Class D BTL LP filter (inductor saturation)

[Bowman1234]

Hi guys,

I am trying to design a LPF for a BTL similar to page 12, figure 10 on the link below

http://www.ti.com/lit/an/sloa119b/sloa119b.pdf

I'm a bit stuck on designing the inductor for the LPF, do you have to take into consideration the lower audio frequencies when trying to calculate the core saturation? Seeing as you have the PWM higher frequency on one leg and the lower frequency on the other?

Any help appreciated

Cheers

 

[DUG]

Hi guys,

...
Seeing as you have the PWM higher frequency on one leg and the lower frequency on the other?

...

Hmmm

The switching frequency is on both output lines of a BTL class D amplifier.

Here is a partial list of inductors that can be used in filter networks.

COILCRAFT RFS1412 10uH 6.2A(10%L drop) RFS1412-103ME
WURTH WE-PD SMD 10uH 10.5A(10%L drop) 7447709100
SUMIDA CDEP 10uH 11.5A(10%L drop) CDEP147NP-100MC-125
COILCRAFT SER2900 10uH 13A(10%L drop) SER2915L-103KL
COILCRAFT SER2900 10uH 18A(10%L drop) SER2915H-103KL
WURTH WE-HCF SMD 10uH 22A(10%L drop) 7443631000
COILCRAFT SER2900 10uH 28A(10%L drop) SER2918H-103KL
WURTH WE-HCF SMD 10uH 34A(10%L drop) 7443641000

Review their data and compare with what you design.

As a rough selection process I take the supply voltage and divide by nominal speaker impedance and then double that current to find the inductor size.

So a 24V supply and 6 ohm speaker gives 4A and doubled would be 8A.

So I pick an inductor that is rated at 10% (or less) inductance drop @ 8A.

The other way is to read through the following and see what others are using.

http://www.diyaudio.com/forums/group-buys/269855-wiener-tpa3118-amplifier-card.html

http://www.diyaudio.com/forums/grou...oup-buy-2-stereo-pbtl-editions-available.html

http://www.diyaudio.com/forums/group-buys/257996-gb-tpa3116-8-pbtl-bare-pcb.html

http://www.diyaudio.com/forums/class-d/237086-tpa3116d2-amp.html

http://www.diyaudio.com/forums/class-d/276982-tpa3251d2.html

Enjoy.

 

[nigelwright7557]

I fell foul of wrong inductors when I first started on class d amps.
I used power supply 10uH inductors and they got to 120 degree C !
I found the answer was to use a t106-2 core with as many turns of 22SWG wire I could get on. After that the inductors barely got warm.

 

[DUG]

... t106-2 core
...

I would like to check it out.

Who is the manufacturer?

 

[FauxFrench]

For smaller class D amplifiers, some frequent types of chokes are used in the output filter. Random checks of choke saturation shows that the specification of maximum amplifier output power is only possible by exceeding the saturation current of the chokes which seriously hampers the functioning of the filter and the class D chip.
I found this very useful table of saturation current for some of the most frequent chokes (scroll down on the page) 10 pcs/lot Blinde SMD inducteur de puissance 2.2UH 470UH Haute qualite vente CD32 CD43 CD54 CD75 4D28 5D28 CDRH104R RH127-in Inductances from Bricolage on AliExpress . Notice that the values given are DC (AC peak current) values. For example, when it is claimed that a PBTL coupled TPA3116 can deliver 100W (in 2 Ohm) it is often not true due to the chokes used on the board. 100W output power would require an RMS current just above 7A (10A peak) which the frequently used RH127 cannot handle without saturation.
Also for small Buck converters, the chokes used often seem insufficient.

 

[mboxler]

So a 24V supply and 6 ohm speaker gives 4A and doubled would be 8A.

Since it's a single supply, BTL, 6 ohm speaker, wouldn't 24V draw 2.8 amps per side (per inductor)?

 

[FauxFrench]

A BTL coupled board, supplied from a single 24V supply, will max draw 4A peak (in a constant 6 Ohm load) per channel. Depending on the inductance value, this is on the limit of what a RH127 can handle.

A BTL coupled board, supplied from a dual +/- 24V supply, will max draw 8A peak (in a constant 6 Ohm load) per channel. This is above what a RH127 can handle.

 

[mboxler]

Sorry, but my math must be wrong somewhere.

1) A single 24v power supply = 8.5v RMS.
2) Each channel driving a 6 ohm speaker single-ended = 12 watts = 1.4 amps.
3) Since 1.4 amps is flowing through each driver, 1.4 amps is flowing through
each inductor.
4) In BTL, Each channel is driving a 3 ohm load = 24 watts = 2.8 amps.
5) Since 2.8 amps is flowing through each side of the driver, 2.8
amps is flowing through each inductor.

Where did I go wrong?

Thanks, Mike

 

[FauxFrench]

Mike, your math is OK.
1-3), you assume single supply, SE coupling and not BTL coupling. The current is RMS current and not peak current.
4-5), you assume single supply and BTL coupling. You arrive at 2.8Arms which corresponds to 4Apeak.
As your calculations use power, you arrive at RMS values. My focus was on saturation of chokes and there the peak value is relevant.