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Question on RFIC and relevant solution

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bigheadfish

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Question on RFIC

"if the TX power amplifier delivers 1W to a 50Ω antenna, the peak to peak voltage swing is equal to 20V. The leakage to the receiver path is on the order of 30mVpp(=-26dBm)"

I think P=Vpp*Vpp/(2*R), then Vpp=10V ,not 20V. Am I right?
30mVpp=-26dBm? How can I get it? Thanks in advance!
 

Re: Question on RFIC

bigheadfish said:
"if the TX power amplifier delivers 1W to a 50Ω antenna, the peak to peak voltage swing is equal to 20V. The leakage to the receiver path is on the order of 30mVpp(=-26dBm)"

I think P=Vpp*Vpp/(2*R), then Vpp=10V ,not 20V. Am I right?
30mVpp=-26dBm? How can I get it? Thanks in advance!
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Hi,bigheadfish
The effictive value is equal to (sqrt(2)/2)*Vp and equal to (sqrt(2)/4)*Vpp. So, 20V is the corret answer.
The dBm is a measurent of power that is relative to 1mW. The equation is dBm=10*log(P/1mW), where P= V^2/R, V is the effective value of the signal as described as above.
sixth
 
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