In this article, after impedance matching, Vin decreases significantly though Vout increases. And Vout/Vin is not equal to 2. Why?
Article DESIGN OF RF TO DC RECTIFIER AT GSM BAND FOR ENERGY HARVESTI...
Dear Pengyu,
welcome,
I will give a conceptual answer to your question and then you can find out the specific answer. Assume you have a source Vs with an impedance Zs and a load with impedance ZL. Then according to the voltage division rule VL= Vs ZL/ ZL+Zs,
If initially before the matching ZL > Zs then the input voltage will be greater than Vs/2. When matching the load, one has to decrease ZL which means that when matching the input the the input voltage which is here the load voltage must decrease to ultimately to Vs/2 when the input is perfectly matched. The opposite is true when the load is initially smaller than Zs.
Best wishes
Dear Pengyu,
If the input source Vs is a pure voltage source then it is impossible to change it by loading. So, i must assume that VS IS NOT A PURE VOLTAGE SOURCE. Please review your circuit!
Best wishes
Hi Ll Lee
Its also important to notice that impedance matching makes the higher transfer of "Power" (which is P = V. I).
If you had an infinite load ZL, your output would be equals to the voltage of the power supply (VL = Vs), but your current would be zero thus the power would be also zero.
Now if you have a short-circuit as a load (ZL = 0), your current will be maximum (IL = Vs/ Zs) but your voltage would be zero and the power transfered would be zero.
At last, if you match the impedances, making ZL = Zs* (* = conjugate), you'll have the maximum power transfered from the supply to the load, but now you have the output voltage equals to half the voltage of the supply (VL = Vs/2).
Cheers!
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