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时间:03-31 整理:3721RD 点击:
I am designing an amplifier at 2 GHz. while designing the matching network, this is how I understand it:
for i/p: Zo has to be transformed to Zs conjugate so plot Zs conjugate on smith chart and move from center of smith chart to tht point.

for o/p: ZL conjugate has to be transformed to Zo so plot ZL conjugate and move to center of smith chart.
Is this right?
'coz when I run the simulation the results are obtained when I make both Zs and ZL conjugate move to the center of the Smith Chart. Similar thing happened when I designed a mixer.anybody could throw more light?

Although I'm new to this, I know that the centre of the Smith chart marks the point of real-only impedence at Zo (for most cases, this is 50+0i ohms).
For maximum power transfer, the input and output (s11 and s22) should be transformed to their conjugates, so you don't want to get to the centre of the chart, you want to get from Γin to Γin* [and also Γout to Γout*]

Now, when you've finished you want to match Γout* to your load - which again, is normally 50-ohm. In that case, you _now_ want a match from Γout* to the centre of the Smith.

Maybe one of the "big kids" here can confirm if that's correct - I think it's true, though. Hope it's helpful.

_______
Agronski.

thnx fr the reply. the output part is fine. I mean u have to move to the center from ZL*. but the input also has to be matched to 50 Ω 'coz thats how you give the input through a 50Ω line. But, I was under the impression that u r supposed to move from the centre (Zo=1) to Zs* so that you transform 50 Ω to the complex impedance. but somehow, that does not seem to work.
If I begin from Zs* and add elements and move towards the center everything is fine. Am I doing the same thing and not gettng it?

Sorry, I got a little confuzzled - since you're matching input and output to 50-ohms, you need to move _from_ your given points to the centre of the Smith - i.e. the impedence looking from the source to the device you're matching to should Γin*.

If you start at the centre and try to move, you won't be able to! The Smith apparently works the same way for inputs and outputs - so for load matching, start at your Γout* point and then move until you hit the centre.

There's a useful tutorial here:
http://www.maxim-ic.com/appnotes.cfm/an_pk/742

Try and match the load given in the exam to 50-ohms, and you're numbers should come out the same.

________
Agronski.

thnx fr the reply. I finally got it cleared up. I was under the impression the at the input side u r supposed to start from the center and move towards Zin* but now I've corrected that. thnx agn

>for o/p: ZL conjugate has to be transformed to Zo so plot ZL conjugate and move >to center of smith chart.
>Is this right?
I think there is something wrong here
It depends in which direction you are going (from source to load or opposite)
you can say:
ZL (and not ZL conjugate) has to be transormed to Z0 conjugate (which is certainly the same as Z0 (when Z0 is real))
or
Z0 has to be transformed to ZL conjugate.

Of course this assumes that you are in a linear mode of operation. If you design a PA, the load pull technique must be applied.

thnx fr the reply. Well, I am moving towards the 50 Ω line starting from ZL i.e. transform ZL to Z0. but for maximum power transfer I am supposed to use the conjugate right? I am using ADS, so I measure the Zs and ZL using smith Chart in data display. so whatever value I get, I am supposed to transform that to Zo. But the value I use on the Smith chart will be the conjugate values for max. power transfer.
If I'm reading this wrong, b glad if u could correct.

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