What is the ideal core for a Class D amplifier with a frequency of 62khz-500khz?
None listed will go that high, and those rated to 200MHz may have larger losses
near that limit. Why do you need 500MHz?
500khz Which ferrite cores in these ranges?
Look at 3F3 ferrite.
https://www.ferroxcube.com/upload/media/product/file/MDS/3f3.pdf
https://www.ferroxcube.com/en-global/download/download/11
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ferrite cores are not available on these links
This is a manufacturer. You can order samples, or purchase through a distributor.
Even though I can't read Turkish, it looks like those cores are powdered type toroids. Whereas you were asking about ferrites in the thread title. Powdered types are likely to have higher losses than ferrite at the frequencies you're interested in.
For 150uH you'd want a gapped ferrite core like this one, it will need about 30 turns : B65815E0160A041 EPCOS / TDK | Mouser
If you're concerned about losses you should use multiple stranded wire, not single core.
For 150uH you'd want a gapped ferrite core like this one, it will need about 30 turns : B65815E0160A041 EPCOS / TDK | Mouser
If you're concerned about losses you should use multiple stranded wire, not single core.
rm14 with 0.80-1mm wire with 10 spiral coil. The value was 150uh. I left the air gap with paper tape for core.
Since the number of spirals is small, will mosfet transistors and ir2110 be damaged? I'm worried?
Since the number of spirals is small, will mosfet transistors and ir2110 be damaged? I'm worried?
Attachments
http://www.farnell.com/datasheets/528673.pdf
IIRC
Nmin = LI^2/Bpk.Ae
I^2 = 10 x 300E-3 x 198E-6/150E-6
I^2 = 3.96
It looks like your inductor will saturate at 2 amps.
More sums... might be wrong.
Work out what your peak audio current will be. Root2 * Irms.
Work out what half the ripple current will be. Vs/4FsL
Add them and call it Ipk
100W into 8R
Irms = 3.54A Peak audio current = 5.
Fs = 100KHz L = 150uH Vs=40
Half ripple = 0.67A
Ipk = 5.67A
Nmin = LIpk^2/Bpk.Ae
Nmin = 150E-6 x 5.67^2 / 300mT x 198E-6
Nmin = 81
Al = L/N^2
Al = 23nH/root turn
Unless there is something about your application that is very different from the approximate numbers I have used the core you have is not suitable for the task.
IIRC
Nmin = LI^2/Bpk.Ae
I^2 = 10 x 300E-3 x 198E-6/150E-6
I^2 = 3.96
It looks like your inductor will saturate at 2 amps.
More sums... might be wrong.
Work out what your peak audio current will be. Root2 * Irms.
Work out what half the ripple current will be. Vs/4FsL
Add them and call it Ipk
100W into 8R
Irms = 3.54A Peak audio current = 5.
Fs = 100KHz L = 150uH Vs=40
Half ripple = 0.67A
Ipk = 5.67A
Nmin = LIpk^2/Bpk.Ae
Nmin = 150E-6 x 5.67^2 / 300mT x 198E-6
Nmin = 81
Al = L/N^2
Al = 23nH/root turn
Unless there is something about your application that is very different from the approximate numbers I have used the core you have is not suitable for the task.
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10 spir will also burn in 2 / amps. If I did what I did, I didn't hold 150. What could be the ideal core?
You have to calculate a core area product. Winding Area x Core Area. Aw.Ae
What is the load of your amplifier?
What is the power of your amplifier?
What is the switching frequency of your amplifier?
What is the supply voltage of your amplifier?
Why have you chosen a 150uH inductor?
Amplifer power supply 2x85vdc.
Min freq ; 62khz - max freq ; 400khz
Mosfet transistor 2x irfp250
Oh.... One of those.
You missed out the load and the power.
schema here. Power is not clear
What power do you want and into what load?
The current power is good. I'm not quite sure but it could be 900w. 4 ohms.
Which core is required for this circuit?
OK. Ignore the rest. 900W 4R 15A RMS 21A PEAK
AwAe = LIrmsIpk/BpkJKcu
AwAe = (150u x 15 x 21)/(300mT x 4E6 x 0.7)
AwAe = 5.625E-8
Assume Aw = Ae
Aw = Ae = 2.37E-4 = 237mm^2
ETD49
Ae = 211mm^2
Aw = 269mm^2
Aw.Ae = 5.676E-8
Nmin = LIpk/BAe
Nmin = (150E-6 x 21)/(300mT x 211E-6)
Nmin = 50
More later.
Attachments
Oh... The above assumes that you want 900W continuous rated power. If you assume music is 1/8 continuous then you can use a higher number for J, current density in the wire, and end up with a smaller inductor. You still have to design to avoid saturating the inductor based on the peak current though.
Once again why 150uH
Once again why 150uH
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