08hw4solutions

08hw4solutions - ECE287A HWK4 Solutions Problem 1 1) Since...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE287A HWK4 Solutions Problem 1 1) Since Class E amplifiers have efficiency of 100%, P out = P dc = VccI 0 = 0.5 ° so I 0 = 0.5 Vcc = 0.5 4 = 0.125A I L = 1.862I 0 = 0.233A & I max = 2.862I 0 = 0.358A P out = 1 2 R L I L 2 = 1 2 R L (0.233) 2 = 0.5W R L = 18.42 Ω R L = 0.18 ω C = 18.42 Ω ω C = 9.77 × 10 3 C = 1.58pF X L = 0.21 ω C = 21.49 Ω L L = 0.21 ω 2 C = 3.42nH V max = 3.6V cc = 3.6 × 4 = 14.4V , so we need to set breakdown voltage higher, to 25V. Add a filter to the load, 0 = j ω L + 1 j ω C = j ω L 1 ω C ± LC = 1 ω 2 = 2.53 × 10 20 Since the conduction angle is ² , the device should have approximately equal time on and off, we could bias the transistor at threshold voltage, but if we do so, current at -3V will be restricted to a large extent, and the transistor can not behave like an ideal switch. After some different trials, choose Vgs=-1.8V. Fig 1. Equivalent circuit of a Class E amplifier without a filter
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/09/2010 for the course ECE 230 at UCSD.

Page1 / 4

08hw4solutions - ECE287A HWK4 Solutions Problem 1 1) Since...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online