Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

Btl class d filter design

Status
Not open for further replies.
bowman1710

bowman1710

Full Member level 3
Joined
Nov 8, 2014
Messages
183
Helped
6
Reputation
12
Reaction score
6
Trophy points
18
Activity points
1,604
Hi guys,

I'm looking at designing a BTL filter for a class D type setup as per attached, links below that are documents I have looked at already.

http://www.ferroxcube.home.pl/appl/info/class D audio amplifier.pdf

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

BTL LOW PASS FILTER.png

Specification:

The PWM signal will be around 300KHz
The sinewave will be 2-20KHz at about 50V p-p
185 ohm load

Really I need the filter to be pretty flat and cut off at about 25-30KHz and really I want as high attenuation as possible at 300KHz, this is where i have a couple of questions

1) What is the best way with this design to achieve this with a high attenuation at 300KHz, would it be a case of higher order filter, if so, how does it work with a BTL design?

2) When designing the inductor how does having a PWM on one leg and a sinewave wave on the other effect the inductors saturation, do you only care about the lower frequency signal and the p-p voltage or is it more complex then that?

Any help would be welcome!
 

Hi,

There are two L in series: L_total = L_BTL + L_BTL = 2x L_BTL.

There are two C_G in series but in parallel with C_BTL.

so C_total = C_BTL + C_G / 2

Now you can calculate it as usual. Known = L, C, R...

Klaus
 

This simulation mimics a class D amplifier (using your specs). It should give an idea what L & C values to choose.



As you can see, L1 and C1 are just a bit smaller than they ought to be. They have a noticeable effect in smoothing the 300 kHz DC pulses.

As you reduce L1's value, the spikes get worse.
As you make L1 larger, it starts to 'choke' the output amplitude. It's hard to be sure what tradeoff point to choose.

C1 can be increased to reduce spikes, although this also causes it to carry more current. This may be of concern to your capacitor size.

This is a second order filter. There is a formula for calculating the values. However the simulator makes it easy to see the effects of different values.

You must watch for resonant action from the LC. Obviously it is best if you make sure the resonant frequency falls outside the frequency band being amplified.
 

There are two L in series: L_total = L_BTL + L_BTL = 2x L_BTL.

There are two C_G in series but in parallel with C_BTL.

so C_total = C_BTL + C_G / 2
Click to expand...

Yes I get that, but can you use this arrangement for 3rd order ,5th order with this design. Does it have its own complications if you do?


As you make L1 larger, it starts to 'choke' the output amplitude. It's hard to be sure what tradeoff point to choose.
Click to expand...

i have been looking at stuff like this in LTspice but I am trying to see the best to simulate it, I will have a look at it in that software too to see what happens. Is there any way to get a higher order? Or any comments on the inductor design in terms of Isat?

BTL LOW PASS FILTERLT.PNG
 

really I want as high attenuation as possible at 300KHz
Click to expand...
Statements like this are pretty usless. Specify your requirements and find the necessary filter order.

For usual class-D audio applications, stop band attenuation requirements are moderate, mainly imposed by EMC regulations. Apparently you have a different application.

There's no substantial difference between a single-ended and a differential-ended ("BTL") LC filter. Any order can be implemented in differential form. Design a single-ended filter and transform to differential.
 

Hi,

i have been looking at stuff like this in LTspice but I am trying to see the best to simulate it, I will have a look at it in that software too to see what happens. Is there any way to get a higher order? Or any comments on the inductor design in terms of Isat?
Click to expand...

I´m sure you did your research.. and you already know that if you want higher order you need extra Ls and Cs.

First order low pass: R-C, or L-R
second order: L-C
4th order: L-C-L-C
... the output of the one stage is the input of the next stage...

Now you have one active L and one active C. With "one" i mean: athough there are three Cs, they work as a singel one. The same is with the Ls.

Adding a new L and a new C. After the first "symmetric" stage you may now use an unsymmetric stage, that means "really one" L and "one" C.

