I think the simulator tools I found had trouble with the big difference between input and output impedance of the filter circuit. Looking at the tables in the book referenced in the Linear Audio article the solution for this filter might get a bit dicey when output impedance is more than ten times less than the input impedance...current drive not voltage drive. I manually calculated the circuit using the method in the book but am not sure I de-normalised things properly at the end.
Downloaded LTSpice and put the original circuit in and it does all work as intended...6th order at 82kHz...just am unable to calculate the matching component values in the first place.
Downloaded LTSpice and put the original circuit in and it does all work as intended...6th order at 82kHz...just am unable to calculate the matching component values in the first place.
Isn't it just a matter of scaling by impedance factor Marcel's value/your value?
"multiply all resistor and inductor values and divide all capacitor values by this".
"multiply all resistor and inductor values and divide all capacitor values by this".
BTW, Marcel - you mentioned filters with input/output impedance equalled being more robust and suggested in input resistor to ground.
I think this input resistor needs to be in series with the filter, as Micro-Cap does automatically when you let it design a filter.
I mucked about with this, found that resistor to ground does not make the filter robust. The serial one would have the added advantage of reducing output voltage for those that need this (like myself).
I maybe wrong in my assumptions and may have made mistakes in using the sim tools, but this was my conclusion.
I think this input resistor needs to be in series with the filter, as Micro-Cap does automatically when you let it design a filter.
I mucked about with this, found that resistor to ground does not make the filter robust. The serial one would have the added advantage of reducing output voltage for those that need this (like myself).
I maybe wrong in my assumptions and may have made mistakes in using the sim tools, but this was my conclusion.
The component values for these filters are based on polynomial equations and are non linear so when you change the ratio of input to output impedance linear scaling will yield incorrect filter values.
Are you sure? I quoted from an Analog Devices White Paper on Filter Design:
https://www.analog.com/media/en/training-seminars/design-handbooks/Basic-Linear-Design/Chapter8.pdf
In it you find the design tables for Marcel's "0.05 degrees" filter besides Butterworth and so on.
https://www.analog.com/media/en/training-seminars/design-handbooks/Basic-Linear-Design/Chapter8.pdf
In it you find the design tables for Marcel's "0.05 degrees" filter besides Butterworth and so on.
I just looked through some old files to see how I came up with the filter component values for the original valve DAC, as I didn't quite remember.
What I did was to take the prototype values from Anatol I. Zverev's Handbook of filter synthesis for an infinite termination resistance ratio, so a pure current source at the input and a termination resistor at the output. Using reciprocity, the filter for current source drive and termination at the output is exactly the same as the filter for termination at the input with an open output, only swapped (input becomes output and output becomes input).
I then used a pole-zero extraction program (a DOS program called Linda from the late Catena Microelectronics B.V.) to check the impact of the imperfect current drive at the input and found it was quite small. I also checked whether increasing the output termination resistor a bit to compensate for the resistor at the input made things better or worse, but that only resulted in poles moving further from the intended locations. Finally I compared various options for rounding the components to practical values (E24 capacitances and multiple of a half turn inductors) and took the one that came closest to the intended pole locations.
I later found a way to interpolate between the ratio of ten and ratio of infinity values from Zverev, but I'll get back to that later.
What I did was to take the prototype values from Anatol I. Zverev's Handbook of filter synthesis for an infinite termination resistance ratio, so a pure current source at the input and a termination resistor at the output. Using reciprocity, the filter for current source drive and termination at the output is exactly the same as the filter for termination at the input with an open output, only swapped (input becomes output and output becomes input).
I then used a pole-zero extraction program (a DOS program called Linda from the late Catena Microelectronics B.V.) to check the impact of the imperfect current drive at the input and found it was quite small. I also checked whether increasing the output termination resistor a bit to compensate for the resistor at the input made things better or worse, but that only resulted in poles moving further from the intended locations. Finally I compared various options for rounding the components to practical values (E24 capacitances and multiple of a half turn inductors) and took the one that came closest to the intended pole locations.
I later found a way to interpolate between the ratio of ten and ratio of infinity values from Zverev, but I'll get back to that later.
Micro-Cap probably assumes that the circuit driving the filter is a voltage source. In this case, it's essentially a current source in parallel with the anode resistors in parallel with the extra shunt resistor, if any(*).
A current source with value i in parallel with a resistor with resistance R and a voltage source with value v = iR in series with a resistor with resistance R are terminal equivalents (behave the exact same at their terminals), see also the Wikipedia pages about Norton and Thévenin equivalents. So all in all, you have to place the resistor in series when the driving circuit is a voltage source and in parallel when it is a current source.
(*): The extra resistor is supposed to be placed after the DC blocking capacitor and the anode resistor is placed before it, but the DC blocking capacitor is essentially a short for AC, so they are still practically in parallel.
