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Magnetic Flux Density Depending on Current

D

D from BC

Jan 1, 1970
0
I don't know what it is about magnetics..it just never sinks in.. :(

Here's what's on my inductor:
Voltage: 240Vpp 40% duty square wave @ 100khz
Current: +/-100mA triangle wave hovering around 1.5amps.
L=1.3mH
u=125

Problem:
1) What's Bmax
2) What's delta B

I thought about solving this is terms of voltage using
Bmax=(Vrms*10E8)/4.44fAN
But the Vrms is supposed to be from a sine wave...I have squareish.

However, it looks easier to solve in terms of current.
Neglecting the ripple current.. (low impact of accuracy)
and just calling it 1.5Amps of DC current and using:

magnetizing force H
expressed by:
H=(0.4*pie*NI)/l
N=turns
I=peak current Amps
l=magnetic path length cm

along with u = B/H
(u=125)

B can be solved. (Gauss)
Hopefully to be less than the datasheet core sat.

Delta B would be from the +/-100mA or 200mApp.
200mA is put into the B=uH equation.
I'd use that to look up the core loss on the datasheet.
(The more Bchange the more core loss.)

Then I found this eqn:
Bdc=(L*Idc*10E8)/(N*Ac)
L=inductance
Idc=DC amps
N=turns
Ac=cross section cm^2
Which should give the same results.

Have I goofed up somewhere?
D from BC
 
J

John Larkin

Jan 1, 1970
0
I don't know what it is about magnetics..it just never sinks in.. :(

Here's what's on my inductor:
Voltage: 240Vpp 40% duty square wave @ 100khz
Current: +/-100mA triangle wave hovering around 1.5amps.
L=1.3mH
u=125

Problem:
1) What's Bmax
2) What's delta B

I thought about solving this is terms of voltage using
Bmax=(Vrms*10E8)/4.44fAN
But the Vrms is supposed to be from a sine wave...I have squareish.

However, it looks easier to solve in terms of current.
Neglecting the ripple current.. (low impact of accuracy)
and just calling it 1.5Amps of DC current and using:

magnetizing force H
expressed by:
H=(0.4*pie*NI)/l
N=turns
I=peak current Amps
l=magnetic path length cm

along with u = B/H
(u=125)

B can be solved. (Gauss)
Hopefully to be less than the datasheet core sat.

Delta B would be from the +/-100mA or 200mApp.
200mA is put into the B=uH equation.
I'd use that to look up the core loss on the datasheet.
(The more Bchange the more core loss.)

Then I found this eqn:
Bdc=(L*Idc*10E8)/(N*Ac)
L=inductance
Idc=DC amps
N=turns
Ac=cross section cm^2
Which should give the same results.

Have I goofed up somewhere?
D from BC

Magnetics datasheets are a pain. For things like pot cores and
toroids, for a given material, why don't they give us gauss per
ampere-turn? That, plus Al which they usually *do* supply, would be a
big help.

John
 
F

Fred Bloggs

Jan 1, 1970
0
Here's what's on my inductor:
Voltage: 240Vpp 40% duty square wave @ 100khz
Current: +/-100mA triangle wave hovering around 1.5amps.
L=1.3mH
u=125


Pick one terminal of inductor as reference so that for flux balance you
must have VH*0.4+VL*0.6=0 and VH-VL=240 making VH=240*0.6=144V and
VL=144-240=-96V. Then you have 144V=N*d/dt(phi)=N*d/dt(Aeff*B) so that
dB=144*dt/(N*Aeff)=144*0.4/(N*Aeff*100E3). Then Bmax=Bdc+dB where
Bdc=N*Idc*u(1.5A) where u(1.5A) denotes permeability of core at 1.5A dc
bias.
 
F

Fred Bloggs

Jan 1, 1970
0
Fred said:
Pick one terminal of inductor as reference so that for flux balance you
must have VH*0.4+VL*0.6=0 and VH-VL=240 making VH=240*0.6=144V and
VL=144-240=-96V. Then you have 144V=N*d/dt(phi)=N*d/dt(Aeff*B) so that
dB=144*dt/(N*Aeff)=144*0.4/(N*Aeff*100E3). Then Bmax=Bdc+dB where
Bdc=N*Idc*u(1.5A) where u(1.5A) denotes permeability of core at 1.5A dc
bias.

Actually that's a bit conservative. Since flux balance requires that
+dB=|-dB|, or equivalently, Bdc=(Bmax+Bmin)/2, then since Bmax=Bmin+db
and vice versa, you have Bmax=(Bmax+Bmin)/2+(Bmax-Bmin)/2=Bdc+dB/2.
 
L

legg

Jan 1, 1970
0
Then I found this eqn:
Bdc=(L*Idc*10E8)/(N*Ac)
L=inductance
Idc=DC amps
N=turns
Ac=cross section cm^2
Which should give the same results.

Have I goofed up somewhere?

B units of Tesla, area units in square meters, inductance units of
henries - knocks out the 10E8 multiplier from this formula, (if it is
the correct multiplier with the original units).

DC bias has no frequency relation, so the earlier sinusoidal formula
is misapplied in the average static case, regardless of its potential
suitability elsewhere..

