2
\$\begingroup\$

I want to determine the resistance \$R_k\$ and capacitance \$C_k\$ of this simple RC network by measuring the total and the shunt voltage:

schematic

simulate this circuit – Schematic created using CircuitLab

Therefore I generate a sine wave \$V_{T(otal)}=10.146 \ \mathrm{V}\cdot\sin(\omega t)\$ and measure the voltage \$V_{S(hunt)}\$. To get the formula, I calculate the current flowing through the lower node: $$ \frac{U_S}{R_S}=\frac{U_T-U_S}{Z_{R_k\parallel C_k}}=\frac{U_T-U_S}{\frac{R_k\cdot\frac{1}{j\omega C_k}}{R_k+\frac{1}{j\omega C_k}}} $$ $$ \rightarrow\frac{U_S}{R_S\cdot(U_T-U_S)}=\frac{1}{R_k}+j\omega C_k $$

Substituting the voltages by its complex pendants (\$U_S=\hat{U}_S\cdot e^{j\omega t}\$, \$U_T=\hat{U}_T\cdot e^{j\omega (t-T)}\$ ) and getting the final equation: $$ \frac{\hat{U}_{S}\cdot e^{j\omega t}}{R_S\cdot(\hat{U}_{T}\cdot e^{j\omega (t-T)}-\hat{U}_{S}\cdot e^{j\omega t})}=\frac{1}{R_K}+j\omega C_k $$ $$ \Rightarrow\frac{\hat{U}_{S}}{R_S\cdot(\hat{U}_{T}\cdot e^{-j\omega T}-\hat{U}_{S})}=\frac{1}{R_k}+j\omega C_k $$

I simulated the circuit in LTSpice and with the following parameters:

\$U_{Tmax}=10.146\ \mathrm{V}\\ R_k=10\ \mathrm{k\Omega}\\ C_k=1\ \mathrm{uF}\\ R_S=146\ \mathrm{\Omega}\\ f=100\ \mathrm{Hz}\$

I measured the following values:

\$U_{Smax}=0.925116\ \mathrm{V}\\ T=2.9\ \mathrm{ms}\$

The given equation results in:

\$R_k=5006\ \mathrm{\Omega}\\ C_k=915\ \mathrm{nF}\$

Why do my calculations differ from the values used in the simulation that much? What's my mistake?

\$\endgroup\$
2
  • \$\begingroup\$ Can you explain what \$\text{T}\$ is? \$\endgroup\$ Jan 5, 2023 at 18:52
  • \$\begingroup\$ T is the delay of the shunt voltage to the total voltage as visible in my 3 equation. $$U_S=U_{Smax}\cdot e^{j\omega t}$$, $$U_T=U_{Tmax}\cdot e^{j\omega (t-T)}$$ \$\endgroup\$
    – GURKE
    Jan 5, 2023 at 19:52

2 Answers 2

2
\$\begingroup\$

All your reasoning and mathematics are sound. It just looks like you mis-measured the time delay between Ut and Us.

I didn't run a simulation but did the math instead and it yelds some 2.105ms which plugged back in your formula returns 9.95kohm and 1.00uF.

In fact 2.9ms and 2.1ms sum up to 5ms which is a semiperiod, this suggests you simply swapped roles of reference voltage Us, and delayed one Ut while measuring the simulation waveforms.

\$\endgroup\$
2
  • \$\begingroup\$ Hey carloc! Thanks for your answer! You are true! I already tried changing \$T=2.9\ ms\$ to \$T=1/f-2.9ms=7.1\ ms\$ but I overlooked that I have compared rising edge to falling edge (zero crossing was calculated by LTSpice automatically), therefore it would be: \$T=1/(2f)-2.9\ ms\$ or just measuring the right T -.- SO, thanks! I would thumbs up your solution, but I have too less votes to do so 😬 \$\endgroup\$
    – GURKE
    Jan 6, 2023 at 7:16
  • 1
    \$\begingroup\$ Glad to help, you can accept my answer though so it all looks like "solved". You may just wait a little longer, waiting for someone else trying to answer. Cheers \$\endgroup\$
    – carloc
    Jan 6, 2023 at 7:26
3
\$\begingroup\$

