# Node Analysis problem

1. Mar 18, 2006

### svenneman

Hi, I was searching google for answers about my problem and stumbled upon this forum. So I'll post my problem here and hope that some friendly soul can help me :)

I need to do a node-analysis of the circuit that I've attached here. The opamp is ideal so Ri=inf and R0=0 and A=inf. I end up with 3 equations for the nodes v1,v2:
eq1: (1/(1/jwC1)+1/R1)v1 = Uin/(1/jwC1)
eq2: (1/R3 + 1/(1/jwC2))v2 = Av1/R3
eq3: Uout=Av2

I am pretty sure I've done wrong in the node analysis, but I am unable to find out what I've done wrong (or right ;)). The equations I get later by solving the equationsystems doesn't add up. Any input or help on this problem is greatly appreciated!

#### Attached Files:

• ###### opamp.PNG
File size:
1.9 KB
Views:
130
Last edited: Mar 18, 2006
2. Mar 18, 2006

### SGT

You did it wrong. Equation 1 should be:
$$U_{in}\dot(j\omega C_1 + \frac{1}{R_1}) = V_1\dot\frac{1}{R_1})$$
or, in simpler form:
$$V_1 = U_in\dot\frac{R_1}{R_1+\frac{1}{j\omegaC_1}$$
Since the amplifier is ideal, there is a virtual short circuit between the + and - terminals and the output voltage is $$V_1$$
Then equation 2 will be:
$$V_2 = V_1\dot\frac{\frac{1}{j\omegaC_2}}{R_2+\frac{1}{j\omegaC_2}$$
and
$$U_{out} = V_2$$

3. Mar 18, 2006

### SGT

You did it wrong. Equation 1 should be:
$$U_{in}\dot(j\omega C_1 + \frac{1}{R_1}) = V_1\dot\frac{1}{R_1}$$
or, in simpler form:
$$V_1 = U_in\dot\frac{R_1}{R_1+\frac{1}{j\omegaC_1}}$$
Since the amplifier is ideal, there is a virtual short circuit between the + and - terminals and the output voltage is $$V_1$$
Then equation 2 will be:
$$V_2 = V_1\dot\frac{\frac{1}{j\omegaC_2}}{R_2+\frac{1}{j\omegaC_2}}$$
and
$$U_{out} = V_2$$

Reposted because I was not able to edit

4. Mar 18, 2006

### SGT

Still unable to write correctly the equations in LaTex I will write it in normal text.
Eq.1: V1 = Uin [R1/(R1+1/jwC1)]
Eq.2: V2 = V1 [(1/jwC2)/(R2+1/jwC2)]

5. Mar 18, 2006

### svenneman

Last edited: Mar 18, 2006
6. Mar 18, 2006

### SGT

Yes, my mistake.

7. Mar 18, 2006

### svenneman

oky, then it makes more sense :P

thanks for the help, I'll sit down with the new equations and see if I can get a result out of it :)