rf lna

Is this statement true ? I am trying to do a differential LNA and am having trouble trying to picture how to simulate it in say SpectreRF.

Simple...

Design a L-C balun centered at the frequency of interest with 50Ohm input and 200Ohm output impedance and place it between your LNA and generator and simulate S-Parameter for s-parameters and noise parameters ( don't forget to open noise box ) and PSS to find IP3 and that's it.
To design balun, use this formula.

L=sqrt(Rs.RL)/omega
C=1/(omega*sqrt(Rs.RL)

Formulaes come from Peter Wizmuller RF Design Guide.

Thank you but are those matching numbers correct ?

Why 100Ohm?

Yes it might be...

The input impedance of a LNA can be differential 200 Ohm and your system characteristic impedance generally 50 Ohm. So you have to convert 200Ohm diff to 50 Ohm single ended impedance.
In fact 100 Ohm single ended input impedance in anyhow symmetry will be 200 Ohm differential.It's very clear...

so when creating a differential LNA like this, your input match is to 200 ohms and not 50 ohms ?

(noise, gain matching)

I'm not convined..

If I was say design a balanced amplifier then I would design a single 50 ohm amplifier then connect it to some sort of combiner to 50 ohms (1:1), not 100 or 200ohms.

In my Diff LNA tutorial the single ended LNA was designed for 50 ohms, then for the differential version I connected a 1:1 Balun - seems to work OK?

odyseus,

i read your tutorial after i saw that and noticed what you had done.

i am very confused as to what to do, since i need to make a differential lna !

any more help appreciated !

On reflection...

One way to look at it is this.. with a 1:1 balun will provide two 25 ohm outputs,
a 1:2 will provide two 50 ohm outputs and 1:4 will provide two 100 outputs.

Now at DC the two LNA inputs will be connected together by the balun coil and will be in parallel! thus their combined impedance will be half.

So with two 100 ohm LNA's and a 1:4 balun you should transform to a source of 50 ohms?

Or design the LNA's to be 50 ohms and use a 1:2 balun?

Maybe Bigboss can comment?

odyseus i think your analysis is right.

but is the ADS block you used in your tutorial a 1:1 balun ?

the tool, as usual, is what confuses me

http://www.designers-guide.com/Analysis/diff.pdf

Odyseus,

I just read your new CMOS Differential Mixer Tutorial on RFIC.CO.UK.

There you made your own baluns.

My question, relevant to this thread is whether the ADS 3 port balun performs this transformation for you (the 4:1 transformation I spoke about). In ADS, you cannot change the ratio or any other parameters of the 3 or 4 port baluns.

In Cadence you can do this.

Did you make those baluns in your paper ? Then used them ?

How do you find the ones in ADS that you used for your differential CMOS LNA ?

Thanks!

diff lna

Be aware that NF is a parameter related to char impedance.
The best way to proceed is to simulate it as explained by bigboss (or, if you want a wideband balun you can use a transformer with proper turn ratio, in the 1:4 impedance ratio you should use 1:2 turn ratio, or you you have Cadence RFLib, there is a balun in it).
The LNA will be measured in this way, so, excluding real balun losses and nonidealities it will be easier to check the results.

Take care of the voltage gain that, in matched conditions, you will have between single end and differential side of the balun: it will hel you in NF, but will degrades the linearity.

I hope it will help.
Mazz

oh,oh! I'm seeing only now that it's an old discussion!

I think it is useful

http://www.designers-guide.com/Analysis/diff.pdf