Well, the total impedance of the circuit is given by:

\begin{equation} \begin{split} \underline{\text{Z}}_{\space\text{i}}&=\text{R}_\text{s}+\left(\frac{1}{\text{j}\omega\text{C}}\space\text{||}\space\text{R}_\text{k}\right)\\ \\ &=\text{R}_\text{s}+\frac{\frac{1}{\text{j}\omega\text{C}}\cdot\text{R}_\text{k}}{\frac{1}{\text{j}\omega\text{C}}+\text{R}_\text{k}}\\ \\ &=\text{R}_\text{s}+\frac{\frac{\text{j}\omega\text{C}}{\text{j}\omega\text{C}}\cdot\text{R}_\text{k}}{\frac{\text{j}\omega\text{C}}{\text{j}\omega\text{C}}+\text{j}\omega\text{CR}_\text{k}}\\ \\ &=\text{R}_\text{s}+\frac{\text{R}_\text{k}}{1+\text{j}\omega\text{CR}_\text{k}}\\ \\ &=\text{R}_\text{s}+\frac{\text{R}_\text{k}}{1+\omega\text{CR}_\text{k}\text{j}}\\ \\ &=\text{R}_\text{s}+\frac{\text{R}_\text{k}}{1+\omega\text{CR}_\text{k}\text{j}}\cdot\frac{1-\omega\text{CR}_\text{k}\text{j}}{1-\omega\text{CR}_\text{k}\text{j}}\\ \\ &=\text{R}_\text{s}+\frac{\text{R}_\text{k}\left(1-\omega\text{CR}_\text{k}\text{j}\right)}{1+\left(\omega\text{CR}_\text{k}\right)^2}\\ \\ &=\text{R}_\text{s}+\frac{\text{R}_\text{k}-\omega\text{CR}_\text{k}\text{R}_\text{k}\text{j}}{1+\left(\omega\text{CR}_\text{k}\right)^2}\\ \\ &=\text{R}_\text{s}+\frac{\text{R}_\text{k}-\omega\text{CR}_\text{k}^2\text{j}}{1+\left(\omega\text{CR}_\text{k}\right)^2}\\ \\ &=\text{R}_\text{s}+\frac{\text{R}_\text{k}}{1+\left(\omega\text{CR}_\text{k}\right)^2}-\frac{\omega\text{CR}_\text{k}^2}{1+\left(\omega\text{CR}_\text{k}\right)^2}\cdot\text{j} \end{split}\tag1 \end{equation}

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

So, we can see that the modulus of the impedance is given by:

\begin{equation} \begin{split} \left|\underline{\text{Z}}_{\space\text{i}}\right|&=\sqrt{\Re^2\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im^2\left(\underline{\text{Z}}_{\space\text{i}}\right)}\\ \\ &=\sqrt{\left(\text{R}_\text{s}+\frac{\text{R}_\text{k}}{1+\left(\omega\text{CR}_\text{k}\right)^2}\right)^2+\left(-\frac{\omega\text{CR}_\text{k}^2}{1+\left(\omega\text{CR}_\text{k}\right)^2}\right)^2}\\ \\ &=\sqrt{\left(\text{R}_\text{s}+\frac{\text{R}_\text{k}}{1+\left(\omega\text{CR}_\text{k}\right)^2}\right)^2+\left(\frac{\omega\text{CR}_\text{k}^2}{1+\left(\omega\text{CR}_\text{k}\right)^2}\right)^2} \end{split}\tag2 \end{equation}

And the argument of the impedance is given by:

\begin{equation} \begin{split} \arg\left(\underline{\text{Z}}_{\space\text{i}}\right)&=\arg\left(\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)\cdot\text{j}\right)\\ \\ &=\arg\left(\text{R}_\text{s}+\frac{\text{R}_\text{k}}{1+\left(\omega\text{CR}_\text{k}\right)^2}-\frac{\omega\text{CR}_\text{k}^2}{1+\left(\omega\text{CR}_\text{k}\right)^2}\cdot\text{j}\right)\\ \\ &=\arg\left(\underbrace{\text{R}_\text{s}+\frac{\text{R}_\text{k}}{1+\left(\omega\text{CR}_\text{k}\right)^2}}_{:=\space\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)\space>\space0}+\left(\underbrace{-\frac{\omega\text{CR}_\text{k}^2}{1+\left(\omega\text{CR}_\text{k}\right)^2}}_{:=\space\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)\space<\space0}\right)\cdot\text{j}\right)\\ \\ &=\frac{3\pi}{2}+\arctan\left(\frac{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\left|\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)\right|}\right)\\ \\ &=\frac{3\pi}{2}+\arctan\left(\frac{\text{R}_\text{s}+\text{R}_\text{k}}{\omega\text{CR}_\text{k}^2}+\omega\text{CR}_\text{s}\right) \end{split}\tag3 \end{equation}


