robismyname
Full Member level 6
- Joined
- Jan 17, 2008
- Messages
- 390
- Helped
- 11
- Reputation
- 22
- Reaction score
- 9
- Trophy points
- 1,298
- Location
- Central Florida
- Activity points
- 4,603
lumped matching
Im following an example of how to design a lumped matching circuit to match 50 ohms in order to ensure minimum noise. The example starts of by implemeting a series inductor.
The example assumes that the data sheet or CAD yields a optimum noise match of 3 + j2.0 ; (normalised to 50-ohms).
If you look at part III page 1 of the article, I noticed that there is green admittance circle (G=0.2). I see how he got the G=0.2 circle. My problem is that underneath the smith chart diagram on page 1 he says that the distance between B and C gives the value of the inductor as shown in part 1.
So I then looked at part I page 10(bottom) he's saying to subtract .2 from .5 to get the reactance (XL) and then the value of L.
My question is since .2 is associated with an admittance circle and .5 is associated with a impedance circle how can you subtract an admittance circle from an impedance circe to obtain (XL) and L?
Im following an example of how to design a lumped matching circuit to match 50 ohms in order to ensure minimum noise. The example starts of by implemeting a series inductor.
The example assumes that the data sheet or CAD yields a optimum noise match of 3 + j2.0 ; (normalised to 50-ohms).
If you look at part III page 1 of the article, I noticed that there is green admittance circle (G=0.2). I see how he got the G=0.2 circle. My problem is that underneath the smith chart diagram on page 1 he says that the distance between B and C gives the value of the inductor as shown in part 1.
So I then looked at part I page 10(bottom) he's saying to subtract .2 from .5 to get the reactance (XL) and then the value of L.
My question is since .2 is associated with an admittance circle and .5 is associated with a impedance circle how can you subtract an admittance circle from an impedance circe to obtain (XL) and L?