I am confused about conversion.

1
http://www.designers-guide.org/Forum/YaBB.pl?num=1122952471/2#2

If I setup E in port, then in reality, the signal at the gate of the mosfet is 2E.

I wonder if the E means PAC mag of RF POIRT in cadence?

Why the gate of the mosfet is 2E? Assume the port is ac coupling to the gate and there is no match net work, which means signal at the right side and left side are different. One side is E, the other side is 2E.

Why？
how can a port with E can generate a 2E at the gate of mosfet?

Then Voltage gain of transient means Vout/Vin= Vout/E.
Why not Vout/2E?
Because the signal at gate is 2E~~

2
There are many voltage definition:
For      PAC, PXF and Pniose, it is Vout/(E/2)
For transient and PSP, it is Vout/Vin, Vin is the signal voltage at a input node of DUT.

I wonder if the input node means the + terminal of port or vsin?

It seems that Vout/(E/2) is used for power based system, while Vout/Vin is for voltage based system or for low frequency.

As mixer is a      RF block, it seems that the voltage conversion gain  Vout/（E/2）is correct. Say, PAC plot in Direct Plot is correct.
But why “http://www.designers-guide.org/Forum/YaBB.pl?num=1237656418/0” says , CG plot from Direct Plot is wrong?

3
I parallel a 50 Ohm resistor with PR port and LO port, respectively.
This is for broadband resistive matching.
But as http://www.designers-guide.org/Forum/YaBB.pl?num=1194006190 says, 50 Ohm does not one part of circuit. So the stimulation results are not correct.
But in reality,without 50 Ohm resisitor how can I realize broadband match?
My mixer is resistive mixer, input signal from tens of MHz to 2GHz. Can I use off-chip match network? But the imput impedance is time variant. How can I match the input impedance to 50Ohm for such a wideband?

pancho_hideboo wrote on Jun 28^th, 2010, 8:06am:


I can not understand.
In my opinion, if a RF port in cadence has E in PAC mag, it mean the port generate E Vp-p voltage to the gate of mosfet through AC capacitor.

In deed , the input impedance is very large. But why it is 2E instead of 3E or 4E. I can not understand it through intuition. Can I get the result from mathematical derivation?

pancho_hideboo wrote on Jun 28^th, 2010, 8:06am:


Which net represents Vin?
I am sorry. I do not know why I can not upload picture. So I can not describe my question clearly.
So I can only describe my circuit in following words.
Apploogized

Commonly, Vsin series a  AC capacitor. And the AC capacitor series the gate of mosfet.
Does Vin net mean the net between AC capacitor and gate of the mosfet?

While port is made of a ideal source and a internal resistor. If PAC mag of the port is E, the internal resistor is 50 Ohm. I series the port and gate of mosfet through a AC capacitor. Then signal at the gate of mosfet is 2E( I can not understand until now).

In these two cases, source signal means the signal at the gate of the mosfet.
And the gate net is the input net. Also the gate signal is the input signal.

pancho_hideboo wrote on Jun 28^th, 2010, 8:06am:



You say Vout/Vin is more useful and practical than Vout/(E/2) in IC chip.

I wonder does the CG in many IEEE paper represent Vout/Vin or Vout/(E/2)?

In my receiver, LNA, mixer and IF filter are all off chip block. The LNA and filter need to buy. The mixer is need to design in IC. So in the case Vout/(E/2) is more useful?



pancho_hideboo wrote on Jun 28^th, 2010, 8:06am:



Sorry PR is a mistake. I means 50 Ohm resistor paralleling with RF PORT and LO PORT.

pancho_hideboo wrote on Jun 28^th, 2010, 8:06am:



You say 50 Ohm resistors are part of my circuit.
Yes, mixer is a wideband mixer.
But I do not want this resistor as possible as I can. It seem than match net work is the only way. But for a wideband, it may be a little difficult work. Any other good method?