B= n . I . ue / le in teslas amps and meters.

For gapped structures, out of saturation:

B = n . I . uo / lg

l in amps
uo = 4 . pi . 10E-7
lg = gap length meters

AC flux, as always, is

deltaB = E . t / ( n . a )

volts, seconds, meters and Teslas,
Half the delta is arithmetically additive/subtractive to the DC flux
value.

RL
 
D

D from BC

Jan 1, 1970
0
B units of Tesla, area units in square meters, inductance units of
henries - knocks out the 10E8 multiplier from this formula, (if it is
the correct multiplier with the original units).

DC bias has no frequency relation, so the earlier sinusoidal formula
is misapplied in the average static case, regardless of its potential
suitability elsewhere..

B= n . I . ue / le in teslas amps and meters.

For gapped structures, out of saturation:

B = n . I . uo / lg

l in amps
uo = 4 . pi . 10E-7
lg = gap length meters

AC flux, as always, is

deltaB = E . t / ( n . a )

volts, seconds, meters and Teslas,
Half the delta is arithmetically additive/subtractive to the DC flux
value.

RL

In this case, just using the nearly DC low ripple current waveform
through the inductor, I could simply use the relation:

B=(n.I.ue)/le
ue is that the permeability?
le is the the magnetic path length?

I guess this would be useful too if I ever needed to make an
electromagnet powered by a DC current source.

D from BC
 
D

D from BC

Jan 1, 1970
0
Actually that's a bit conservative. Since flux balance requires that
+dB=|-dB|, or equivalently, Bdc=(Bmax+Bmin)/2, then since Bmax=Bmin+db
and vice versa, you have Bmax=(Bmax+Bmin)/2+(Bmax-Bmin)/2=Bdc+dB/2.


Oops my OP is a little fuzzy...

There's no current bias through the inductor. The inductor current
waveform is the result of being in a continuous mode converter. The
converter doesn't let the inductor current drop to zero.
Using the inductor voltage waveform in a B calculation and then adding
on B from the inductor current waveform doesn't seem right.
But ok for dc bias as described.

Bdc=N*Idc*u
Is this a lean equation?..
No other inductor parameters like Across and lmag path length??
But then, I'm ok with B error up to 20%.

Note: I'm watching the u and aware it can be current dependent.
D from BC
 
L

legg

Jan 1, 1970
0
In this case, just using the nearly DC low ripple current waveform
through the inductor, I could simply use the relation:

B=(n.I.ue)/le
ue is that the permeability?

yes - the permeability of the medium used to store the energy.
le is the the magnetic path length?

In gapped media, the gap will tend to dominate, hence dominance of the
gap length.

In homogenous media the full path length carries the permeability
of the material used. This may degrade with direct current bias.
I guess this would be useful too if I ever needed to make an
electromagnet powered by a DC current source.

An electromagnet develops a useful external field in the (varying)
gap.The field does not have to be DC; is often more effective with AC
due to the potential reduction in residual magnetism.

RL
 
T

Tim Williams

Jan 1, 1970
0
D from BC said:
1) What's Bmax
2) What's delta B

I wish I knew. Just what *IS* the purpose of quoting Bmax for ferrite,
anyway? That doesn't help me a damn when I want to measure the fucker.
Geometry is fixed, give me amp-turns, vs. gap if need be!

Tim
 
D

D from BC

Jan 1, 1970
0
[snip]
I guess this would be useful too if I ever needed to make an
electromagnet powered by a DC current source.

An electromagnet develops a useful external field in the (varying)
gap.The field does not have to be DC; is often more effective with AC
due to the potential reduction in residual magnetism.

RL

I'm thinking of the electromagnet used by the auto wreckers to pick up
cars.. AC or DC?
So an electromagnets core might stay a little magnetized after a shot
of DC.
Wouldn't an AC powered electromagnet just be kinda jiggy when
attracting say some iron..
Reminds me of those buzzing solenoids for doors.
D from BC
 
F

Fred Bloggs

Jan 1, 1970
0
D said:
Bdc=N*Idc*u
Is this a lean equation?..
No other inductor parameters like Across and lmag path length??
But then, I'm ok with B error up to 20%.

Not exactly, I was thinking of the case where the manu plots B vs H at
DC for you.
 
L

legg

Jan 1, 1970
0
[snip]
I guess this would be useful too if I ever needed to make an
electromagnet powered by a DC current source.

An electromagnet develops a useful external field in the (varying)
gap.The field does not have to be DC; is often more effective with AC
due to the potential reduction in residual magnetism.

RL

I'm thinking of the electromagnet used by the auto wreckers to pick up
cars.. AC or DC?
So an electromagnets core might stay a little magnetized after a shot
of DC.
Wouldn't an AC powered electromagnet just be kinda jiggy when
attracting say some iron..

The function is the same for a shaped field, and the average forces
applied average out.
Reminds me of those buzzing solenoids for doors.

There are many ways to configure electronic noise makers, not all of
them intentional. No doubt German-made models would be configured to
produce a more 'noble' sound, if noise was actually considered to be
neccessary to do the job of unlatching an electronic door.

RL
 
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