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Inductance and capacitance from complex impedance solution in RF module
Posted 20.07.2016, 15:23 GMT-4 RF & Microwave Engineering Version 4.4 1 Reply
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I was wondering if there is any way of backing out a value for inductance and capacitance between two ports/ lumped ports in the RF module (specifically, in the emw frequency domain solver). The module can calculate a complex valued impedance, however, I can't figure out how to back out specific values of capacitance and inductance from the imaginary component of the impedance.
I can provide a file if necessary, but the setup is very simple. Just some arbitrary geometry separating two lumped ports from which I retrieve an impedance value. The impedance has some imaginary component with a negative part associated with capacitance, and a positive part associated with inductance, but since they are summed together I am not sure how/if I can extract those values.
Any help would be greatly appreciated.
Thanks,
I can provide a file if necessary, but the setup is very simple. Just some arbitrary geometry separating two lumped ports from which I retrieve an impedance value. The impedance has some imaginary component with a negative part associated with capacitance, and a positive part associated with inductance, but since they are summed together I am not sure how/if I can extract those values.
Any help would be greatly appreciated.
Thanks,
1 Reply Last Post 21.07.2016, 08:34 GMT-4
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Posted:
8 years ago
Are you asking if there is a built in way to do this? I don't think so with comsol but if you have the real/imag parts you can come up with equivalent R+jX and R-jX circuits, both series and parallel representations.
For instance if the program gives you an impedance of say 11.3 + j56 ohms at 400MHz, you can directly calculate the series RL circuit by using XL=2*pi*f*L and backing out L by direct substitution. In this case you would get an answer of: re = 11.3 im = 22.28nH
In order to come up with a parallel representation of the same circuit you can use the following formulas:
1) Rp = Rs * (1+Q^2)
2) Xp = Xs * (1/(1/Q^2))
where Q = abs(im)/re
Using the above example of 11.3 + j56 @ 400Mhz gives a parallel circuit: 289ohms || 23.2nH
These formulas can be entered into comsol. Also, there is a program called appcad (developed by HP/Agilent but now looks to be maintained by Avago) and it has a very nice calculator that does what I showed above. In the program go to ENG tools > complex math RF calculator. Just pay attention to the frequency being used. Does quite a few neat calculations.
For instance if the program gives you an impedance of say 11.3 + j56 ohms at 400MHz, you can directly calculate the series RL circuit by using XL=2*pi*f*L and backing out L by direct substitution. In this case you would get an answer of: re = 11.3 im = 22.28nH
In order to come up with a parallel representation of the same circuit you can use the following formulas:
1) Rp = Rs * (1+Q^2)
2) Xp = Xs * (1/(1/Q^2))
where Q = abs(im)/re
Using the above example of 11.3 + j56 @ 400Mhz gives a parallel circuit: 289ohms || 23.2nH
These formulas can be entered into comsol. Also, there is a program called appcad (developed by HP/Agilent but now looks to be maintained by Avago) and it has a very nice calculator that does what I showed above. In the program go to ENG tools > complex math RF calculator. Just pay attention to the frequency being used. Does quite a few neat calculations.