Page 120 - Complete Wireless Design

Amplifier Design

Amplifier Design 119

10 S 21 2

a. MAG 10 log 10 log |K ± K 1|
10
S
12
2
2
2
b. B 1 |S | |S | |D | 0.717
1 11 22 S
2.5 2
c. MAG 10 log 10 log (1.1 (1.1) 1 10.63 dB
0.139
Do not use vector quantities to calculate MAG and B , use magnitudes
1
only. Since B 0.717, then the sign between 1.1 and the square root in
1
step 2c is negative. 10.63 dB MAG is all right for our needs, so we continue.
3. Calculate (load reflection coefficient) required for a conjugate match for
L
the transistor. As stated above, the and are the values that the tran-
S L
sistor must have at its input and its output for a perfect match:
a. C S (D S
) (0.508 32°) [(0.25 61.4°) (0.195 167.6°)]
2 22 S 11
b. C 0.555 33.5°
2
2
2
2
2
2
c. B 1 |S | |S | |D | 1 (0.508) (0.195) (0.25) 2
2 22 11 S
d. B 1.157
2
e. Therefore, since B equals , the sign equals in the equation
2 L
below:
B ± (B ) 4|C | 2 1.157 (1.157) 4 (0.555) 2
2

2
2
2
2
0.748
L 2|C | 2 (0.555)
2
So the answer for the magnitude of is:
L
0.748
L
f. Now find the angle of : The angle equals the same value as in C
L 2
(0.555 33.5°), but is opposite in sign. Therefore, the angle of
L
33.5°.
g. Our complete answer is 0.748 33.5°
L
Use magnitudes, not vectors, to calculate B and .
2 L
4. Calculate the source reflection coefficient ( ):
S
S S S L
a. S
12
21
1 ( S )
11
L 22
S (0.139 61.2°) (2.5 62.4°) (0.748 33.5°)
b. 0.195 167.6°
1 (0.748 33.5°) (0.508 32°)
(0.61 160.8°)
c. 0.61 160.8°
S
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