power losses across inductive chokes

"Totally_Lost"

The RMS current value may consist of a DC plus and an AC RMS component - which if known separately are combined by taking the square root of the sum of the squares of each value.

In the case of an inductor with a magnetic core, there are additional heat losses in the core and the R value of the copper wire will increase at high frequencies due to skin effect and of course increase with temperature by

The all up loss is by no means a trivial thing to compute in a real situation.

..... Phil

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Ok, that is a better definition of how to compute the resistive losses for the AC component, that is super imposed on the DC offset voltage/ current.

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In this particular case, the inductor is an air core with a lot of turns, so there are no eddy current losses in an iron core. The ripple frequency is between 600 to 3000 Hertz. Is skin effect really a problem at these low frequencies?

So, I still do not have an answer to the problem originally posted.

I suppose I should have put it more correctly as, are the actual choke heating losses I^2 * Z, where Z =3D sqrt (R^2 + Xl^2) across the inductor? This is implied by the AC voltage drop across the inductor as the frequency increases. Not that different than the forward conduction voltage drop across a diode the results in V*I losses.

Thanks!

"Totally_Lost"

Ok, that is a better definition of how to compute the resistive losses for the AC component, that is super imposed on the DC offset voltage/ current.

In this particular case, the inductor is an air core with a lot of turns, so there are no eddy current losses in an iron core. The ripple frequency is between 600 to 3000 Hertz. Is skin effect really a problem at these low frequencies?

So, I still do not have an answer to the problem originally posted.

I suppose I should have put it more correctly as, are the actual choke heating losses I^2 * Z, where Z = sqrt (R^2 + Xl^2) across the inductor?

The formula is: I squared R .

where I = amps RMS.

.... Phil

So, IOW, the only thing that counts is the resistive (R) component. And the only tricky part may be measuring or computing the true RMS current if the waveform is complex.

I suppose at higher frequencies there might be interwinding capacitance that could cause additional current that would not be included in the heating losses formula.

Paul

Thanks Phil.

So, the bottom line, is that the V*I energy in the voltage drop across the choke is returned to the circuit out of phase, and doesn't create any significant heat as long as the coil resistance is relatively low.

So that a sine voltage source at 100VAC RMS 3000Hz, feeding an inductor with Xl=10mH and R=0.001 ohms, in series with a resistive load of 99.999ohms (and no DC component), will produce 99.999% of the heating losses in the resistor, even though the inductor is creating the majority of the voltage drop with Z approximately equal to Xl =

So, in effect the inductor drops the voltage into the resistive load by about 67V with a .346621^2*0.001= .000120W power loss in the inductor.

"Paul E. Schoen"

Any current flowing via parallel capacitance cause no energy loss.

..... Phil

"Totally_Lost"

None is lost in inductive or capacitive reactances.

I squared R rules.

.... Phil

I guess the part I'm struggling with is apparent power supplied by the sine wave generator is 100*.346621 =3D 34.66W. Power dropped by the resistor appears to be .3466^2*99.999 =3D 12.014W. So, I'm trying to figure out why this is off by 22W. Somewhere I got something very wrong.

There are some things that need to be corrected in the earlier calculations.

The impedance is sqrt((99.999 + 0.001)**2 + 188.5**2) = 213.4 ohms. The value previously calculated (188.5 + 99.999) ignored the fact that the reactive impedance of the inductor is 90 degrees out of phase with the resistance of the resistor.

The phase angle will be arctan(188.5/100) = 62.05 degrees.

The current is 100 / 213.4 = 0.4686 amps.

The power lost in the inductor is I**2 * 0.001 = .0002196 watts ( 0.2196 mW).

The power lost in the 99.999 resistor is 0.4686**2 * 99.999 = 21.96 watts.

The power from the power supply is v * i * cos(angle) = 100 * .4686 * cos(62.05) = 21.96 watts.

Please note that the cosine of the phase angle in the previous calculation. This value is called the 'power factor'. This factor is needed because we have two things happening. We have real power being dissipated in the resistance and we have reactive power as energy is being loaded and unloaded twice per cycle in the magnetic field of the inductor.

The reactive power in the inductor is (0.4686)**2 * 188.5 = 41.39 watts.

The volt * amps from the power supply is 100 * 0.4686 = 46.86 voltamps.

This same value is given by sqrt(21.96**2 + 41.39**2)

Mucho thanks Dan ... now things make sense :)

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To find the average power loss, plot the voltage drop across the component times the current through the component as a function of time and average the result.

Resistance causes a component of the product of current and voltage drop that goes as 1 + cos(2*pi*f*t); reactance causes a component that goes as cos(2*pi*f*t).

So what you were told is true, the energy from reactance just gets stored in the inductor for a while then returned to the circuit later, while the energy from resistance gets burned up.

"Tim Wescott"

.... Phil