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Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

The half-wave rectiﬁer as shown is operating at a frequency of 60Hz, and the rms value of the transformer output voltage v_I is 12.6V±10%. What are the nominal and worst case values of the dc output voltage V_O if the diode voltage drop is 1V?

Expert Solution

Step 1

Given data,

The value of rms voltage is V_rms = 12.6 $\pm$ 10% V.
The value of operating frequency is f = 60 Hz.
Thee value of voltage in on-state is V_ON = 1 V.

The expression for nominal dc output voltage is given as,

$\begin{array}{rcl} V_{r m s} & = & \frac{- V_{d c}^{n o m} + V_{O N}}{\sqrt{2}} \\ V_{d c}^{n o m} & = & - (\sqrt{2} V_{r m s} - V_{O N}) \\ = & - (\sqrt{2} \times 12.6 - 1) \\ = & - 16.8 V \end{array}$

Step 2

The expression for maximum rms value of transformer output voltage is,

$\begin{array}{rcl} V_{r m s}^{m a x} & = & (1 + \frac{10}{100}) V_{r m s} \\ = & 1.1 \times 12.7 \\ = & 13.97 V \end{array}$

Hence the expression for maximum value of dc output voltage is,

$\begin{array}{rcl} V_{d c}^{m a x} & = & - (\sqrt{2} V_{r m s}^{m a x} - V_{O N}) \\ = & - (\sqrt{2} \times 13.97 - 1) \\ = & - 18.74 V \end{array}$

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