Using this, you can see that:

$$\hat{\text{V}}_{\space\text{shunt}}=\frac{\hat{\text{V}}_{\space\text{source}}}{\left|\underline{\text{Z}}_{\space\text{i}}\right|}\cdot\text{R}_\text{s}\space\Longleftrightarrow\space\left|\underline{\text{Z}}_{\space\text{i}}\right|=\frac{\text{R}_\text{s}}{\hat{\text{V}}_{\space\text{shunt}}}\cdot\hat{\text{V}}_{\space\text{source}}\tag4$$

Which gives:

$$\sqrt{\left(146+\frac{\text{R}_\text{k}}{1+\left(200\pi\text{CR}_\text{k}\right)^2}\right)^2+\left(\frac{200\pi\text{CR}_\text{k}^2}{1+\left(200\pi\text{CR}_\text{k}\right)^2}\right)^2}=\frac{146}{0.925116}\cdot10.146\tag5$$

Solving \$(5)\$ gives:

$$\text{C}=\frac{\sqrt{\left(77093\text{R}_\text{k}-112187422\right)\left(77093\text{R}_\text{k}+134698578\right)}}{\text{R}_\text{k}\cdot29200\pi\sqrt{708926919351}}\tag6$$

With \$\displaystyle\text{R}_\text{k}>\frac{112187422}{77093}\approx1455.22\space\Omega\$.

In order to solve for \$\text{C}\$ and \$\text{R}_\text{k}\$ we need to find the argument of the shunt voltage (assuming that the argument of the source is zero):

\begin{equation} \begin{split} \arg\left(\underline{\text{V}}_{\space\text{shunt}}\right)&=\arg\left(\underline{\text{I}}_{\space\text{shunt}}\cdot\text{R}_\text{s}\right)\\ \\ &=\arg\left(\underline{\text{I}}_{\space\text{shunt}}\right)+\underbrace{\arg\left(\text{R}_\text{s}\right)}_{=\space0}\\ \\ &=\arg\left(\underline{\text{I}}_{\space\text{shunt}}\right)\\ \\ &=\arg\left(\underline{\text{I}}_{\space\text{in}}\right)\\ \\ &=\arg\left(\frac{\underline{\text{V}}_{\space\text{source}}}{\underline{\text{Z}}_{\space\text{in}}}\right)\\ \\ &=\underbrace{\arg\left(\underline{\text{V}}_{\space\text{source}}\right)}_{=\space0}-\arg\left(\underline{\text{Z}}_{\space\text{in}}\right)\\ \\ &=-\arg\left(\underline{\text{Z}}_{\space\text{in}}\right)\\ \\ &=-\left(\frac{3\pi}{2}+\arctan\left(\frac{\text{R}_\text{s}+\text{R}_\text{k}}{\omega\text{CR}_\text{k}^2}+\omega\text{CR}_\text{s}\right)\right) \end{split}\tag7 \end{equation}

So, we get:

$$\text{V}_\text{shunt}\left(t\right)=\hat{\text{V}}_{\space\text{shunt}}\sin\left(\omega t+\arg\left(\underline{\text{V}}_{\space\text{shunt}}\right)\right)\tag8$$

Which is the same as:

$$\text{V}_\text{shunt}\left(t\right)=\frac{\hat{\text{V}}_{\space\text{source}}}{\left|\underline{\text{Z}}_{\space\text{i}}\right|}\cdot\text{R}_\text{s}\cdot\sin\left(\omega t-\left(\frac{3\pi}{2}+\arctan\left(\frac{\text{R}_\text{s}+\text{R}_\text{k}}{\omega\text{CR}_\text{k}^2}+\omega\text{CR}_\text{s}\right)\right)\right)\tag9$$

And in order to solve for \$\text{T}\$ you need to solve \$(8)\$ when \$\text{V}_\text{shunt}\left(t\right)=0\$ for the first time with \$t>0\$.

I did the math for you and found, using the data provided:

$$\text{C}=\frac{1}{\frac{210477565549508 \pi \sec \left(\frac{2 \pi }{25}\right)}{651821315}-58400 \pi \tan \left(\frac{2 \pi }{25}\right)}\approx9.99760\cdot10^{-7}\space\text{F}\tag{10}$$ $$\text{R}_\text{k}=\frac{103503330225246}{77093 \left(845500 \sin \left(\frac{2 \pi }{25}\right)-77093\right)}-292\approx9789.36\space\Omega\tag{11}$$

Which is consistent with the chosen values.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Thanks Jan for this comprehensive reply and your different approach to this problem! I understand it and checked the math. I still wonder myself, what's wrong with my approach? Do you have an idea? \$\endgroup\$
    – GURKE
    Jan 5, 2023 at 20:28

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.