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Marios

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source voltage
Sep 01^st, 2022, 4:59am

Dear all,

I have a general question with regards to the PORT instance found in analogLib of cadence. At AC magnitude section of the PORT I set a unit magnitude. Then the port internally will have a magnitude of 2. Initially I thought that this "doubling" of magnitude is performed internally so that the port behaves as a transmission line and take care of the effect of impedance mismatch. Asking around I was then told that the "doubling" is so that when I calculate voltage gain (say I drive an LNA with that port), I present my LNA with unit magnitude at its input. So my question is this. Does that mean that a real antenna that has an internal resistance of 50 ohms, delivers a unit magnitude to an LNA? In other words, the open circuit voltage of the antenna has a magnitude of 2? Because if not, then there is a discrepancy between how I model my circuit and the reality.

Thanks in advance
Marios

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Ken Kundert

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Re: source voltage
Reply #1 - Sep 1^st, 2022, 4:20pm

Your transmission line explanation is correct.
Your colleagues explanation is a bit off: they are explaining why you specify the port voltage magnitude to be unity, not why there is a doubling from the specified value to the internal drive voltage.

Basically, when you specify the port voltage, you are specifying the that will result when the port is loaded with a matched load. This is the exactly the same thing you do when you specify the voltage on a high-frequency signal generator. If you connect an hp8640 signal generator to an oscilloscope and set it to deliver 1Vp, you will see a 2Vp signal on your oscilloscope until you remember to enable the 50Ω terminating resistor on the scope, at which point you see a 1Vp signal.

The same would be true from a 1Vp signal coming from an antenna.

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Marios Junior Member Offline Posts: 15	Re: source voltage Reply #2 - Sep 5^th, 2022, 2:44am Ok thank you for your detailed answer! Marios
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