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What would the magnitude response curve for a resistor, inductor and capacitor look like with parasitic effects not ideal components? From what I have gathered they would look something along the lines of:

These are log-log graphs.

Resistor: enter image description here

Inductor:

enter image description here

Capacitor:

enter image description here

Resistor: At high frequency the inductor characteristic dominates jwl hence impedance increase, capacitance would go to zero and R would become negligible.

Inductor: there is the peak there due to the resonance otherwise straight line gradient 1.

Capacitor: straight line gradient -1, not sure of any parasitic effects here?

Are these correct? If not I would love to hear your contributions. If my explanations are lacking anything please add on for better understanding.

To add on to this, if I were to draw the circuit symbols of these non-ideal passive components what would they look like?

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  • \$\begingroup\$ This would probably be better as three separate questions. \$\endgroup\$
    – The Photon
    Commented Dec 8, 2018 at 1:06
  • \$\begingroup\$ What do you mean by magnitude response? On their own they are just two terminall components and, without other components, there is no input and output to differentiate. So, what do you mean? \$\endgroup\$
    – Andy aka
    Commented Dec 8, 2018 at 9:50
  • \$\begingroup\$ Not using Logarithmic graphs a Reactive Inductance increases linearly with frequency so the Voltage peaks of an AC waveform decreases with increasing Frequency. Reactive Capacitance decreases Exponentially with increases in Frequency so an AC Voltage will increase in Peak to Peak voltage as the frequency increases if passed through a capacitor. A resistor has series reactive inductance and parallel reactive capacitance. It has 1/f noise so only the inductance has a useful frequency response as the capacitance is over-run by 1/f noise \$\endgroup\$ Commented Dec 8, 2018 at 11:13
  • \$\begingroup\$ You write '\$ R \$' and '\$ |Z| \$' at your graphs. If this applies to the capacitor graph also then the concave decay in impedance isn't so strange as it's \$ |Z| = (\omega L)^{-1} \$. \$\endgroup\$ Commented Dec 26, 2018 at 6:06
  • \$\begingroup\$ For the capacitor: if you increase the frequency enough, the package inductance will come to dominate, and the Z curve will change from (roughly) linear (on a log/log scale) decline to a linear incline (shape of the letter "v"). The low point in the curve is where the capacitance and inductance have the same Z value. As an example: a 10PF 0402 SMD cap will reach this low point at around 2GHz. \$\endgroup\$
    – Troutdog
    Commented Jun 7, 2023 at 18:32

3 Answers 3

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I think the graphs you have are not correct in general. You can't draw generalizations without discussing the frequency range of interest and physical construction of the components.

Essentially every real component can be modeled as a combination of ideal components. The more ideal components you keep adding the higher order and more precise your model becomes. Once you do the transfer function of this "real approximation" model using ideal components you will get a bode plot which will show you how the part will behavior under various frequencies.

It is true that often a resistor will first start looking like an inductor, a capacitor first like an inductor and an inductor first like a capacitor as you get outside of their expected performance range, but this is very much a rule of thumb to be taken with a grain of salt.

For example below are some first order approximations that are good enough for most EE work and rooted in physics...but precise applications may require more ideal components to model various behavior than is even shown here.

Here is a resistor:

enter image description here

Here is a capacitor:

enter image description here

Here is an inductor:

enter image description here

Here is a decent intro to bode plots.

Addendum

Here is a Vishay study on frequency response of thin-film resistors

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    \$\begingroup\$ Thank you for your response. Considering a series network for wire wound resistors, if the frequency was to increase to very high levels, would the overall impedance of the circuit increase? and would also the phase of the circuit begin to change from 0-90 as it starts to act like an inductor. \$\endgroup\$
    – fred
    Commented Dec 8, 2018 at 15:48
  • \$\begingroup\$ @fred I don't have the data and what is "very high frequency" will depend on the particular component. There are multiple wiring methods too (eg simple vs bifilar) which would impact my answer. In general for a simple wirewound resistor I would expect it to first behave as a resistor, then as an air core inductor and at even higher frequencies the inter-winding capacitance would probably start to dominate. Signal phase will follow the rules for what the apparent impedance of the circuit is at that frequency. PS Also added study on thin film resistors FYI as an example. \$\endgroup\$
    – EasyOhm
    Commented Dec 8, 2018 at 21:04
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Here's some general points;

  • tracks and axial parts are 1nH/mm +/-50% typ depending on geometry
  • shunt capacitance depends on surface area/gap ratio
  • crosstalk can be computed as coplanar stripline in nH/cm and pF/cm for some geometry
    • transfer function depending on input/path impedance ratio.
  • Common Mode coupling depends on stray EMF and MMF , imbalanced impedances, proximity
  • CM noise is common and is reduced with balanced low impedances, CM choke (= "Baluns") to raise and balance the CM Z
    • then add shunt ground caps as Pi filters to attenuate with 2nd order CM noise reduction.
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In general, refer to the component datasheets or characteristics, when available.

There are too many types to go through in detail, but this is an illustration of what to expect from chip resistors:

From Frequency Response of Thin Film Chip Resistors - Vishay:

enter image description here

Generally speaking, the larger the component, the lower the cutoff/break frequencies will be. This shows 0201 outperforming 0402/0603, and it seems there's some competition between the latter, but 0402 generally comes out on top.

For resistors of small values (under 100 ohms or so), inductance dominates first, and the curve bends upward. Larger, capacitance dominates first, and the curve bends down.

The curve diverges sooner, the farther the value is from that point -- that is, a 10Ω or 1k diverge 10 times lower than 100, etc.

The "flattest" resistance (the value for which flat/resistive bandwidth is maximum) varies with construction type. We can expect spiral-cut and wirewound types to have a higher "best" impedance.

Fundamentally, what this impedance is, is the characteristic impedance of the assembly. That is, there are leads with some stray inductance, end caps or pads with some capacitance, and a distribution of both along the resistive element itself. The characteristic impedance is given by \$Z_0 = \frac{L}{C}\$.

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