### :: Toroids / magnetism / permeability / capacitances

this article is about getting to make inductors/transformers on toroids, and how to get to an approximate estimation of saturation just by using mini ring calculator values.

we refer to a very useful piece of datasheet from fair-rite

in the datasheet they describe flux vs Oersteds required to reach saturation, and we can infer from it :

relative permeability : 125, 800, 5000, 50000?
oersteds to Bsat      : 15 ,  10,    5,     1?

the higher the permeability, the lower the Oe needed to saturate a core.

we then refer to the AC equation for B(peak)

B(peak)= (Vrms*10)/(4.44*F*N*A)

V = AC volts
F = operating frequency (kHz)
N = number of turns (inductor of course)
A = cross section area (in cm2)

H = (m.N.I)/2.pi.r
so calculating for H
(if m =  0.002266Wb/A.m-1, N = 20, 1 = 5A, 10x5x5mm toroid)
H = 0.721Oe

and the simple equation which gives us permeability variables

m (permeability) = B(flux density) / H(magnetic field/oersteds)

if we use the mini ring calculator to estimate an unknown core say 10mm x 5mm x 5mm, with 20turns measured in with 500uH. the calculator gives us

AL = 1250nH/N^2
ui = 1803.4

actual permeability = 1803.4 * 4(pi) * 10^-7
= 0.002266Wb/A.m-1

using m = B/H, and having arbituary H =10
Bsat = m / H
= 0.02266 / 10
we get an estimated Bsat = 0.02266T

using the AC B(peak) equation we find that @ 10V, 20 turns, the minimum operating frequency to avoid Bsat is

4.44*F*N*A = Vrms*10/B(peak)
F = (10 * 10) / (0.02266 * 4.44 * 20 * 0.25)
F = 199kHz

if we increase the number of turns to 50

F = (10 * 10) / (0.02266 * 4.44 * 50 * 0.25)
F = 79khz

if we revise H = 5, F will double up. double voltage = double F ... and etc etc relatively

so there we go, we have a method of estimating what to do with the inductor in terms of voltage loading, operating frequency, size and number of turns.

approx inductance of toroid
L = (u * N^2 * A/100)/(2 * pi * r)
= 0.002266 * 2500 * (0.25/100) / (2*pi*r)
= 0.003H (3mH)

*additionals 28 dec 2014

the wonderful constant (e)

e = 1 - 1/(exponential(1))
= 0.632121

when charging a capacitor to 63.2121%of total charge
time required = resistance x capacitance

when introducing a current into a inductor energizing the inductor
time required = inductance / reactance

@ 100kHz, time constant = 10us, for an ESR of 0.01Ω
C = 10/0.01
= 1000uF

scenario 1
energy charged (assuming 1volt, R = 0.01Ω, 63.21%)
Q = C * V

= 1000uF * 0.632121
= 0.6321mC
RC constant = 0.01Ω x 1000uF = 10us

current for 0.6321mC in 10us
= 0.6321mC / 0.01mS
= 63.21mA (so the 1000uF capacitor reaches 0.6321v)
peak current = 1/0.01 = 100A

scenario 2
energy charged (assuming 10volt, R = 0.01Ω, 63.21%)
Q = C * V

= 1000uF * 6.321
= 6.321mC

current for 6.321mC in 10us
= 6.321mC / 0.01mS
= 0.6321A (so the 1000uF capacitor reaches 6.321v)
peak current = 10/0.01 = 1000A

scenario 3
item RNU1H470MDN1PH 47uF 50v NICHICON
RC constant = 47uF x 0.024Ω = 1.13u
Q = C * V

= 47uF *31.6
= 1.49mC

scenario4
item 330uF 25v NICHICON
RC constant = 330uF x 0.014Ω = 5us
Q = C * V

= 330uF *15.8
= 5.2mC
I = 5.2 / 0.005 = 1040mA

V(ripple) = V(peak) / (2*F*C*R)
= 25/(2*200000*0.00033*0.014)
= 13.5v

**edit 10 Jan 2015
discharging capacitive energy into LED
energy = 1/2 .C.V^2

eg : capacitor 100uF charged to 50v = 1/2 x 100uF x 2500
= 0.125J

peak power curve discharged in 10ms burst = 0.125 / 0.01 = 12.5watts

via a power burst ratio of 1:1, we string a series of LED with total power capacity of 12watts, each LED cell is rated for about 2.7v (12 cells = 32.4v)

work in progress ...