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

 

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.

 

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.

 

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.

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.

\(I_R_1 = \frac{V_S - 0}{R_1}\)

 

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 ?

 

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 ?

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.

 

Are the first R in the feedback loop and the R connected to the ground in parallel ?

 

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.

 

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 ?

 

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 ?

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.

 

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 ?

 

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

 

All you need is to find equivalent resistance of this

That's what I don't know how to do.

 

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.

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 ?

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.

 

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.