BTL CLASS D FILTER (high impedance load)

Hi Guys,

I am designing a high current BTL filter with a high impedance load, whereas in normal designs you would expect a low impedance of say 6ohm I am looking at an impedance level of 3333ohm. My issue is that to give the filter a decent Q the inductor would have to be massive in size say 150mH, I cannot use something that large because of size constraints. The filter I have at the moment to give acceptable losses is severely under-damped (as per picture)



Is there any way to reduce this peak (Q factor), or given the constraints of the design is this something I will have to live with?

Thanks in advance

A class-D power amplifier usually has an 8 ohm or 4 ohm speaker as its load. Your load resistance is so high that the LC filter is resonating and has almost nothing to reduce its Q.

Instead of relying on a specific real load impedance, you could add intentional loss to the filter. In my view it's needed for most filters that can't be designed for a fixed load impedance. You'll usually connect a RC series circuit parallel to the shunt C of a low-pass filter.

Is there any way to reduce this peak (Q factor), or given the constraints of the design is this something I will have to live with?

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Placing damping across the output capacitor as suggested above is certainly possible and will work, but it does impose a very wasteful dissipative loss across the output to achieve any worthwhile damping.

I have had slightly more success by placing a damping resistor directly across the inductor.
And a workable resistor will probably be a lot lower in resistance than you might expect.
But it is probably the most power efficient way.

Only simulation or a practical test will tell you if the results are satisfactory.
Problems like this tend more towards "try it and see" rather than a design from first principles.

I don't suppose you could use an audio transformer to transform that high impedance into something lower?

I would use an LC filter with the appropriate resonant frequency and add series resistance to the capacitor to reduce gain peaking.

Can you use a couple of series RC lowpass filters in series? They do not resonate.

Yup, damping resistors across the L (or L's if you want a balanced o/p) AND RC across the C, this will damp the Q, can all be modelled pretty easily on LT spice so you can see the effect if the switching frequency is changed, unfortunately for hi o/p Z, you do tend to get solutions with high Henry's and lower C's, as the surge impedance is SQRT(L/C) and to match your 3k33 L must be bigger and C smaller, usually the fo of the LC combo should be at least 10x lower that the switching fundamental...

I have had slightly more success by placing a damping resistor directly across the inductor.
And a workable resistor will probably be a lot lower in resistance than you might expect.
But it is probably the most power efficient way.

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AND RC across the C

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I don't suppose you could use an audio transformer to transform that high impedance into something lower?

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I think with a RC over the cap may be a good way forward. The damping resistor over the inductor may be a bit lossy. I think an audio transformer with a high turns ratio might be the answer to obtain low losses and bring it down do 10ohm for e.g, then the need for a higher inductance will be eliminated, thus reducing the Q.

Unless you have a lossy transformer the Q will be un-affected, you did not say in your original post that you wished to reduce the o/p voltage by use of a filter, as implied by your now stating you are happy to use a step down transformer, assuming you still wish to filter the square waves from your BTL amp, you will still need a choke and a cap, the ratio of L/C will be smaller now due to the Tx, but you now need to build a higher current choke, for the same power out the choke will be a similar physical size to the high voltage one you needed as I^2.L will be the same, the caps will need to be bigger uF now (but lower voltage rating) if the step down Tx has significant leakage you can use this as part of the filter inductance, however the total volume of Tx + filter L will be larger than just a filter L.

The damping resistor over the inductor may be a bit lossy.

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Not really, because at the much lower frequency of the reconstructed PWM output, the choke will (or should have) negligible series impedance.
Shunting that with a resistor will have very little power loss.

At the much higher PWM switching frequency the inductor will have a very high impedance, and the damping effect of the resistor will be very significant, and its a good solution to cure output ringing with zero load.

Just about all the burned up watts in the resistor will be the higher harmonic frequencies you are trying to eliminate anyway.

Pretty much the same is true of shunting the output capacitor with a CR network.
The watts burned up by the resistor will be mostly the higher frequencies that you are trying to get rid of.

Which system is more efficient depends on the normal output load.
If the normal load is zero, or a very high impedance, then shunting the L will offer minimal loss under those conditions. That might be a big advantage with a mains power inverter inverter that normally runs very lightly loaded.

If the output is always fairly heavily loaded, as with an audio amplifier, then the CR shunting network will work fine, maybe even better.

It depends what all this is for.

Shunting that with a resistor will have very little power loss.
At the much higher PWM switching frequency the inductor will have a very high impedance, and the damping effect of the resistor will be very significant, and its a good solution to cure output ringing with zero load.

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Yes. A certain disadvantage of a parallel resistor is the reduction of high frequency filter attenuation. I found a C series resistor in the previously suggested LCRC topology sufficient to adjust the filter Q in all applications. But can't exclude that a L parallel resistor might be helpful for certain load situations.

For filter resonance considerably below PWM frequency, I would expect however higher PWM losses of the parallel resistor than of an equivalent (same resonant Q) series resistor, due to the effect of the additional parallel C. To be checked in a calculation.

Yup, there are quite a few very diverse factors to take into consideration.

Its probably best to try out few different ideas either simulated or real, and see which gives the best results in the particular circumstances.

getting back to your original graph, any unloaded LC filter will give that same peaky response, try adding the typical load and re-plotting...

Another consideration is the bandwidth of the reconstructed PWM signal.

If this is a power inverter it can likely start rolling off above maybe 80Hz.
If its an audio power amplifier it might need to be pretty flat right up to 20 Khz, which makes things a bit more difficult.

Warpspeed said:

Another consideration is the bandwidth of the reconstructed PWM signal.

If this is a power inverter it can likely start rolling off above maybe 80Hz.
If its an audio power amplifier it might need to be pretty flat right up to 20 Khz, which makes things a bit more difficult.

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The load is 3333 ohms which is not a loudspeaker.

Audioguru said:

The load is 3333 ohms which is not a loudspeaker.

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But we do not know what the load is, or what the signal driving it is supposed to be.