Getting back to the 250 V or 400 V DC blocking capacitor:
With the present resistor power ratings, you need 400 V to be sure that the capacitors can still block the DC voltage when there is a short between the cathode and heater of one of the upper valves. Immediately after the short has occurred, the voltage will still be below 250 V, but R141 and the anode resistors will dissipate much more than they normally do. Depending on which resistor fails first, you could get a too high voltage across the capacitors.
When you would use 2 kohm +/- 1 % 4 W wirewound resistors for R14, R25, R26, R27, R103, R104, R105 and R106 and 2.2 kohm, 10 W for R141, they won't fail (at least not according to my calculations) and 250 V DC blocking capacitors should then survive the fault. Needless to say, the PCB was not designed for such resistors, so I don't know if there is a way to squeeze them in.
R141 could also be a fusible 2 W resistor, but fusible resistors with 2 W power rating and a value of 2.2 kohm appear to be nonexistent.
With the present resistor power ratings, you need 400 V to be sure that the capacitors can still block the DC voltage when there is a short between the cathode and heater of one of the upper valves. Immediately after the short has occurred, the voltage will still be below 250 V, but R141 and the anode resistors will dissipate much more than they normally do. Depending on which resistor fails first, you could get a too high voltage across the capacitors.
When you would use 2 kohm +/- 1 % 4 W wirewound resistors for R14, R25, R26, R27, R103, R104, R105 and R106 and 2.2 kohm, 10 W for R141, they won't fail (at least not according to my calculations) and 250 V DC blocking capacitors should then survive the fault. Needless to say, the PCB was not designed for such resistors, so I don't know if there is a way to squeeze them in.
R141 could also be a fusible 2 W resistor, but fusible resistors with 2 W power rating and a value of 2.2 kohm appear to be nonexistent.
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What sort of timeframe are you talking here before the mains fuse blows to cut power?
Slipping a 10w or 12w resistor into R141 is doable.
Slipping a 10w or 12w resistor into R141 is doable.
I've figured out this filter calculation thing and have made a spreadsheet for it. Looks like the filter was originally calculated for an 800r termination impedance and 750r (as used in the ValveDac) still works well but at 600r termination things get a bit wobbly around the cutoff frequency. At least that is according to my LTSpice simulation which may or may not be the be-all-and-end-all.
The spreadsheet allows me to quickly calculate the ideal filter components for varying cutoff frequencies and termination impedances...so I can now answer my own questions in this regard. Thanks Marcel for your patience.
The spreadsheet allows me to quickly calculate the ideal filter components for varying cutoff frequencies and termination impedances...so I can now answer my own questions in this regard. Thanks Marcel for your patience.
So, with a 600r filter termination impedance (to lower the dac output to 1Vrms and improve noise) and to retain all the filter caps at the original easy to source sizes, the inductors need to be changed to 517.8uH, 365.9uH and 107.9uH respectively. This is a sixth order 0.05 degree Linear Phase Low Pass filter with the corner frequency set at 95kHz.
Marcel, I'm trying to spec a custom wound output transformer and need to set the current rating. I have the idea that the dac output is 2.5mA...is this correct?
I haven't a clue. If the anode resistors go open-circuit before the mains fuse does, it may not blow out at all. Even when the resistors don't blow out at all, I'm not sure if the excess current is enough to blow out the mains fuse - fuses tend to be rather inaccurate and there is only about 40 mA extra drawn from the -300 V.
If your custom-wound output transformer has enough insulation and keeps enough insulation when there is an abnormally high current flowing through its primary windings, it may not be an issue for you at all.
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2.5 mA peak is the normal 0 dBFS level, yes. In the raw-DSD version, it might end up somewhere between 2.5 mA and 5 mA peak when the DSD stream is not Scarlett Book compliant. In the normal version, it might end up somewhere between 2.5 mA and 5 mA peak during intersample overshoots. As always, on sine waves, the RMS levels are the peak levels divided by the root of two.
I would like to have an idea what a super DAC does reproduce a burst of 20khz of 1ms. Can you picture a scope image?
No, as I never measured my DAC's response to 20 kHz tone bursts. If I should find the time to measure it, at what sample rate do you want it measured and with or without anti-aliasing filtering? When you switch a 20 kHz carrier on and off with a 500 Hz square wave (my scope has no single-shot function), there will be sidebands at odd multiples of 500 Hz above and below 20 kHz. Normally the anti-aliasing filter would suppress some of these during recording, as otherwise they alias.
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The impulse response of the digital part of the original valve DAC in the steep mode is just a long sinc-like response. The analogue reconstruction filter will cause a very small asymmetry, as it smears out the end of it a bit. The response to anything else than an impulse is then the convolution of the signal and the impulse response.
In the apodizing mode, which is only meant to be used at high sample rates, the impulse response is as shown in the attachment, but then with smooth lines between the points and with a bit of smearing out due to the analogue reconstruction filter.
In the apodizing mode, which is only meant to be used at high sample rates, the impulse response is as shown in the attachment, but then with smooth lines between the points and with a bit of smearing out due to the analogue reconstruction filter.
