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Coil Impedance

by FredGarvin
Tags: coil, impedance
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FredGarvin
#1
Jun8-06, 07:06 AM
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I'm straying off the path by having to look at an electrical problem. I was hoping you guys might be able to point me in the right direction since it has been a long time since I have done anything like this. I'm hoping I'm having a serious brain fart and it's nice and easy.

What I have is a coil of some set resistance (the coil is being used as a resistive heater element). I have a DC power source feeding the coil. Dealing with the calculations in the DC realm is straight forward. No issues there. However, I was asked about the coil's impedance as well. So, I am assuming that, because the word "impedance" was used, that they are refering to a probable AC component coming from the PS. Is this a safe assumption?

Also, if my PS has a ripple spec of, say 10%, how would one calculate the AC component of the output based on the 10%? It's a 3 AC input.

Thanks for the help
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berkeman
#2
Jun8-06, 11:07 AM
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The magnitude of the impedance of an inductor is [tex]|Z| = 2 \pi f L[/tex]
where f = frequency and L = inductance of the coil. It takes a lot of inductance to give you much impedance at 50/60Hz powerline frequencies. Power transformers, for example, use many turns around ferrous iron plate cores to get the inductance up enough to work at AC powerline frequencies. An air core (or other non-ferrous core) coil will have a low inductance, and hence a low impedance at powerline frequencies.

If the DC power supply spec is for a ripple voltage of 10% or less, that means that the AC ripple waveform that is superimpressed on the DC will have 10% or less of the DC value. So the output would look something like this in the worst case:

[tex]V = V_{dc} (1 + 0.1 sin(\omega t))[/tex]
FredGarvin
#3
Jun8-06, 03:08 PM
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Excellent. I was hoping I could do something like that along those lines. Thanks for the assist.


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