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− | [[Category: | + | [[Category:ECE201]] |
+ | [[Category:ECE]] | ||
+ | [[Category:ECE201Spring2015Peleato]] | ||
+ | [[Category:circuits]] | ||
+ | [[Category:linear circuits]] | ||
+ | [[Category:problem solving]] | ||
− | = | + | =Summing Amplifier Practice= |
+ | <center><font size= 4> | ||
+ | '''Practice question for [[ECE201]]: "Linear circuit analysis I" ''' | ||
+ | </font size> | ||
+ | By: Chinar Dhamija | ||
+ | Topic: Summing Op Amp | ||
− | + | </center> | |
+ | ---- | ||
+ | ==Question== | ||
+ | Find the output voltage, Vout. | ||
+ | [[File:ECE201_P4.png|700px|center]] | ||
+ | ---- | ||
+ | ---- | ||
+ | ===Answer === | ||
+ | First we know that <math>V_-</math> = <math>V_+</math><br /> | ||
+ | <math>V_-</math> = 0V = <math>V_+</math><br /> | ||
+ | The op amp can be ignored for this particular problem.<br /> | ||
+ | We can apply the formula we learned in class to solve for Vout.<br /> | ||
+ | <math>\begin{align} | ||
+ | Vout = (\frac{\frac{-R_f}{R_1}})V_1 + (\frac{\frac{-R_f}{R_1}})V_2 | ||
+ | Vout = -24V | ||
+ | \end{align} | ||
+ | </math> | ||
+ | ---- | ||
+ | ==Questions and comments== | ||
+ | If you have any questions, comments, etc. please post them below | ||
+ | *Comment 1 | ||
+ | **Answer to Comment 1 | ||
+ | *Comment 2 | ||
+ | **Answer to Comment 2 | ||
+ | ---- | ||
+ | [[2015 Spring ECE 201 Peleato|Back to 2015 Spring ECE 201 Peleato]] | ||
− | [[ | + | [[ECE201|Back to ECE201]] |
Revision as of 16:54, 26 April 2015
Summing Amplifier Practice
Practice question for ECE201: "Linear circuit analysis I"
By: Chinar Dhamija
Topic: Summing Op Amp
Question
Find the output voltage, Vout.
Answer
First we know that $ V_- $ = $ V_+ $
$ V_- $ = 0V = $ V_+ $
The op amp can be ignored for this particular problem.
We can apply the formula we learned in class to solve for Vout.
$ \begin{align} Vout = (\frac{\frac{-R_f}{R_1}})V_1 + (\frac{\frac{-R_f}{R_1}})V_2 Vout = -24V \end{align} $
Questions and comments
If you have any questions, comments, etc. please post them below
- Comment 1
- Answer to Comment 1
- Comment 2
- Answer to Comment 2