# Resonant inductor and saturation?

Discussion in 'General Electronics Chat' started by electronice123, Oct 12, 2014.

1. ### electronice123 Thread Starter Senior Member

Oct 10, 2008
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I am trying to design an inductor which will be part of a series LC resonant circuit.

At resonance the voltage across the inductor is higher than the applied voltage

Do I use the applied voltage or the magnified voltage when calculating flux density of the inductors magnetic core?

B=V*t/N*Ae

Just want to be sure...

Also, if the core hits saturation at resonance won't that force the circuit out of resonance since the inductance will drop?

2. ### Alec_t AAC Fanatic!

Sep 17, 2013
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Does this help?

3. ### electronice123 Thread Starter Senior Member

Oct 10, 2008
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I know the formulas already. Thank you though.

What I'm wondering is if the reactive voltage is used to calculate flux density. If it is then I need many many more turns of wire to make sure I don't saturate the core.

4. ### Alec_t AAC Fanatic!

Sep 17, 2013
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That's the voltage the coil experiences, so I would have thought so.

5. ### crutschow Expert

Mar 14, 2008
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It's the inductor current that generates the magnetic flux, not the voltage.

6. ### electronice123 Thread Starter Senior Member

Oct 10, 2008
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Ok, but I thought the volt-turns determined the flux density?

That's why I'm wondering, at resonance the voltage across the inductor is the product of Vsand the q factor. So it seems like at resonance the core flux density will be higher?

7. ### alfacliff Well-Known Member

Dec 13, 2013
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its ampere turns.

8. ### crutschow Expert

Mar 14, 2008
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The flux density will be higher at resonance but it's due to the higher inductor current at resonance (which, of course, is related to the inductor voltage and frequency).

9. ### MrAl Distinguished Member

Jun 17, 2014
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Hello,

The flux density can be gotten from the voltage if the voltage is pure AC because there is a relationship between current and voltage in an inductor.

The formula used is sometimes called the "Transformer Equation" and it is used for the primary winding but there are other equations also called that. The one here would be:
B=E*10^8/(4.44*F*A*N)

where
B is the flux density in Gauss,
E is the AC voltage in volts rms,
F is the frequency in Hertz,
A is the core cross sectional area in square centimeters,
N is the number of turns.

The current affects the inductance because the effective permeability starts out low for very low current, then quickly goes higher, stays somewhat higher, then eventually comes back down again as the current increases. The most important point is usually that if the current goes too high the inductance falls too much and so the inductance falls too low and causes problems. When the inductance falls too low in a converter circuit the current shoots up too high and can blow out transistors, and for a resonant circuit the inductance value will detune the circuit causing the resonant point to shift.

So for resonant circuits the target flux density is kept low so that the inductor current can never get too high, so it is best to keep the operating point somewhere in the middle of the BH curve which would be somewhere around half the saturation flux density for the material type.

For more reading on this though and a better understanding, i'll refer you to Magnetics Inc. which has a lot of design information on their site...
http://www.mag-inc.com/

They do have a section on designing inductors for resonant circuits somewhere in their documentation.

electronice123 likes this.
10. ### t_n_k AAC Fanatic!

Mar 6, 2009
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It is possible to determine the inductor operating flux conditions without reference to current. The reasoning is at least consistent if one considers the case of an ideal inductance in the first instance and then extends the idea to the case where saturation might be avoided.

It must be true that for a sinusoidal voltage e(t) across an inductor of N turns that

$e(t)=N \frac{d \Phi}{dt}$

Where $\text{\Phi }$ is the sinusoidal flux in the magnetic core

If we define e(t) as

$\text{e(t)=V_{m} \cos{\omega t}}$

it is readily shown that

$\text{\Phi=\frac{V_m}{\omega N} \sin{\omega t}}$

If the inductor magnetic circuit cross sectional area A is known and the magnetic core saturation flux density $\text{B_{sat} }$ is also known, one may reason that the inductor will not saturate provided

$\text{B_{sat} \gt \frac{V_{m}}{\omega AN}}$

Last edited: Oct 13, 2014
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11. ### MrAl Distinguished Member

Jun 17, 2014
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Hi tnk,

That looks good, and that's where the 4.44 comes from in the denom in post #9:
sqrt(2)/(2*pi)=1/(sqrt(2)*pi)=~=1/4.44