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
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