1
\$\begingroup\$

I am calculating transfer function with some parallel caps in the circuit, same ones. I have parallel 4 capacitors with 100 microF each then total capacitance is 400 microF and it is presented as

\$ Z_{tot} = \dfrac{1}{s \cdot 400 \times 10^{-6}} \$

But if I calculate it this way then I got different result: One capacitor:

\$ Z = \dfrac{1}{s \cdot 100 \times 10^{-6}} \$

4 caps in parallel:

\$ Z_{tot} = \dfrac{Z \cdot Z \cdot Z\cdot Z }{Z + Z + Z + Z} = \dfrac{Z^3}{4} = \dfrac{1}{s^3 \cdot 4 \times 10^{-12}} \$

which is totally different result, and wrong I guess, but I don't see where the error is... I would appreciate some help for this basic question.

\$\endgroup\$
  • 4
    \$\begingroup\$ ‘Product over sum’ only works for two impedances \$\endgroup\$ – Chu Aug 30 '19 at 13:17
  • \$\begingroup\$ The dimension of your equation says it all \$\endgroup\$ – Bart Aug 30 '19 at 14:31
1
\$\begingroup\$

As mentioned in comments the formula for \$ n \$ impedances in parallel is:

\$ Z_{tot} = \dfrac{1}{\dfrac{1}{Z_1} + \dfrac{1}{Z_2} + \dfrac{1}{Z_3} + \dfrac{1}{Z_4}} \$ etc. Extend to any number of parallel impedances.

This only simplifies to \$ Z_{tot} = \dfrac{Z_1 \cdot Z_2}{ Z_1 + Z_2} \$ If there are exactly two impedances, not more.

\$\endgroup\$
  • \$\begingroup\$ Thank you for your answer, and to others also.Thanks everybody for answers. It was really about basics, but sometimes you get stuck and nothing works except someones help. \$\endgroup\$ – krg Sep 1 '19 at 20:48
1
\$\begingroup\$

The math doesn't work that way. In general, be careful not to assume equations can just be extended by throwing in more variables in a form that looks similar, especially when it is just tossing things into more than one place )in this case, both numerator and denominator). Fractions just don't work that way. It doesn't make any intuitive mathematical sense if you actually think about it.

\$ Z = \frac{Z_1Z_2}{Z_1+Z_2} \$ is a shortcut formula for only two resistances.

The real base formula from which it is derived is:

\$\frac{1}{Z} = \frac{1}{Z_1} + \frac{1}{Z_2} +...+\frac{1}{Z_n}\$

I don't even remember or use the simplified two-Z equation. I only ever use the base form.

\$\endgroup\$
  • \$\begingroup\$ This is somewhat unrelated, but involves the same kind of mindset when manipulating numbers and equations so don't do this either. It involves converting complex impedances to admittance that have both real and imaginary parts versus a shortcut when the impedance ONLY has a real or imaginary part. Just like this case, you can't use the shortcut beyond its bounds as it is not the general case and it doesn't make intuitive sense if you think about it. electronics.stackexchange.com/questions/449982/… \$\endgroup\$ – DKNguyen Aug 30 '19 at 14:49
0
\$\begingroup\$

for parallel capacitors, you really need to treat each capacitor and the traces connecting them as an RLC model with inductance, and resistance, without this extra information it will likely not reflect how it reacts in the real world, as multiple capacitors can actually increase impedance in spikes around certain frequencies.

e.g. https://youtu.be/BcJ6UdDx1vg?t=1465

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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