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IIP3 calculation: Pin or Pout?

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rghrob

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Hi,
I've seen a few sources post the following and needed some clarification. Thanks in advance:
1. IIP3=Pin+(Pout-IM3)/2
2. IIP3=Pin+(Pin-IM3)/2
Is #2 valid only if the gain is unity (Pin=Pout)?
Or is there something else I'm missing?

-Robert
 

OIP3(dBm)=Pout(dBm)+[Pout(dBm)-IM3(dBm)]/2
IIP3(dBm)=OIP3(dBm)-Gain(dB)
where Gain is the gain of the amplifier
 

rghrob said:
I've seen a few sources post the following and needed some clarification. Thanks in advance:
1. IIP3=Pin+(Pout-IM3)/2
2. IIP3=Pin+(Pin-IM3)/2
Is #2 valid only if the gain is unity (Pin=Pout)?
Or is there something else I'm missing?
Click to expand...
IIP3 is calculated by #1.
I've never seen #2.
 

Topic of the quoted lection is different, digital receiver not amplifier. There's neither a gain nor Pout in this case.
 

In #2 they may be discussing this IM3 at the input.

For example,
OIP3 = Pout + (Pout - Po,IM3)/2
Pout = G + Pin
Po,IM3 = G + Pi,IM3
OIP3 = IIP3 +G

therefore,
IIP3 + G = Pin + G + (Pin + G - P,IM3 - G)/2
IIP3 = Pin + (Pin - Pi,IM3)/2

where (Pin - Pi,IM3) = (Pout - Po,IM3). Theoretically, this is because the input and output powers are just differentiated by the gain (or loss) of the system but you can't measure intermods at the input I don't think. Normally there's enough isolation that you don't have measurable IMDs seen at the input.
 
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