⚠️ WARNING ⚠️ THE CHOKE MUST BE MADE OF *IRON POWDER* not *farrarite* and the inductance will be 300uH not 3000uH . *Farrarite* =3000uH ❎ *Iron Powder* =300uH ✅
If you use a farrite core then the inductance will be 10 × higher . I made the same mistake using Green Cores . Which i got 10 × the inductance of what it should. I got 1000uH instead of 100uh . So i strongly recommend using an iron powder core . From the computer power supply . If you don't have a store to buy
You deserve it bro♥️..thank you for your suggestion, I actually need to make a 8.45mH, so the number of turns will be much less in farrite core, but I have heard that iron powder works best at this frequency range, that is why I was confused.
Maybe you are doing something wrong! 8.45mH is huge inductance 8450uH I think You need like 100uH inductance for the 25-50khz range . 8mH will reach saturation. Then it will drow continuous current and blow up everything.
Ok, let's do the calculation for inductor #1 Specific inductance 18 turns, 2.6mH, Al = L/N^2 = 2.6e-3H / 18^2 = 8uH/N^2 ur ~ 12,000! This is a high permeability core with no air gap, probably ferrite. No air gap, very low saturation current Magnetic cross section Ae = (Da-Di)/2*h = (31-18)/2*6 = 39mm^2 = 39e-6m^2 average magnetic path length le = (Da+Di)/2*Pi = (31+18)/2*Pi = 77mm relative permeability ur = L * le / (N^2 * A * u0) = 2.6mH * 77e-3m / (18^2 * 39e-6m^2 * 1.257e-6) = 12,600 Saturation current Is = Bmax * N * Ae / L = 0.3T * 18 * 39e-6m^2 / 2.6e-3H = 81mA "Slightly" below 30A . . .
⚠️ WARNING ⚠️
THE CHOKE MUST BE MADE OF *IRON POWDER* not *farrarite* and the inductance will be 300uH not 3000uH .
*Farrarite* =3000uH ❎
*Iron Powder* =300uH ✅
very good project to make inductor high current
Thanks
Very nice 👍👍👍
thanks
Nice 👌🏻
Thank you! Cheers!
mantap 👌🏻ungkal 😄
Nice video broo ☺️👍
Make an experimental video link
How much current flows in solt water and in regular water
Nice Idea 🤔
Thanks for the idea .. i will definitely try 👌🏻
@@Electronic_For_You 😁👍🔥🔥🔥🔥🔥🔥
😁
👍👍👍👍👍👍
Friend how to calculate turns
I didn't have any formula to calculate
😊😊😊
you just earned a sub, I am also making a inductor for my sepic, 20A rms, 25 kHz, If Iron powder is not available in my country, should I use farrite?
If you use a farrite core then the inductance will be 10 × higher . I made the same mistake using Green Cores . Which i got 10 × the inductance of what it should. I got 1000uH instead of 100uh . So i strongly recommend using an iron powder core . From the computer power supply . If you don't have a store to buy
And
Thanks for subscribing
:-)
You deserve it bro♥️..thank you for your suggestion, I actually need to make a 8.45mH, so the number of turns will be much less in farrite core, but I have heard that iron powder works best at this frequency range, that is why I was confused.
Maybe you are doing something wrong!
8.45mH is huge inductance
8450uH
I think You need like 100uH inductance for the 25-50khz range . 8mH will reach saturation. Then it will drow continuous current and blow up everything.
how do you calculation,..??
What?
What its for?
For boost Converter
@@Electronic_For_You
Ok
what is the formula for Minimum inductor in design 55v current 30A frequency 39kHz ??
How to measure the size and turn? Could I know the theory?
You can use an online calculators
Name of inductor tester pls?
Multi Transistor Tester
😂
These will saturate at a few amps at most, nowhere near 30A!
Core material is the main problem, I think sendust will work , btw I made it for a buck boost converter!
@@Electronic_For_You Yes, and you should actually show saturation current measurement please.
This is why I warned in the first comment about core material
Why not talk? An instructional video with no instruction is useless.
Ok, let's do the calculation for inductor #1
Specific inductance
18 turns, 2.6mH, Al = L/N^2 = 2.6e-3H / 18^2 = 8uH/N^2
ur ~ 12,000!
This is a high permeability core with no air gap, probably ferrite.
No air gap, very low saturation current
Magnetic cross section
Ae = (Da-Di)/2*h = (31-18)/2*6 = 39mm^2 = 39e-6m^2
average magnetic path length
le = (Da+Di)/2*Pi = (31+18)/2*Pi = 77mm
relative permeability
ur = L * le / (N^2 * A * u0) = 2.6mH * 77e-3m / (18^2 * 39e-6m^2 * 1.257e-6) = 12,600
Saturation current
Is = Bmax * N * Ae / L = 0.3T * 18 * 39e-6m^2 / 2.6e-3H = 81mA
"Slightly" below 30A . . .
Thanks
greate infomation about inductor, thanks.