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#1 |
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diyAudio Member
Join Date: Jun 2004
Location: Edmonton area, Alberta
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In a >=2nd order sigma-delta amp, where do the extra lowpass sections go? I assume in the feedback loop and not in between the first integrator and the comparator. And how does one choose the corner frequencies for each extra lowpass/integrator?
Sorry for the newbish question, and thanks in advance for tolerating it. Tim |
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#2 | |
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diyAudio Member
Join Date: May 2002
Location: Switzerland
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What loop order are you looking at ? Regards Charles |
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#3 | ||
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diyAudio Member
Join Date: Jun 2004
Location: Edmonton area, Alberta
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If you know of any websites or readily available books that would give me some introductory loop theory (beyond gain vs bandwidth and the sort) that would be helpful. |
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#4 |
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diyAudio Member
Join Date: Apr 2004
Location: BE/NL/RW/ZA
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Normally the forward path of the loop is a chain of integrators. Lowpass filters can be used, but normally you'll only find this in cases when the loop is constructed using passive components (mostly passive controlled self-oscillating class D amps) or when the designer intentionally wanted to limit loop gain at lower frequencies (e.g. UcD).
The traditional method for synthesising a deltasigma loop is thus: 1) One takes a chain of n integrators. Go easy on your math skills and start with a time constant of unity. This realises a loop function 1/s^n. This loop function is obviously not going to make a stable loop. 2) To make the loop stable, the transfer function must fold over from an nth order function to a 1st order at some frequency before the unity gain point. The loop gain looks like a steep downward slope of -n*20dB/dec, changing to a less steep slope of 20dB/dec shortly before the unity gain point. It is common to make this crossover using a butterworth function. This means that to the loop function a set of n-1 zeros is added distributed evenly along a circle of radius Wchangeover. 3) The zeros are practically realised by making a weighted sum across all integrator outputs. The loop function is of the form (a1+a2*s+a3*s^2+ ... +an*s^(n-1))/s^n. Symbolically or numerically determine a[n] to get the zeros where you wanted them. 4) scale the whole function by a constant such as to get the unity point somewhere in the 1st order region as planned. Individual internal gains can be scaled (with retention of overall gain of course!) to optimise signal swings. 4) becomes moot when the circuit is a 1-bit converter. Gain will be determined automatically by the probability density function of the noise that comes out of the loop. When you're designing a PWM system, usually the equation runs backward. First you set the switching frequency, from this follows the unity gain point, and from this the changeover frequency. When you're designing a 1-bit converter, the choice of changeover frequency determines the maximum stable modulation index. My own method produces somewhat more optimal coefficient sets, and works equally well with loop functions with complex poles (and can be used to include output filters), but is quite complicated to explain here. |
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#5 |
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diyAudio Member
Join Date: May 2002
Location: Switzerland
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I once derived the maths to determine the coefficients for a given NTF for loop-orders up to five. I know that some programs can do this easily but I don't own one.
It is difficult to determine the cutuff frequency and order of an optimal NTF however but the calculation of the coefficients for a given NTF is trivial. Something that isn't trivial (at least to my knowledge) is the determination of optimal integrator clipping levels for 1-bit modulators. If someone is interested I can post my findings here. Regards Charles |
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#6 |
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Banned
Join Date: Nov 2004
Location: Paris
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Bruno, when you talk about PWM I assume you talk about clock based amps. In that case, as you suggest, you determinte the sw. freq. and then the 0dB point.
I suppose you are trying to say that the 0dB point in the open loop function must be placed at <Fs/2, right? Thanks |
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#7 | ||
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diyAudio Member
Join Date: Apr 2004
Location: BE/NL/RW/ZA
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#8 |
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diyAudio Member
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I don't know if higher order self-oscillating sigma-delta will work. If the whole thing is a system is a 2nd order integrator for example, there will be a phase shift of 180° at all frequencies above unity gain, so the whole thing can oscillate at any frequency (and will do so).
Also adding zeros will not change that for frequencies below the zero frequencies, so the oscillation frequency will just be kept below the frequencies of the zeros. In my opinion, self oscillating converters will only work as a system of order 1. digi |
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#9 | |
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diyAudio Member
Join Date: Apr 2004
Location: BE/NL/RW/ZA
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#10 | |
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diyAudio Member
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I will refine my argument as such: The better noise suppression from a higher order sigma delta is because of the increased gain in the audio band. In the case 40db/dec is reached (against the 20db/dec of a first order system), the phase shift will be enough to start it oscillating. If it is lower (due to some pole/zero arragment and you are on the save side, there is hardly more gain compared to a straight first order system). So, you might end up like something that looks like its second order, but in reality has not much more gain compared to a first order system (and will also have noise figures like a first order system) For example: Real integrator (pole at zero freqency) + zero a little above the audio band: The phase shift at lower frequencies (nearly 180°) + a little delay (even jsut 100ns) will be enough to make it oscillate in the audio band, also the zero takes away some gain and makes it more like a first order system Or a clever pole/zero arragment that will make a phase shift of 120° + a ripple around that value will also not work, because a ripple in the frequency domain is a delay in the time domain -> very bad. On another note: I looked at the Ucd patent and i don't like what i see there. Unless something has been done about this i see a great problem: If a load is connected, everything is fine. But if there is no load present, the Q of the output filter will get very high, causing a huge phase shift at the pole frequencies and bring down the oscillation frequency. So, there will be a quite high output voltage and high reactive energy transfer in a no load no input condition, which is not nice at least. Especially in high power amps output voltages can be dangerously high when you don't expect anything on the output. Of course this only applies to real amps not active systems with a direct connection to the loudspeaker. digi |
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