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I sort of understand the formulas for inductive reactance. Do they apply to a LC filter?

I have this, inductor at 700nH and caps at 120pF for a 20MHz cutoff emi filter:

schematic

simulate this circuit – Schematic created using CircuitLab

The specs for my EMI filter say max 8 ohms DC resistance. If I solve for inductance in the reactance formula for 8 ohms I get much higher inductance levels than 700nH. Is this happening, or am I missing a key concept here.

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  • \$\begingroup\$ The inductance is 700 nH. If you have conspired to somehow calculate inductance based on something that isn't stated in your question then, I repeat, the inductance is 700 nH (as stated). \$\endgroup\$
    – Andy aka
    Aug 5 '16 at 20:54
  • \$\begingroup\$ Inductance is the value of an inductor. Impedance is the "resistance" value which a certain component (like an inductor) has at a certain frequency. Reactance is the real part of the impedance, caused by only the resistors. You might want to ignore reactance for now until you fully understand impedance. Read some explanations on inductors and capacitors and their impedance because you are confused. \$\endgroup\$ Aug 5 '16 at 20:56
  • \$\begingroup\$ @FakeMoustache Does that mean you don't get to solve for inductance in a circuit like this? That if there is any extra impedance it is not a function of increased inductance but something else? \$\endgroup\$
    – RobC
    Aug 5 '16 at 20:58
  • \$\begingroup\$ I'm being pushed to show that given the chip has 8ohms max dc resistance, then given a 1KHz signal for example, that the inductance is calculated much higher... that sounds just wrong and the inductance is just 700nH and whatever else is happening has nothing to do with increased inductance. \$\endgroup\$
    – RobC
    Aug 5 '16 at 20:59
  • \$\begingroup\$ What "chip" are you talking about? You don't mention any chip in your question. Please, provide all the details of your setup, otherwise we may not be able to help you. \$\endgroup\$ Aug 5 '16 at 21:42
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Inductance is, in general, a fixed quantity dictated by the component construction, etc. Under some cases as pointed out by @N. G. near there can be parasitic effects that change this value. However, at 20 MHz, I doubt you will see any of these effects on your 700nH inductor. My guess is that the EMI spec is written so that at DC you don't have an inductor of more than 8 Ohms. Such a resistance would come from the wire windings since they are not perfect conductors. As long as your inductor of choice has a DC resistance lower than 8 Ohms, your filter should pass the EMI requirement you mention.

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Not so fast! Inductors do exhibit some resistance due to being wound from wire that has a finite conductivity. Also they possess some capacitance due to the windings being side by side in close proximity. These parasitic quantities are not inconsiderable at high frequencies - indeed an inductor will become self resonant at some frequency - this is often quoted by the manufacturer for RF inductors.

Also, you clear state '8 ohms at DC' - there is zero reactance at DC;

ZL = 2 * PI * f * L (DC equates to f = 0)

Shown below is a model of a typical inductor (source - http://www.coilcraft.com )

enter image description here

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  • \$\begingroup\$ 8ohms DC increasing from there, using 1 hertz and the ohms to solve for inductance gets crazy numbers, when i think the answer here is the inductance is just fixed at the value it claims to be. \$\endgroup\$
    – RobC
    Aug 8 '16 at 17:49
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This question is a logical falacy. You can't solve for inductance after you fix the inductance value.

The overall circuit reactance may change, but that is due to the frequency change OR relationship between the capacitors and the inductor, not the inductance changing.

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    \$\begingroup\$ 8 Ω is the DC resistance. At 0 Hz the impedance is 8 Ω. At any frequency above 0 the impedance is \$ 8 + j 2 \pi f L~ \Omega \$. \$\endgroup\$
    – Transistor
    Aug 8 '16 at 18:01
  • \$\begingroup\$ @Transistor It was assumed there was variable inductance due to a 'fixed' ohm rating at low frequencies when this is more a behavior of the overall 'complex' circuit and not the inductance itself changing. \$\endgroup\$
    – RobC
    Aug 8 '16 at 23:12

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