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Unformatted text preview: R 1 = R f R 2 = 4 k v s = 3 V . Hence, v o = p R f + 4 k 4 k P 3 = ( . 75 R f + 3 ) V, with R f in k . P L = v 2 o R L = v 2 o 3k = 75 10 3 W , or v o = 15 V, which is less than V cc = 16 V. Solving for R f : 15 = . 75 R f + 3 , which gives R f = 16 k . All rights reserved. Do not reproduce or distribute. 2009 National Technology and Science Press...
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This note was uploaded on 04/19/2010 for the course EE 220 taught by Professor Rawat during the Spring '09 term at Nevada.
- Spring '09