficiency_pae := (P_out - P_in)/P_dc

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

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.

Reply to
Simon S Aysdie

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

Reply to
Tom Bruhns

No reply, total silence. Another thread stopped dead. :(

--
Tony Williams.
Reply to
Tony Williams

Steve is thinking hard.

Reply to
Winfield Hill

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.

--
Tony Williams.
Reply to
Tony Williams

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

Cheers, Tom

Reply to
Tom Bruhns

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:

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Resonance is full of surprises for the uninitiated.

--
 JosephKK
 Gegen dummheit kampfen die Gotter Selbst, vergebens.  
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Reply to
joseph2k

What in the pluperfect hell do you think you are saying? Are you advocating serious mismatches in the kilowatt, megawatt and above ranges?

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

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

--
 JosephKK
 Gegen dummheit kampfen die Gotter Selbst, vergebens.  
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Reply to
joseph2k

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.

What is it about a smith chart that makes it not work at 100kHz (or

1Hz for that matter)?

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

Reply to
Simon S Aysdie

...

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

Reply to
Tom Bruhns

What generator resistance Tom?

With a zero ohm generator LTspice shows a 900KHz resonant circuit being shock-excited every 5uS. You can see that it is not a pickoff of the 9th harmonic in the square wave because a similar shock-excitation happens if it is resonated at the 10th instead (8.44uH). There is also a smoother transition at every incoming edge because the 10th harmonic has a whole number of resonant cycles every 5uS.

With a 50 ohm source resistance the Q drops so low that there is a just slight peaking of the 100KHz square wave. Experimenting with LTspice shows that 50 ohms and 2.5uH in series would look to be a useful combination.... 50 ohms takes the direct reactive load off the amplifier and the 2.5uh almost halves the risetime and reduces the peak current requirement from 400mA to 270mA.

This is all based on a WAG'd load, but maybe a pointer to the OP if he gets some real numbers.

--
Tony Williams.
Reply to
Tony Williams

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