# Understanding op amp solution

Discussion in 'Homework Help' started by Hitman6267, Jun 2, 2010.

1. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0

I also don't know V1 V2... are the voltages at what resistors.

2. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
Nodal analysis is typically used to analyze an op amp with a single voltage source on the input.

You need to remember the ideal attrributes of an op amp:

• Infinite Gain
• Infinite Input Impedance
• Zero Output Impedance
• V+ = V-
• I+ = I- = 0
Consider the resistor in series with Vs. Since the V+ input has infinite input impedance, the current flow theoretically can't go into your V+ input. By KCL, the current has to go somewhere, so it goes through the feedback network. Based on those hints, see if you can solve it now.

3. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
Could you tell me the effects of having the grounds the way we do ? Thats what is throwing me off. I have no idea what they do.

4. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
Take a look at your first attachement. You see how V+ is connected to GND? Therefore, at that point, it is 0V. Also keep in mind that V+ = V-, so V- = 0 as well. I'll show a quick example of solving for the current in the first resistor in series with Vs.

$I_R_1 = \frac{V_S - 0}{R_1}$

5. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
I get that, here's where I'm facing trouble.
After the current passes the first R in the feedback cable does it divide itself between the R linked to the ground and the other ?

V2= -Vs -2 isrc R , is relative the the ground linked to the Vs. The V0 they calculated is relative to which ground ?

6. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
Think of KCL, all your resistors are equal to 0.5kΩ. So after the first feedback resistor, the current splits equally into each of the next resistors.

Try taking the op amp out of your drawing and just simplify all the resistors. Remember that the first resistor in series with Vs is also in series with the first feedback resistor.

7. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
Are the first R in the feedback loop and the R connected to the ground in parallel ?

8. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
Didn't I say to take your op amp out of your drawing and solve for the resistor network?

Look at the attachement, this can be solved by very simple source transformations.

Or look at the Vo resistor, it's in parallel with one of the R's. You could simplify the feedback network that way also.

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9. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
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First thank you for the effort of drawing it, second I can't do that unless I know if they're considered to be in parallel . In your drawing would the resistor that would result from the transformation of Vs and R1 be considered in parallel with R2 ?

10. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
It's almost like a puzzle, there's a certain way you have to solve it and you won't be able to solve it otherwise. In this case you'll have to work backwards, because R4 and R5 are in parallel with -8V across each of them. Or, do the source transformations yourself and you'll see which are in parallel or not.

Last edited: Jun 2, 2010
11. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
In your drawing between R4 and R5 there is a resistor missing. Did you remove it somehow ? If it still were there would R5 and R4 still be considered in parallel ?

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
All you need is to find equivalent resistance of this circuit :

Rz = 1.6R = 800Ω

Then we need to find the voltage on on this node

So if we find V1 then we know the current that is flow through Vs

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Last edited: Jun 2, 2010
13. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
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That's what I don't know how to do.

14. ### The Electrician AAC Fanatic!

Oct 9, 2007
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339
All the grounds are connected to the same point. Imagine that the circuit was sitting on a copper sheet; all the grounds could be connected to that sheet of copper. Ground is the reference for all the voltage measurements in your circuit.

15. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
We can try solve this circuit without finding equivalent resistance.
We draw this

Is = I1
I3 = I1 + I2
I5 = I3 + I4
V1 = I2*R2
V2 = I3*R3 + I2*R2
Vout = I5*R5 + I4*R4

Vs/R = (0 - V1) / R1 ----> V1 = -R1/R * Vs

I2 = V1/R2 = Vs/R2*R1/R

I3 = I1 + I2 = Vs/R + Vs/R2*R1/R = Vs/R *( (R1+R2) )/R2) =
Vs/R*(1+R1/R2)

V2 =
V2 = I3*R3 + I2*R2 = Vs*(R1/R + R3/R + R3/R*R1/R2)

I4 =
(Vs*(R1/R + R3/R + R3/R*R1/R2))/R4 = Vs/R4 * (R1/R + R3/R + R3/R*R1/R2) = Vs/R4 * [ (R3*(R2+R1))/(R*R2) + R1/R ]

I4*R4 = (R1/R + ((R1 + R2)*R3)/(R*R2)) *Vs

I5 = I3 + I4 and so on
We finally find the voltage gain:
Au = - (R1*R5*( R2 + R3 + R4 ) + R1*R4*( R2 + R3 ) + R2*R5*( R3 + R4 ) + R2*R3*R4 ) / (R*R2*R4)

It would be much faster if we substitute the nambers in the first place.

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