Ch_4prob_0009

# Ch_4prob_0009 - R 1 = R f R 2 = 4 k Ω v s = 3 V Hence v o...

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Sections 4-3 and 4-4: Ideal Op-Amp and Inverting Amp Assume all op amps to be ideal from here on forward. Problem 4.9 The supply voltage of the op amp in the circuit of Fig. P4.9 is 16 V. If R L = 3 k , assign a resistance value to R f so that the circuit would deliver 75 mW of power to R L . R L R f V cc = 16 V 3 V 50 4 k + _ + _ Figure P4.9: Circuit for Problem 4.9. Solution: Per Table 4-2, this is a noninverting ampliﬁer circuit for which v o = Gv s = p R 1 + R 2 R 2 P v s . In this case,
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Unformatted text preview: R 1 = R f R 2 = 4 k Ω 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 Ω . 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 Ω . All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press...
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