IIP3 of differential pair in Cadence

[Julian18]

Hi, there

I want to see the IIP3 of a differential pair amp, the circuit is shown below, I put a parallel resistor to the port and give the same value to both this resistor and the resistance associated with the port, but when I changed this resistance value, the IIP3 also changed, and the values are btw -2dbm to 29dbm (1K Ohms to 0.1Ohms ) , so how can i choose the right value for the IIP3 of my ckt? ( The load of differential pair is resistor.)

By the way, I also have simulated the situation where the resistor in parallel to the port is cut, and this time i also got different IIP3 values as I changed the resistance value associated with the port.

So what is the right setup and what is the right value of IIP3? Thanks

 

[hr_rezaee]

Hi
As I know, resistors are determined by output resistance of previous stage and input resistance of next stage.
I'm not sure, mut I think :
If you change the resistor, you changed the voltage that is delivered to your circuit, but power is constant yet. so you have different results.
regards

 

[Julian18]

Hi, I set up the simulation this way, put a parallel resistor to the port and give the same value to both this resistor and the resistance associated with the port, then, i change this value, and different IIP results come out.

 

[hr_rezaee]

Hi
I think there is another problem.
use two VCVS (voltage controled voltage source) as input source.
in your circuit, there is no path from input source to gnd.
regards

 

[fanshuo]

you know the unit of IIP3 is power, if you change the resistor value, the power changed
it is required that the input resistor value and the output resistor value to be the same

one way to resolve it is to use VCVS at the output, connect it with the required resistance and select it as the output port