{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Ch_4prob_0008

# Ch_4prob_0008 - ©2009 National Technology and Science...

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

Problem 4.8 The op-amp circuit shown in Fig. P4.8 has a constant dc voltage of 6 V at the noninverting input. The inverting input is the sum of two voltage sources, a 6-V dc source and a small time-varying signal v s . (a) Use the op-amp equivalent-circuit model given in Fig. 4-5 to develop an expression for v o . (b) Simplify the expression by applying the ideal op-amp model, which lets A , R i and R o 0. R L v o v s 6 V 6 V + _ + _ _ _ Figure P4.8: Circuit for Problem 4.8. Solution: (a) Replacing the op amp with its equivalent circuit: R L R o R i v o v n v p v s _ _ 6 V _ 6 V + _ A ( v p - v n ) For node v o , v o R L + v o A ( v p v n ) R o = 0 . Also, v p = 6 V , v n = 6 V + v s . Solving for v o , we have: v o = Av s R L R L + R o . (b) For R L R o , v o = Av s . All rights reserved. Do not reproduce or distribute.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ©2009 National Technology and Science Press Hence, the output will be ampliﬁed by A , but inverted. If | Av s |≥ V cc , then the op amp will saturate. Hence, v o = − Av s , for | v s | ≤ V cc A − V cc , if | v s | > V cc A and v s is positive + V cc , if | v s | > V cc A and v s is negative. All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

Ch_4prob_0008 - ©2009 National Technology and Science...

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

View Full Document
Ask a homework question - tutors are online