This preview shows page 1. Sign up to view the full content.
Unformatted text preview: compared to C .
b. For L = 500 nH, C = 1 pF, and R = 1 Ω the term R2 /L2 = 4(1012 ) whereas
is an excellent approximation. Hence
LC = 2(1018 ) so ωp √1
= 225.1 MHz
2π LC c. At fp the impedance is purely resistive and equal to Z 1
ω2 C 2 R = 500 kΩ. d. At 0.75fp direct calculation shows that the reactance is 1212 Ω. The eﬀective inductance is therefore
Lef f = 2π(0.75)(225.1 MHz) = 1143 nH which is more than twice the low-frequency value of 500 nH. More
generally, assuming that R2
Lef f = X (ω )
1 − f2
fp which means that the eﬀective inductance will increase rapidly as f approaches fp from below. 4...
View Full Document
- Spring '08