2
\$\begingroup\$

Consider the following the frequency response of the impedance of a resistor component : enter image description here

We know that a resistor has a parasitic capacitance and inductance due to its leads. So the impedance of the resistor = \$Z_{inductor} + \frac{Z_{resistor} \cdot Z_{capacitor}}{Z_{resistor} +Z_{capacitor}}\$.

To find this parasitic capacitance we should use the point at \$f = 1.592\cdot 10^3\$ in such a way that \$100 = \frac{1}{\omega\cdot C}\$.

So my question is, why did we equate Z=R=100 to \$\frac{1}{\omega\cdot C}\$ only?

Why not to \$Z_{inductor} + \frac{Z_{resistor} \cdot Z_{capacitor}}{Z_{resistor} +Z_{capacitor}}\$?

One more question, why did the impedance decrease after this mentioned point? According to my understanding, the impedance before this point =100=R, if the capacitor effect will be effective after this point then this effectiveness should be added to R (i.e Z= R+1/jwc) and therefore the curve should not decrease?

\$\endgroup\$
3
\$\begingroup\$

So my question is why did we equate Z=R=100 to \$ \big( \frac{1}{j\omega C} \big)\$ only? Why not to \$Z_{inductor} + \frac{Z_{resistor} \cdot Z_{capacitor}}{Z_{resistor} +Z_{capacitor}}\$?

You're right. You should use the latter to calculate it correctly, but the first approximates it very well and computes more easily.
The effect of the inductor is present over the whole frequency range. But it starts to be significant at higher frequencies, so, at lower frequencies its impedance can be approximated by zero.

Why did the impedance has decreased after this mentioned point?

The impedance of the capacitor decreases with increasing frequency \$ \big( \frac{1}{j\omega C} \big)\$. It is in parallel with the resistor. So the effective impedance of the will decrease with increasing frequency. The effect of the capacitor is also present over the whole frequency range. But it starts to be significant around this mentioned point.

\$\endgroup\$
2
\$\begingroup\$

Realize that this is a really bad academic question.

You will never find a resistor with these breakpoints.
Also, the graph misrepresents the numbers The 4 digits of values is another clue this is also a poor question for stray capacitance.

Choose R=Zc(f) gives the reduction in impedance at the low frequency of 1.6kHz results in C= 1uF of "parasitic capacitance" from basic LPF from RC values.

The minimum impedance at 5MHz only comes from a series LC "Series Resonant Frequency (SRF)" using C, from above.

The minimum impedance needs to be 5 decades down to have these breakpoints at 100 Ohms 3 decades apart meaning the notch at resonance is at 10 milliohms.

But with R=100, C= 1uF then L= 1nH approx. This is NOT a RESISTOR. But it could be massive parallel plates of lossy thin film plastic many meters wide with about 0.1mm gap to a ground plane to achieve these values.

So throw out the window all your assumptions about parasitics when given such an academic question.

strong text

Note the weak similarity with the given asymptotic graph.

\$\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.