In my attached circuit, Pin is varied from -10 to 20 dBm. There are two ways to calculate RF-to-DC conversion efficiency, and I do not know which one is correct. This is because I still do not understand whether the power received by the antenna is Pin or the power measured by Probe ( Pin_Probe.p ).
Pdc_Watt = 0.5 * real(Vout[0] * conj(Iout.i[0]))
1/ PCE1 = Pdc_Watt / dbmtow(Pin) * 100
2/ PCE2 = Pdc_Watt / Pin_Probe.p[1] * 100
Any suggestion would be highly appreciated. Thank you very much for your time.
Dear minh,
If you calculate the efficiency as the ratio of DC output power output Podc to the input probe rf power input, then this will be the AC to DC conversion efficiency of the diode. On the other side if you calculate the efficiency of the whole circuit containing the antenna, this will be the overall efficiency of the rf power harvesting. In every case you have to define your performance parameters precisely.
Best wishes
Dear Mr. Zerky,
Thank you very much for your answer. So I think the correct term would be the overall efficiency in your answer, which reflects the amount of power delivered to the rectifier.
Can I ask you one more question? Most papers in this field model the antenna as a power source or a voltage source with a series resistance, usually 50 ohm (the input impedance of the antenna). In this model, compared with a practical case when the power received at the antenna (so-called P_Rx) is measured by a spectrum analyzer, I think P_Rx is the var Pin (not Pin_Probe) in my simulation. But I'm not quite sure.
Dear Minh,
welcome again, wish you success!
When you use a spectrum analyzer to measure the received power from the antenna, the spectrum analyzer has 50 ohm impedance on which it acquires the received voltage on it. So, this power is called the maximum available power from the source and it is equal to one half the antenna source power.
The antenna power will be double the power of the analyzer.
If and only if the input impedance of the rectifier circuit is equal 50 ohm , then the power measured by the probe will be equal to the power by the analyzer.
Best wishes
The efficiency of the RF to DC conversion is defined as the ratio of the output dc power (Pout) and input RF power(Pin).
Where, Pout = Vout,DC / Rload , Vout = Output DC Voltage, Rload = Resistive load present at the output.
The following article may be useful:
Article A Passive Wake-Up Circuit for Event Driven Wireless Sensor N...
Dear Mr. Zerky, Mr. Hassouni and Mr. Shekhar,
Thank you very much for all of your valuable explanations. So we cannot map the power measured by a spectrum analyzer with the power in the simulation, unless the impedance of the rectifier is 50 ohm. This is then the goal of the impedance matching network, another problem I have to solve.
sir can you please suggest that when i connect a 50 ohm load after the power meter at the input of the circuit just after the power source then i get full power in the reading. here then why i am not getting half the power as per the power transfer theorem
Sumit Varshney
I am not able to understand your question. Are you connecting a 50 ohm load in series with rectifier?
Dear Mr. Varshney and Mr. Shekhar,
I'm asking about the correct equation to calculate the power conversion efficiency, an important metric to evaluate the performance of converting RF power to DC power. The problem in Mr. Varshney's simulation might be because of the harmonics. Could you please set the order to be 1 and try again?
I've found one big mistake in my equations. Pdc_watt is not the RMS value, because the DC output is not a sinusoidal waveform. It should be calculated as normal: Pdc = real(Iout.i[0]*Iout.i[0]*RL) or real(Vout[0]*Iout[0]).
Dear Mr. Chandra Shekhar,
Perfect. Thank you very much for your great material. So, the followings are my understandings. Please correct them if I'm wrong.
1/ The available power at the antenna, Pav, is Pin in my circuit.
2/ The relationship between the power density, S, of the incident RF waves, and my Pin is:
Pin = S . Ae, with Ae is the effective area of the antenna.
S is what I can get from an spectrum analyzer, despite the matching conditions between the analyzer and the antenna. Is it correct?
Yes, The available power at the antenna, Pav, is Pin in your circuit (if antenna is perfectly matched with your circuit). Minh Lam
Dear Chandra Shekhar and Smail Hassouni
Thank you very much for your valuable answers. Can I use the equation Pav = S . Ae to find Pav, with S the measured value from the spectrum analyzer? And after that, the predicted Pin can be determined using Mr. Smail Hassouni's answer?
Check my recent paper titled "
A Reconfigurable Passive RF-to-DC Converter for Wireless IoT Applications" published recently.
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