Calculate Power conversion Efficiency of RF rectifer in Cadence virtuoso

Hi,

Generally Efficiency = (Pout/Pin)*100, here i give the input power using port (-20dbm = its equal to 0.01mw power ). So i know the input power, and output power calculated using P= V*I formula. If in this method i calculated the efficiency become too low.. I would like to know, whether this method is right or wrong...?

Thanks.

Wrong.
You know available power not input power to rectifier.

If conjugate matching is satisfied, input power to rectifier will be equal to available power.
Otherwise, input power to rectifier < available power.

Yes., I agree. But it is matched (s11=-16db) at operating frequency. Beyond this, is that method right..!

Show me plot of Efficiency v.s. Pin.
Input impedance of rectifier is dependent on Pin.

You ignore one important thing..
You can not calculate the power as in Ohm's law because it's not %100 real signals and you forget reactive power.In order to calculate the power you should use this..

Power=1/2*Re{V.I*} or 1/2Re{V*.I}

Wrong.
Rectifier output is not sinusoidal.
It is almost DC signal if large capacitor is attached.

See https://www.edaboard.com/thread354542.html

Here is graph attachment. I used two methods,
1. plot the magnitude input power, then plot magnitude output power. Then calculate efficiency.
2. Plot the magnitude input power, then plot the output voltage, current. Last calculate (P= I*V), efficiency.

These both methods are same result. But it comes at very low efficiency.

../imgqa/eboard/Antenna/rf-rwvzivq0jz0.jpg
../imgqa/eboard/Antenna/rf-0wy203gvq25.jpg

../imgqa/eboard/Antenna/rf-rwvzivq0jz0.jpg
Why do you apply dbm() for input and output powers ?
Don't apply dbm().

Efficiency is defined like following.
pvi('pss, "/net03" 0, "/IPRB0/PLUS" 0, 0) / pvi('pss, "/net4" 0, "/IPRB1/PLUS" 0, 1) * 100

Here
pvi('pss, "/net4" 0, "/IPRB1/PLUS" 0, 1) is an input power in Watts.

pvi('pss, "/net03" 0, "/IPRB0/PLUS" 0, 0) is an output power in Watts.

Strictly speaking, above defintion of input power is wrong since you do floating driving of rectifier.

You have to calculate both input powers of plus and minus to rectifier.

Above pvi('pss, "/net4" 0, "/IPRB1/PLUS" 0, 1) is plus input power.

Actual input power to rectifier is:
pvi('pss, "/net4p" 0, "/IPRB1p/PLUS" 0, 1) + pvi('pss, "/net4m" 0, "/IPRB1m/PLUS" 0, 1)

or

pvi('pss, "/net4p" "/net4m", "/IPRB1p/PLUS" 0, 1)

Is this because of Differential end circuit..! If in-case of single end circuit, the equation will be
pvi('pss, "/net4p" 0, "/IPRB1p/PLUS" 0, 1). Is it right..!

Right.

Answer my following question.
Show us correct efficiency plot.

Here is the efficiency plot
../imgqa/eboard/Antenna/rf-aubnncaooxm.jpg

It seems efficient curve itself is reasonable.
However efficient value is still very low.

Show us output voltage and load current of your rectifier.

When I see your load in ../imgqa/eboard/Antenna/rf-rwvzivq0jz0.jpg
it is Rload=10Mohm.

Pout = Vout**2/Rload.
So Pout will be very small.
This is a reason why your efficiency is very small.