After designing an optical splitter using photonic crystals with four equal and symmetric ports using ring resonator, I am getting output to be greater than input voltage provided (output is the addition of voltages obtained from the fur ports of splitter). How is that possible? Can anyone explain that, Please?
If your input and output ports have the same impedance, and the power is spread equally to each of 4 output ports, each port will have 1/4 of the power. The voltage in each port will be 1/2 of the input voltage, and so will the current.
When 4 of these are added together you can get twice the input voltage. This is equivalent to transforming to four times the impedance. Presumably you only get 1.2 volts because of attenuation through your device, 1.2/2 is about -4 dB.
Thank you Zbigniew Motyka, Malcolm White and Saleh Chebaane .
Regarding my input and outputs... incase of 1*2 equal and symmetrical splitter, i have provided with 1V/m as input amplitude and the output at port 1 is 0.6 and port 2 is 0.61 (relative power) ...
You are being inconsistent with your use of voltage, electric field, and power.
1 v and 1.2 v are voltages and do not say how much power (watts) there is unless the impedance is known. The power is proportional to the square of the voltage. 1 v/m is an electric field and does not specify power by itself. It needs an impedance to give a power density (watts per square metre), and an area distribution to give a total power (watts). You can never have 0.6 and 0.61 relative power out of two output ports for an input of 1, for a passive device. You can have relative voltages of 0.6 and 0.61, which means relative powers of 0.36 and 0.3721 if the impedances are all the same. Then the total output power is 73.21% of the input power, which means about 1.5 dB attenuation.
Malcolm even if i check with the power level it being higher than the input power. I am using optiFDTD software.
The output power will never be higher than the input power. The numbers I gave in my previous answer {You can have relative voltages of 0.6 and 0.61, which means relative powers of 0.36 and 0.3721 if the impedances are all the same. Then the total output power is 73.21% of the input power, which means about 1.5 dB attenuation.} are almost certainly right. Why do you think that these are not the correct powers?
Otherwise, there are several explanations for your results. Three of them might be
1. The software is wrong - not very likely but possible.
2. The problem is not suitable for the software - also possible.
3. You have not understood the outputs from the software - I think this is the most likely reason. I suggest that you should read the help files carefully, and perhaps do a similar run for a problem where you know the answer, so you can see whether what you get is what you expect.
even while considering as you said Malcolm, the output for some simulation is like 1.1 and 0.9. In such case power also exceed input power. where the input is 1V.
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