[moved] How to simulate the conversion efficiency of the rectifier circuit by ADS

Dear everyone

I am now doing the study of 2.45GHz rectenna circuit.
I am now simulating the conversion efficiency of a half-wave rectifier in ADS.
I want to calculate the power conversion efficiency by the function of : efficiency=Pload/Pin

I use the LSSP simulation and add to P_probe in the output port and input port to measure the Pload and Pin, but seems some warning happens.
Can anyone help to solve the problem?
The warning message is :" P_Probe1:Unable to resolve variables or functions in expression p=real(deltav*conj(sh1.i))"
Thank you very much!

Show me schematic.

Hi I cannot upload pic through lab's pic, so I draw it bymyself.
You can have look at it, I just want to calculate the conversion efficiency

Add HB Analysis Controler.

Thanks for your reply. I change he LSSP to HB simulation, then it's ok to probe the power. Now i get twp questions about this:

1. Why the P_probe cannot work in the LSSP simulation and the erro message occur
2. In HB simulation, the input power P_probe consist of several parts: P_probe[0](DC power), P_probe[1](Power in fundamental freq) and P_probe[2], it can be found that P_probe[0], P_probe[2] are both negative, why the value is negative?
when we calculate the power efficiency, what kind of equation should i use? Eff=P_probe2/P_probe1 or EFF=P_probe/P_probe[1]

You don't have to remove LSSP Analysis Controler.
HB Analysis Controler and LSSP Analysis Controler can coexist.

LSSP Analysis is no more than HB Analysis, however it does not save any voltage and current.

Are their absolute values very small ?
Show me results of P_probe[0], P_probe[1] and P_probe[2].
http://edadocs.software.keysight.com...ageId=82848108

Eff=P_probe1[0] / P_probe2[1]
or Eff=P_probe1[0] / sum(P_probe2).

Thank you very much for your reply.

1. Yes, the value of P_probe2[0] and P_roble2[3] is negative and the absolute values are very samll comparing with P_probe2[1]
Would you please help to explain, why the value is negative and why the values are very small?

2. In your equation "Eff=P_probe1[0] / sum(P_probe2)", what does sum(P_probe2) means, is it equal to P_probe2[1]?

3. When i use the he function Eff=P_probe1[0] / P_probe2[1] to calculate the efficiency, does this function include the impedance mismatch? or the function Eff=P_probe1[0] / P_probe2[1] is an ideal equation without return loss(S11)?

Surely respond to my question.
Show me numerical values of P_probe[0], P_probe[1] and P_probe[2].

I think it is due to numerical error.
Order of your 1-Tone HB Analysis is 3, it is too small.
Increase Order, e.g. Order=15, then show followings.

(1) Order=3
P_probe[0], P_probe[1], P_probe[2], P_probe[3]

(2) Order=15
P_probe[0], P_probe[1], P_probe[2], P_probe[3], ...., P_probe[15]

Numerical Lists as Watt
and
Plot dBm(P_probe) as Y-axis with X-axis as index or frequency.

They are not same strictly.
However they are almost same, I think.

No.
Use Eff=P_probe1[0] / dbmtow(PindBm) instead.

Here PindBm is a available power of Port1.
P_probe2[1] = dbmtow(PindBm) * (1 - |S11|**2)

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