Voltage multiplier PCE calculation in ADS

[sizo0708]

Hey everyone, I am trying to calculate the PCE (power conversion effeciency) of a 5 stage low power voltage multiplier using the HSMS-2850 Scottky diode and I have some kind of error because the value goes beyond 100% after a certain input power value.

I tried to use 2 different methods to calculate the PCE and they are both problematic (They can be seen in the images).
A transient simulation of input and output voltage is shown at an input power or -20dBm.

Also, voltage multipliers with fewer stages should have a higher value of PCE because there is less power loss in the diodes, but I get the opposite when I simulate voltage multipliers with less stages...

What may be the problem? Is it the sceme or is it the calculation method?
Kindly help.

 

[BradtheRad]

With AC applied to all those capacitors, you may be seeing power factor issues. Current travels in a series of bursts from your supply. These bursts occur almost immediately after zero voltage crossings. It affects power readings.

I'm not sure how L1 C16 affect performance. I also unsure about the hookup of your load to your multiplier.

 

[BigBoss]

Your Output Measurements are NOT Correct. "P_Probe" element measures Complex Power RESPECT to GND.
But your Output is floating/differential. You have to convert this differential/floating (non grounded ) signal to single ended mode.
You do not need use Current Meter or Voltage Node definitions. They make you confuse.
Measure the Complex I/O Powers only.

 

[sizo0708]

Your Output Measurements are NOT Correct. "P_Probe" element measures Complex Power RESPECT to GND.
But your Output is floating/differential. You have to convert this differential/floating (non grounded ) signal to single ended mode.
You do not need use Current Meter or Voltage Node definitions. They make you confuse.
Measure the Complex I/O Powers only.

Thank you very much for the response but I am not sure I fully understand what youre trying to say...I should measure the input and output power from a different location on the circuit? If so, I would like you to clarify these locations.

 

[BigBoss]

Your Output Voltage is the voltage across the resistor R1 so You have to measure this voltage.
And this voltage has no GND Reference so it's floating.
All you have to do is to convert this differential voltage across R1 into single one by inserting simply an ideal 1:1 transformer then measure this new single ended voltage ( one pin of the transformer is grounded) and compute the power.

 

[BradtheRad]

This reply doesn't specifically address your query, but your schematic has the load in an odd position. You appear to have half of a Cockcroft-Walton (positive polarity only). It's made up of several Villard cells. (In that sense it's a Villard multiplier rather than half-wave parallel multiplier.) Articles depict the load R1 having the same ground as the power supply.

In that way the bottom row of series capacitors add their charge together, to deliver to the load. Did you try connecting the left terminal of R1 to gnd, and retain its right terminal where it is?