# Which book is this problem from?

Discussion in 'Homework Help' started by jimmeh, Dec 2, 2011.

1. ### jimmeh Thread Starter New Member

Dec 2, 2011
5
0
Hey,

I just wanted to ask which book is this question from.

Thanks

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2. ### Gump Member

Jun 7, 2010
57
1
I don't have a clue, but to help others narrow things down:

1) When you say book are you meaning a printed book or one of AAC e-books/worksheets?
2) Where did you get the figure from?
3) What is the actual question, as that's just a schematic, the question could be anything.

3. ### jimmeh Thread Starter New Member

Dec 2, 2011
5
0
a) I mean Textbook.
b) From the quiz given by my instructor.
c)It wants you to find an expression for Vo the resistance values in variable terms.

4. ### Vahe Member

Mar 3, 2011
75
9
You should be able to write an expression for Vo in terms of the resistances and the voltage sources Va and Vb. One approach is to take one voltage source at a time while turning the other off (short circuit). This is basically the application of superposition.

--Vahe

5. ### Vahe Member

Mar 3, 2011
75
9
Sorry, it looks like I tried to give you clue to solve the problem rather than answering your inquiry -- which was about the source of the figure. I am not sure what course you are taking to be able to have a guess. If you taking an electronics course, I would look at Sedra and Smith or something like that. If this is your first elementary circuits course you may want to look at Alexander/Sadiku, Nillson, Hayt/Kemmerly, etc.

--Vahe

6. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
It looks like a Sedra/Smith image, but I can't find it in my copy. Maybe I'm not looking hard enough, or I have a different edition.

7. ### praondevou AAC Fanatic!

Jul 9, 2011
2,936
489
It's from "Electrical-Engineering-Principles-and-Applications 5th edition" from Allan R. Hambley, chapter 14, page 706, figure P14.21.

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jimmeh likes this.
8. ### jimmeh Thread Starter New Member

Dec 2, 2011
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@praondevou

Thank you so much man !!!!

9. ### jimmeh Thread Starter New Member

Dec 2, 2011
5
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This is the solution but, I cant seem to reach it in any way..

I did

vp-va/ra + vp-vb/rb = o

Tried to isolate vo.. but couldnt reach the right answer.

Help Anyone?

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10. ### praondevou AAC Fanatic!

Jul 9, 2011
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The origine of the left part of the formula you posted you will find on Wikipedia under "non-inverting amplifier".
R1 and R2 determine the voltage gain of the amplifier.

The other part of the formula is simply giving you the voltage at the non-inverting input (where Ra and Rb are tied together).

Both values are then multiplied.

Separate both parts and try to resolve for the voltage at the noninverting input first.

11. ### Vahe Member

Mar 3, 2011
75
9
You can use superposition to do this problem, the total output voltage $v_o$ is due to $v_A$ alone with $v_B$ turned off (short circuit) plus due to $v_B$ alone with $v_A$ turned off. Start with $v_A$ alone and $v_B=0$ (short circuit).

Under this condition, $R_A$ and $R_B$ form a voltage divider to produce the voltage at the noninverting terminal

$
V_+ = \frac{R_B}{R_A + R_B} v_A
$

Now, you can note that $R_1$ and $R_2$ form an noninverting amplifier with gain of $1 + R2/R1$ from $V_+$ to $v_o$ and therefore

$
v_o = \left( 1+ \frac{R_2}{R_1} \right) V_+ = \left( 1+ \frac{R_2}{R_1} \right) \frac{R_B}{R_A + R_B} v_A
$

You can also find the relation between $V_+$ and $v_o$ by noting that $V_-=V_+$ and $I_-=0$ and writing a KCL equation at the inverting terminal.

Now consider $v_B$ alone and $v_A=0$ (short circuit).

Under this condition, $R_A$ and $R_B$ once again form a voltage divider to produce the voltage at the noninverting terminal (however in this case $R_A$ is grounded instead of $R_B$)

$
V_+ = \frac{R_A}{R_A + R_B} v_B
$

As in the previous case, $R_1$ and $R_2$ form an noninverting amplifier with gain of $1 + R2/R1$ from $V_+$ to $v_o$ and therefore

$
v_o = \left( 1+ \frac{R_2}{R_1} \right) V_+ = \left( 1+ \frac{R_2}{R_1} \right) \frac{R_A}{R_A + R_B} v_B
$

The total output voltage is the sum of the individual output voltages computed above; therefore,

$
v_o = \left( 1+ \frac{R_2}{R_1} \right) \frac{R_B}{R_A + R_B} v_A + \left( 1+ \frac{R_2}{R_1} \right) \frac{R_A}{R_A + R_B} v_B = \left( 1+ \frac{R_2}{R_1} \right) \left( \frac{R_B}{R_A + R_B} v_A + \frac{R_A}{R_A + R_B} v_B \right)= \left( \frac{R_1+R_2}{R_1} \right) \left( \frac{R_B v_A + R_A v_B}{R_A + R_B} \right)
$

The last form matches the answer that you provided. Hope this explanation helps you out.

Best regards,
Vahe

Last edited: Dec 3, 2011