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PA output matching

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junhao

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I'd like to confirm a theory. It's about PA output matching, after getting PA's Zopt, should I translate from 50 ohm to Zopt or Zopt*(conjugation of Zopt). THANK YOU.
 

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No doubt that maximum power transfer is achieved if a PA of output impedance Z1 is terminated with conjugate impedance Z1*.

In usual RF termnology Zopt is understood as an external impedance. But in your drawing, Zopt is an impedance seen into the PA output, in other words its output impedance.

To avoid misunderstandings, you should correct the drawing annotation.

In usual terms, if you achieve maximum power transfer with load impedance Zopt, the output impedance of the PA is Zopt*.
 

No doubt that maximum power transfer is achieved if a PA of output impedance Z1 is terminated with conjugate impedance Z1*.

In usual RF termnology Zopt is understood as an external impedance. But in your drawing, Zopt is an impedance seen into the PA output, in other words its output impedance.

To avoid misunderstandings, you should correct the drawing annotation.

In usual terms, if you achieve maximum power transfer with load impedance Zopt, the output impedance of the PA is Zopt*.
Got it. So if the loadpull gets the impedance assuming Z, then in order to get the largest output power, I should ensure the 50ohm transfers to Z*. Am I right?
 

\[Z_{opt} \neq Z_{conj} \]
Optimum Loading Condition is particularly different for Power Amplifiers.Conjugate Matching is necessary and Sufficient Condition for Small Signal Amplifiers, this is true but the case diverges for Power Amplifiers.
You have to-absolutely-design a Matching Circuit so that while one Port having \[Z_{opt}\]
the Other Port will be simply 50 Ohm.Forget about \[Z_{conj}\] for Large Signal Amplifiers.

Definition : However there isn't Official Definition for Power Amplifiers, 0.25W Delivered Power and beyond can easily be accepted being as Power Amplifier.
 

\[Z_{opt} \neq Z_{conj} \]
Optimum Loading Condition is particularly different for Power Amplifiers.Conjugate Matching is necessary and Sufficient Condition for Small Signal Amplifiers, this is true but the case diverges for Power Amplifiers.
You have to-absolutely-design a Matching Circuit so that while one Port having \[Z_{opt}\]
the Other Port will be simply 50 Ohm.Forget about \[Z_{conj}\] for Large Signal Amplifiers.

Definition : However there isn't Official Definition for Power Amplifiers, 0.25W Delivered Power and beyond can easily be accepted being as Power Amplifier.
Yeah, I know the difference between conjugation matching and power matching. I mean, after getting the optimum impedance, should I design the output matching network with the conjugation of the optimum inpedance or just optimum impedance? The picture below comes from "RF and mm-Wave Power Generation
in Silicon" written by Hua Wang.
 

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Yeah, I know the difference between conjugation matching and power matching. I mean, after getting the optimum impedance, should I design the output matching network with the conjugation of the optimum inpedance or just optimum impedance? The picture below comes from "RF and mm-Wave Power Generation
in Silicon" written by Hua Wang.
But the author talks about Input Impedance Matching.As he said, Conjugate Matching is good enough for Input.
But !
For an Optimum Performance ( Power or Efficency ) Input Impedance should also be examined in order to find Optimum Input Impedance ( Source Pull )
 

But the author talks about Input Impedance Matching.As he said, Conjugate Matching is good enough for Input.
But !
For an Optimum Performance ( Power or Efficency ) Input Impedance should also be examined in order to find Optimum Input Impedance ( Source Pull )
The last sentence means the output matching, not input matching. Also, source pull is important.
 

for theoretical max power transfer, Zout = Zload, but for the most efficient operation of a power amplifier;

it should be designed with a low Zout ( resistive ) compared to the load - which could be 50 ohm ( resistive )

Therefore if the PA Zout is 0.5 ohm resistive, there is only a small loss in the PA and all the intended power ends up in the load.
 

for theoretical max power transfer, Zout = Zload, but for the most efficient operation of a power amplifier;

it should be designed with a low Zout ( resistive ) compared to the load - which could be 50 ohm ( resistive )

Therefore if the PA Zout is 0.5 ohm resistive, there is only a small loss in the PA and all the intended power ends up in the load.
Low Zout may cause the current increase and the PAE will decrease.
--- Updated ---

OK, Author's statement is wrong.Conjugate Matching is not considered in Power Amplifiers.
Wrong information..
I have seen both matching method in different books, really confused me.
 

this comment is non sequitur - can you offer background explanation ?
I'm not sure, just according to the loadline. Assuming the DC point dosen't change. If Zout<Zopt, the swing of current could achieve Imax while the voltage COULDN'T achieve the Vmax. In the same time, the output power doesn't equal to (Vmax*Imax)/8, which should be (Vmax*Imax*Zout/Zopt)/8, then the PAE also decrease.
 

Dear Junhao..
When you design a Power Amplifier, you can find 3 \[Z_{opt}\] for;
-Max Power Gain
OR
-Max. Efficiency
OR
-Max. Delivered Power

So you do your choice and the compromise between these 3 point ( and of course their contours ) will give you an insight about what'll happen.
 
Dear Junhao..
When you design a Power Amplifier, you can find 3 \[Z_{opt}\] for;
-Max Power Gain
OR
-Max. Efficiency
OR
-Max. Delivered Power

So you do your choice and the compromise between these 3 point ( and of course their contours ) will give you an insight about what'll happen.
Yeah, always should trade off between these three contours.
 

With regard to the conjugate vs no-conjugate question, I've seen it described two different ways:
1. The optimal load impedance, seen from the drain (or other reference plane) of the power FET. In this case, drawings show Zopt as just being a property of the load.
2. The conjugate of the optimal load impedance. In this case, drawings show Zopt being a property of the amplifier device itself, and the load impedance is chosen to be a "conjugate match" to Zopt.
Both of these actually say the same thing, but #2 is far more confusing because it seems to imply that Zopt is the actual output impedance, and conjugate match exists on the output.

Unfortunately there's so much variability in how optimal power matching is described that engineers are going to have to think on their feet.
for theoretical max power transfer, Zout = Zload, but for the most efficient operation of a power amplifier;

it should be designed with a low Zout ( resistive ) compared to the load - which could be 50 ohm ( resistive )

Therefore if the PA Zout is 0.5 ohm resistive, there is only a small loss in the PA and all the intended power ends up in the load.
For linear amplifiers (class A/AB/B/C) this makes no sense. Dissipation in the PA is determined primarily by bias conditions, not output impedance.
 

When you design a Power Amplifier, you can find 3

Zopt​

Z_{opt} for;
-Max Power Gain
OR
-Max. Efficiency
OR
-Max. Delivered Power

A possible fourth parameter to be taken in account and which can interfere with other PA optimum loads: max stability impedance.

If not doing conjugate match can things go wrong if effect of feed transmission line impedance is taken in account, if transmission line have a decent length relative wavelength.
To make thing simpler, assume no reactance in below example:

transformer-2.png


It is a bit worst case with for example a 20 mm long PCB transmission line at 2.4GHz, but 50% loss can be bad enough for its power economy and battery weight if it is a battery powered wearable of some kind compeating with concurrent to have smallest IoT thing and longest working time and best RF coverage for a single battery charge, such as a inear headset.
 

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