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  1. #1
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    About Impedance matching

    Hello everyone,

    In order to have maximum power transfer, source and load impedances must be matched => results in 50% efficiency.

    On the other hand, if we would increase the Load impedance with respect to the source impedance, efficiency is increased. This is the reason SMPS work, because the output impedance is small compared to the loads they power. Why isn't applied the same principle to RF PA ? Why don't just increase the load to say 100 times bigger than the source ?

    Consider example:
    Source (Vsource) --> source resistance (Rsource) --> load resistance (R)

    efficiency (eta)=R/(R+Rsource) = x/(1+x) where x=R/Rsource. For x=100, efficiency close to 100%. Why don't apply a "lossless matching network" that increases the load resistance ?

    Output Power = Vsource2*R/(R+Rsource)2 => if R>>Rsource, then all the power goes to the load, same as in SMPS. Why not apply this to RF PA ?

    Thank you for your time !

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    Re: About Impedance matching

    When the impedances aren't matched some of the source power can come back from the load. This is called reflection.
    So reflected power won't be radiated, it will be dissipated on the transferring elements, and on the source impedance.
    Reflection is negligible if the transmission line dimensions are much smaller than wavelength of the operating frequency.
    In high power high frequency circuits (RF PA) the power device, filter element and transmission line sizes are comparable with the wavelength, so matching is very important to minimalize reflection.
    "Try SCE to AUX." /John Aaron/


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    Re: About Impedance matching

    Further, if you have an impedance mismatch, aside from loss of power, you may get distortion as the reflected signal collides with the source signal.


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    Re: About Impedance matching

    RF PAs don't necessarily implement impedance matching on the source side, otherwise they won't achieve efficiency like 70 or 80 percent. In so far your reasoning isn't wrong. The backside is that you need a circulator to absorb load reflections.


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    Re: About Impedance matching

    The final output of the PA isn't matched sometimes? I mean where the antenna or transmission line (cable) is connected?
    I haven't got too much experience with PA and circulator, but circulator's terminations are matched as I know.
    "Try SCE to AUX." /John Aaron/



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    Re: About Impedance matching

    A class-C PA driving 50 ohm load often has not 50 ohm source impedance.


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    Re: About Impedance matching

    Quote Originally Posted by CataM View Post
    Hello everyone,

    In order to have maximum power transfer, source and load impedances must be matched => results in 50% efficiency.

    On the other hand, if we would increase the Load impedance with respect to the source impedance, efficiency is increased. This is the reason SMPS work, because the output impedance is small compared to the loads they power.
    No, this is not correct. Efficiency cannot increase. Efficiency is highest at maximum power transfer == Rload=Rsource.

    Sketch a simple circuit and calculate the power that the load sees at the different instances of Rload>Rsource, Road=Rsource and Rload<Rsource and you'd realize that maximum power is transfered at Rload=Rsource.
    -------------
    --Akanimo.



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    Re: About Impedance matching

    Quote Originally Posted by CataM View Post
    Hello everyone,

    In order to have maximum power transfer, source and load impedances must be matched => results in 50% efficiency.

    On the other hand, if we would increase the Load impedance with respect to the source impedance, efficiency is increased. This is the reason SMPS work, because the output impedance is small compared to the loads they power. Why isn't applied the same principle to RF PA ? Why don't just increase the load to say 100 times bigger than the source ?

    Consider example:
    Source (Vsource) --> source resistance (Rsource) --> load resistance (R)

    efficiency (eta)=R/(R+Rsource) = x/(1+x) where x=R/Rsource. For x=100, efficiency close to 100%. Why don't apply a "lossless matching network" that increases the load resistance ?

    Output Power = Vsource2*R/(R+Rsource)2 => if R>>Rsource, then all the power goes to the load, same as in SMPS. Why not apply this to RF PA ?

    Thank you for your time !
    Your first sentence is wrong. To simplify the discussion, assume only resistances. It is true that max power transfer is obtained by adjusting the load resistance to the source resistance, if the source resistance must remain constant. But, more power at higher efficiency can be obtained by decreasing the source resistance, if you can do it. Increasing the load resistance greater than the source resistance increases the efficiency, but lowers the amount of power transferred. Assume a source voltage of 12 volts and a source resistance of 6 ohms. Start with a load resistance of 6 ohms and compute the amount of power transferred and the efficiency. Then lower the source resistance and notice of both the amount of power and efficiency increases. Do the same after increasing the load resistance. Notice how the efficiency increases, but the power transferred goes down.

    Ratch
    Hopelessly Pedantic



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  9. #9
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    Re: About Impedance matching

    Quote Originally Posted by frankrose View Post
    Reflection is negligible if the transmission line dimensions are much smaller than wavelength of the operating frequency.
    Yes. In cases where reflection is negligible (i.e. wavelenght is much higher than the T line length), it is not necessary to have matching. I mean, you can design your PA to deliver the necessary power the the load, having Rload>>Rsource without any drawback in return.

    Quote Originally Posted by Akanimo View Post
    you'd realize that maximum power is transfered at Rload=Rsource.
    Max power transfer does not mean max efficiency. Equations already provided in OP.

    ----
    Ok, so the point of this thread is "what approach to take for new design". So I see it is already done, as referenced by FvM. The other backside is that you need to design your output voltage for the load you need to drive, which increases as you increase the load and you need high power... unless the source resistance is very very small.

    Quote Originally Posted by Ratch View Post
    But, more power at higher efficiency can be obtained by decreasing the source resistance, if you can do it.
    Yes. That is the point for a new design. That is what I was thinking for the approach to take.

    Increasing the load resistance greater than the source resistance increases the efficiency, but lowers the amount of power transferred.
    Yes, that is true for ready made designes ! But it can be the approach for new ones, if you can deal with the other drawbacks. In other words, you do not follow the Max power transfer theorem because your design is new and specifically made to transfer the necessary power to the load. However, for a ready made one, you need to follow Max power transfer.

    Quote Originally Posted by FvM View Post
    A class-C PA driving 50 ohm load often has not 50 ohm source impedance.
    How small is the source impedance typicall ? What is the Load to source ratio usually ?
    Last edited by CataM; 7th February 2019 at 20:27.



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