# LP filter calculation problem

1. Oct 24, 2013

### rogerk8

Hi!

I wonder why attached LP filter calculation does not compute.

I am considering the node P1 as a complex node yet my approach is obviously wrong.

The only way to solve this problem is with the use of good old KCL which makes

$$I_3-I_2-I_1=0$$
You would however also need to consider that the differential input is zero and that

$$U_-=U_o/Av$$
Yet I think my approach is more intuitive.

What am I missing?

Best regards, Roger

#### Attached Files:

• ###### LP.PNG
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2. Oct 24, 2013

### The Electrician

You haven't provided an overall expression for Uo/Uin. You seem to be headed that way, but you need to finish up.

3. Oct 25, 2013

### rogerk8

Hi!

Thanks for your interest in my problem!

Continueing my faulty approach...

$$U_+=U_o/A_v=(U_{in}+\frac{R_1}{R_1+1/sC_1}(U_o-U_{in}))\frac{1/sC_2}{R_2+1/sC_2}$$

which gives

$$U_o/A_v=U_{in}(\frac{1/sC_2}{R_2+1/sC_2}-\frac{R_1/sC_2}{(R_1+1/sC_1)(R_2+1/sC_2)})+ \frac{R_1/sC_2}{(R_1+1/sC_1)(R_2+1/sC_2)}U_o$$

thus

$$U_o/U_{in}=\frac{1/sC_2(R_1+1/sC_1)-R_1/sC_2}{(R_1+1/sC_1)(R_2+1/sC_2)-A_vR_1/sC_2}A_v$$

multiplying top (denominator?) and bottom (nominator?) with

$$s^2C_1C_2$$

gives

$$=\frac{A_v}{s^2R_1R_2C_1C_2+sR_1C_1+sR_2C_2+1-A_vsR_1C_1}$$

or

$$=\frac{A_v}{s^2R_1C_1R_2C_2+s(R_2C_2+R_1C_1(1-A_v))+1}$$

which is wrong.

$$=\frac{A_v}{s^2R_1C_1R_2C_2+s(C_2(R_1+R_2)+R_1C_1(1-A_v))+1}$$

I thus somehow lose part of the s-prefix which should be the sum of R1 and R2 multiplied with C2.

Yet I think my approach should give a correct answer because I have not "cheated" anywhere and even written the complete complex expression for the P1 node.

This is very strange to me. And a kind of disappointment because I have always felt that using voltages (KVL) is much easier than using currents (KCL).

Best regards, Roger

4. Oct 25, 2013

### The Electrician

You haven't shown how you derived equation 2 in your attached image, but I think I see what you're trying to do. You appear to treating the C1-R1 pair as a voltage divider which divides the difference between Uo and Uin, with the result of the division appearing across R1, and that voltage added to Uin.

The problem with this approach is that you haven't accounted for the loading effect of R2 and C2 on the divider. You can't just ignore that.

5. Oct 26, 2013

### rogerk8

I see what you mean.

It all comes down to the simple fact of R2-C2 loading on the P1 node.

And you can't really ignore that while the potential isn't a pure non-resistive Thevenin source, right?

Thank you for clearifying this to me!

Best regards, Roger