power losses across inductive chokes

[Totally_Lost]

The DC component losses from I^2 * R end up as heat in a choke. I've
received some mixed signals about the AC reactance component of the
impedance of the choke. One theory, is the energy in the reactance is
simply phase shifted, and returned to the circuit out of phase, and
the other is that in an AC circuit the full impedance is subject to
I^2 * R losses in the choke. Which implies the choke power losses
increase with the frequency and ripple currents of the choke input
waveform.

Can someone provide some a better answer one way or the other?

Thanks!

 

[Phil Allison]

"Totally_Lost"

The DC component losses from I^2 * R end up as heat in a choke. I've
received some mixed signals about the AC reactance component of the
impedance of the choke. One theory, is the energy in the reactance is
simply phase shifted, and returned to the circuit out of phase, and
the other is that in an AC circuit the full impedance is subject to
I^2 * R losses in the choke. Which implies the choke power losses
increase with the frequency and ripple currents of the choke input
waveform.

Can someone provide some a better answer one way or the other?

** The heat loss in any *resistance* will be given by the formula I squared
R, where I is the RMS value of the current.

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
0.4% for each degree C rise.

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



...... Phil

 

[Phil Allison]

"Totally_Lost"

"Phil Allison"

** The heat loss in any *resistance* will be given by the formula I
squared
R, where I is the RMS value of the current.

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.

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 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
0.4% for each degree C rise.

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?

** Nope.

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


** I think you do.


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?

** That has been answered.

The formula is: I squared R .

where I = amps RMS.


..... Phil

 

[Paul E. Schoen]

"Phil Allison" <phil_a@tpg.com.au> wrote in message
news:7d4dc8F29cqkvU1@mid.individual.net...

"Totally_Lost"

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?

** That has been answered.

The formula is: I squared R .

where I = amps RMS.

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

 

[Phil Allison]

"Paul E. Schoen"

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.

** The current flowing in the coil is all that matters.

Any current flowing via parallel capacitance cause no energy loss.



...... Phil

 

[Phil Allison]

"Totally_Lost"

Thanks Phil.

So, the bottom line, is that the

** There is only energy loss in a *resistance*.

None is lost in inductive or capacitive reactances.

I squared R rules.


..... Phil

 

[Totally_Lost]

On Jul 26, 6:04 pm, "Phil Allison" <phi...@tpg.com.au> wrote:

** The heat loss in any *resistance* will be given by the formula I squared
R,  where I is the RMS value of the current.

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.

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 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
0.4% for each degree C rise.

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?

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

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? 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]

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 =
188.5 ohms., limiting the total current to 100/(188.5+99.999) = .3466A
RMS.

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.

 

[Totally_Lost]

On Jul 26, 8:06 pm, Totally_Lost <air_b...@yahoo.com> wrote:

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 > 188.5 ohms., limiting the total current to 100/(188.5+99.999) = .3466A
RMS.

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.

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

 

[Dan Coby]

Totally_Lost wrote:

On Jul 26, 8:06 pm, Totally_Lost <air_b...@yahoo.com> wrote:
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 =
188.5 ohms., limiting the total current to 100/(188.5+99.999) = .3466A
RMS.

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.

I guess the part I'm struggling with is apparent power supplied by the
sine wave generator is 100*.346621 = 34.66W. Power dropped by the
resistor appears to be .3466^2*99.999 = 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)

 

[Totally_Lost]

Mucho thanks Dan ... now things make sense


On Jul 27, 1:07 am, Dan Coby <adc...@earthlink.net> wrote:

Totally_Lost wrote:
On Jul 26, 8:06 pm, Totally_Lost <air_b...@yahoo.com> wrote:
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 > >> 188.5 ohms., limiting the total current to 100/(188.5+99.999) = .3466A
RMS.

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.

I guess the part I'm struggling with is apparent power supplied by the
sine wave generator is 100*.346621 = 34.66W. Power dropped by the
resistor appears to be .3466^2*99.999 = 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)

 

[Tim Wescott]

On Sun, 26 Jul 2009 17:49:29 -0700, Totally_Lost wrote:

On Jul 26, 6:04 pm, "Phil Allison" <phi...@tpg.com.au> wrote:
** The heat loss in any *resistance* will be given by the formula I
squared R,  where I is the RMS value of the current.

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.

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 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 0.4% for each degree C rise.

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?

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

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?
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!

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.

--
www.wescottdesign.com

 

[Phil Allison]

"Tim Wescott"

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.

** LOL - only a demented code scribbler would say such trash.



..... Phil