4
\$\begingroup\$

I have the following circuit with resistors and capacitors in parallel (see Picture 1):

Picture 1

I have to determine the capacitance \$ C_1 \$ in such way that \$\frac{u_e}{u_a} = 1 + \frac{Z_1}{Z_2} \$ is not dependent on the frequency \$ \omega \$.

I know that \$ Z_1 = Z_{R_1} || Z_{C_1} = \frac{R_1}{1+ j \omega R_1 C_1} \$ and \$ Z_2 = Z_{R_2} || Z_{C_2} = \frac{R_2}{1+ j \omega R_2 C_2} \$.

Then I get \$ \frac{Z_1}{Z_2} = \frac{\frac{R_1}{1+ j \omega R_1 C_1}}{\frac{R_2}{1+ j \omega R_2 C_2}} = \frac{R_1 + j \omega R_1 R_2 C_2}{R_2 + j \omega R_1 R_2 C_1} \$.

But from here on I fail to simplify this to calculate \$ C_1 \$. I tried to multiply this expression with the complex conjugate \$ \frac{R_1 - j \omega R_1 R_2 C_2}{R_2 - j \omega R_1 R_2 C_1} \$ but I just get a "mess" that I can't simplify

The solution should be \$ C_1 = 2,22 pF \$

Any help would be appreciated!

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Collect R1 on the numerator and R2 on the denominator. Does this help? \$\endgroup\$ Commented May 9, 2023 at 20:50
  • \$\begingroup\$ @SredniVashtar sorry, but not really.. I get \$ \frac{R_1(1+ j \omega R_2 C_2)}{R_2 (1 + j \omega R_1 C_1)} \$.. and then? \$\endgroup\$
    – syphracos
    Commented May 9, 2023 at 20:56
  • 2
    \$\begingroup\$ Good. What happens to the w-dependent factors when R1C1 =R2C2? \$\endgroup\$ Commented May 9, 2023 at 20:58
  • 1
    \$\begingroup\$ @SredniVashtar oooh, now I get it!! Thank You very much!! I tried so much with simplifications of the expression after multiplying it with the complex conjugate of the denominator that I didn't see that it's so "trivial" haha :) \$\endgroup\$
    – syphracos
    Commented May 9, 2023 at 21:00

2 Answers 2

1
\$\begingroup\$

Well, notice that the transfer function of this circuit is given by:

$$\underline{\mathscr{H}}\left(\text{j}\omega\right)=\frac{\displaystyle\text{R}_2\space\text{||}\space\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}_2}}{\displaystyle\left(\text{R}_1\space\text{||}\space\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}_1}\right)+\text{R}_2\space\text{||}\space\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}_2}}=\frac{\displaystyle\text{R}_2\left(\text{C}_1\text{R}_1\omega-\text{j}\right)}{\displaystyle\text{R}_1\text{R}_2\left(\text{C}_1+\text{C}_2\right)\omega-\left(\text{R}_1+\text{R}_2\right)\text{j}}\tag1$$

Where \$\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.

Now, what happens when \$\text{C}_1=\frac{\displaystyle\text{C}_2\text{R}_2}{\displaystyle\text{R}_1}\$:

\begin{equation} \begin{split} \underline{\mathscr{H}}\left(\text{j}\omega\right)&=\frac{\displaystyle\text{R}_2\left(\frac{\displaystyle\text{C}_2\text{R}_2}{\displaystyle\text{R}_1}\cdot\text{R}_1\omega-\text{j}\right)}{\displaystyle\text{R}_1\text{R}_2\left(\frac{\displaystyle\text{C}_2\text{R}_2}{\displaystyle\text{R}_1}+\text{C}_2\right)\omega-\left(\text{R}_1+\text{R}_2\right)\text{j}}\\ \\ &=\frac{\displaystyle\text{R}_2\left(\text{C}_2\text{R}_2\omega-\text{j}\right)}{\displaystyle\text{R}_2\left(\frac{\displaystyle\text{C}_2\text{R}_1\text{R}_2}{\displaystyle\text{R}_1}+\text{C}_2\text{R}_1\right)\omega-\left(\text{R}_1+\text{R}_2\right)\text{j}}\\ \\ &=\frac{\displaystyle\text{R}_2\left(\text{C}_2\text{R}_2\omega-\text{j}\right)}{\displaystyle\text{R}_2\left(\text{C}_2\text{R}_2+\text{C}_2\text{R}_1\right)\omega-\left(\text{R}_1+\text{R}_2\right)\text{j}}\\ \\ &=\frac{\displaystyle\text{R}_2\left(\text{C}_2\text{R}_2\omega-\text{j}\right)}{\displaystyle\text{C}_2\text{R}_2\left(\text{R}_1+\text{R}_2\right)\omega-\left(\text{R}_1+\text{R}_2\right)\text{j}}\\ \\ &=\frac{\displaystyle\text{R}_2\left(\text{C}_2\text{R}_2\omega-\text{j}\right)}{\displaystyle\left(\text{R}_1+\text{R}_2\right)\left(\text{C}_2\text{R}_2\omega-\text{j}\right)}\\ \\ &=\frac{\displaystyle\text{R}_2}{\displaystyle\text{R}_1+\text{R}_2}\cdot\frac{\displaystyle\text{C}_2\text{R}_2\omega-\text{j}}{\displaystyle\text{C}_2\text{R}_2\omega-\text{j}}\\ \\ &=\frac{\displaystyle\text{R}_2}{\displaystyle\text{R}_1+\text{R}_2} \end{split}\tag2 \end{equation}

\$\endgroup\$
1
  • \$\begingroup\$ That's a lot of unnecessary math :-) \$\endgroup\$ Commented May 11, 2023 at 23:55
4
\$\begingroup\$

Here’s a hint. At low frequencies the resistors dominate. Ask yourself why. At high frequencies the resistors become insignificant. Ask yourself why. Should the low frequency ratio be the same as the high frequency ratio?

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.