0
$\begingroup$

A certain RF transistor has a source - reflection coefficient with a magnitude of .105 and a phase angle of 160°. When plotted on the Smith Chart, the normalized impedance is .825 +j .06 Ω. What is the actual or unnormalized impedance? What does it depend on? Is it 50(.825 +j .06 Ω) or 41 +j 3 Ω? Is the Smith Chart automatically normalized to 50 when dealing with RF transistor reflection coefficients that were derived from S - parameter equations and when determining the actual impedance the transistor wants to see?

$\endgroup$
  • 1
    $\begingroup$ maybe you could link the datasheet? $\endgroup$ – Phil Frost - W8II Aug 12 '18 at 2:10
2
$\begingroup$

Since you know the complex reflection coefficient, you can simply use this formula to calculate the complex impedance:

$$Z=Z_O\frac{1+\rho}{1-\rho} \tag 1$$

Assuming the characteristic impedance is 50+j0 ohms, your transistor has an impedance of ~40.9+j3 ohms. You will notice that this is the normalized Smith chart impedance times 50 ohms (assuming it is normalized to 50 ohms). This is true since the normalization process simply divides the actual impedance by the normalization impedance.

To achieve maximum power transfer, the load on the transistor would be the complex conjugate of this or 40.9-j3 ohms.

Its S11 parameter may be computed as:

$$S11=20\log{(|\rho|)} \tag 2$$

This would result in an S11 value of -19.6 dB in your case.

$\endgroup$
  • $\begingroup$ With respect to the Smith Chart, how do you know what its' characterization is? $\endgroup$ – Thomas L Aug 14 '18 at 6:44
  • $\begingroup$ @ThomasL You choose the characteristic impedance to which you wish to normalize. In this way, the Smith chart is usable for any impedance ratio. When you are presented a Smith chart that contains normalized plots, the presenter needs to state the normalization value (typically the characteristic impedance) in order to be able to arrive at any useful conclusions from the presented chart. $\endgroup$ – Glenn W9IQ Aug 14 '18 at 14:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.