Klaus
 

Adding a new L and a new C. After the first "symmetric" stage you may now use an unsymmetric stage, that means "really one" L and "one" C.
Click to expand...
That's an option to save a few components. The differential topology is used to suppress radiated interferences. If you have no speaker cables, you could also use a pure single ended filter. Or take the opposite way, make the whole filter fully differential.
 

Also it there any reason why you cannot use a common mode choke for the two inductors rather then two separate ones?
 

Hi,

you can´t use a common mode choke here, because the useful audio currents will compensate each other.
Thus it can´t filter the PWM frequency.

Klaus
 

1) Look at binary (AD) and tertiary level (BD) Class D and choose BD for better filtering.

This TI doc. tells you everything you need to know
https://goo.gl/68lS1E

2) not seen that config. but anything that improves margin on I peak from saturation is a good thing at the expense of efficiency though.
 
You might also consider adding a pair of 300 Khz notch filters (traps) in addition to the usual LPF.

The switching frequency is far enough away from the LPF cutoff that this should offer significant extra attenuation at the problem frequency, without significant interaction.
 

you can´t use a common mode choke here, because the useful audio currents will compensate each other.
Thus it can´t filter the PWM frequency
Click to expand...

Thanks, that makes sense really.

1) Look at binary (AD) and tertiary level (BD) Class D and choose BD for better filtering.
Click to expand...

I have been but it doesnt have any information about higher order filters in this format

You might also consider adding a pair of 300 Khz notch filters (traps) in addition to the usual LPF.
Click to expand...

The only issue is this will be high current so won't having series or parallel R in the filter cause a lot of power loss?

- - - Updated - - -

Or I could use something like this i guess

LC NOTCH FILTER.PNG
 

Can anybody help with calculating the Saturation current for the inductor with a 300KHz signal on one leg and the audio on the other with 50V P-P with 10A flowing through it, I'm a bit confused in how the two different frequencies present affect the inductors saturation?
 

bowman1710 said:
300KHz signal on one leg and the audio on the other
Click to expand...

I believe one leg is switched while the other is grounded. No need to apply audio.

- - - Updated - - -

Then reverse the situation when you have a polarity change.

Ground the first leg. Switch the second.
 

The low frequency current is the only relevant contribution to core magnetization, or the filter design is flawed.
 

If you want to calculate the losses, then you need a precise equivalent circuit of all parts including ESR, eddy current losses and non-linear saturation effects and temperature factors. Or use the suggested 4th order filter in the TI link I posted for 4 Ohms. THen model it.
 

I believe one leg is switched while the other is grounded. No need to apply audio.

- - - Updated - - -

Then reverse the situation when you have a polarity change.

Ground the first leg. Switch the second.
Click to expand...

So if the other leg is grounded and switching the other leg, then you do have to take the audio in consideration then?


The low frequency current is the only relevant contribution to core magnetization, or the filter design is flawed
Click to expand...
.

So with a BTL you need to use the lowest frequency it sees i.e 2KHz then for calc?
 

bowman1710 said:
So if the other leg is grounded and switching the other leg, then you do have to take the audio in consideration then?
Click to expand...

From what I understand, DC pulses are the only thing that is applied to the output stage. The pulses provide a path to either 50V or 0V ground.
 

Distilled version of class D action. (Only the positive waveform is shown.)

a) Signal and carrier are compared by op amp, creates PWM output.
b) Op amp drives a switch.
c) The switch goes to either 50V, or 0V ground.

 

From what I understand, DC pulses are the only thing that is applied to the output stage
Click to expand...

Is the signal seen by the left hand side on the inductor a 300Khz switched audio signal, as in isn't it basically the 300Khz signal being switched to produce the audio signal??? Below is the signal on the inductor

BTL FILTER.PNG


Isnt this why the LP filter is used to remove this 300KHz?
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top