Vin – Vout Vout – VD.on - R1 R2 How to solue for Vout It follows that R2 Vin + V D.on R1 R2 1+ R1 Vout
Vin – Vout Vout – VD.on - R1 R2 How to solue for Vout It follows that R2 Vin + V D.on R1 R2 1+ R1 Vout
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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How did they solve for Vout? What are the steps they used?
![## Solving for Output Voltage (\( V_{\text{out}} \))
The equation given is:
\[
\frac{V_{\text{in}} - V_{\text{out}}}{R_1} = \frac{V_{\text{out}} - V_{D,\text{on}}}{R_2}.
\]
The annotation next to this equation asks, "How to solve for \( V_{\text{out}} \)?"
### Derivation:
To solve for \( V_{\text{out}} \), it follows that:
\[
V_{\text{out}} = \frac{\frac{R_2}{R_1} V_{\text{in}} + V_{D,\text{on}}}{1 + \frac{R_2}{R_1}}.
\]
This formula helps in evaluating the output voltage based on the input voltage, \( V_{\text{in}} \), diode forward voltage, \( V_{D,\text{on}} \), and the resistances \( R_1 \) and \( R_2 \).
Please note that the explanation is augmented with basic algebraic manipulation to isolate \( V_{\text{out}} \) from the initial equation, following the rules of proportion and fraction arithmetic.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5362cbea-9427-4b2b-99cb-57a208edd853%2Fa7a9d3fd-2a13-4c9d-8e18-8be21331173f%2Fy654g2o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Solving for Output Voltage (\( V_{\text{out}} \))
The equation given is:
\[
\frac{V_{\text{in}} - V_{\text{out}}}{R_1} = \frac{V_{\text{out}} - V_{D,\text{on}}}{R_2}.
\]
The annotation next to this equation asks, "How to solve for \( V_{\text{out}} \)?"
### Derivation:
To solve for \( V_{\text{out}} \), it follows that:
\[
V_{\text{out}} = \frac{\frac{R_2}{R_1} V_{\text{in}} + V_{D,\text{on}}}{1 + \frac{R_2}{R_1}}.
\]
This formula helps in evaluating the output voltage based on the input voltage, \( V_{\text{in}} \), diode forward voltage, \( V_{D,\text{on}} \), and the resistances \( R_1 \) and \( R_2 \).
Please note that the explanation is augmented with basic algebraic manipulation to isolate \( V_{\text{out}} \) from the initial equation, following the rules of proportion and fraction arithmetic.
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