I start a new thread with details of my Transformer/Tube Output Stage, not wanted to spoil the "Passive-Active I/V stage" thread any longer.
The topology is based on the SRPP (Shunt Regulated Push Pull) arrangement. Both tube halves operate in grounded cathode mode, and are driven push-pull. The low input impedance of the cathodes seen by the secondary is transformed back to the primary. The impedance transformation is the turns ratio squared, so the DAC output can see very low load impedance (around 5 ohms).
The circuit is designed for two TDA1541A DACs, one is driven with inverted binary input, so they give push-pull output currents of 8 mA peak-to-peak.
The bias currents in the primary windings flow in opposite direction, so the transformer core will not be magnetized. Same is true for the secondaries.
I built the circuit with Lundahl LL1678 transformers and Mullard E88CC tubes. The measurement results at 2 V RMS output signal are (Left and Right channel):
THD = 0.03% / 0.018%
IMD (250 Hz + 7 kHz, 4:1) = 0.038% / 0.02%
S/N linear = 69 dB / 62 dB
S/N A weighted = 75.3 dB / 72.6 dB
Frequency response:
10 Hz = -0.4 dB
100 kHz = -1.5 dB
Only one half of the primary is connected to a single DAC currently (there is 2 mA bias current flowing through the coil). The 0 dB output signal from a test CD is 2.5 V RMS with the DAC connected. The listening tests are very positive so far.
The topology is based on the SRPP (Shunt Regulated Push Pull) arrangement. Both tube halves operate in grounded cathode mode, and are driven push-pull. The low input impedance of the cathodes seen by the secondary is transformed back to the primary. The impedance transformation is the turns ratio squared, so the DAC output can see very low load impedance (around 5 ohms).
The circuit is designed for two TDA1541A DACs, one is driven with inverted binary input, so they give push-pull output currents of 8 mA peak-to-peak.
The bias currents in the primary windings flow in opposite direction, so the transformer core will not be magnetized. Same is true for the secondaries.
I built the circuit with Lundahl LL1678 transformers and Mullard E88CC tubes. The measurement results at 2 V RMS output signal are (Left and Right channel):
THD = 0.03% / 0.018%
IMD (250 Hz + 7 kHz, 4:1) = 0.038% / 0.02%
S/N linear = 69 dB / 62 dB
S/N A weighted = 75.3 dB / 72.6 dB
Frequency response:
10 Hz = -0.4 dB
100 kHz = -1.5 dB
Only one half of the primary is connected to a single DAC currently (there is 2 mA bias current flowing through the coil). The 0 dB output signal from a test CD is 2.5 V RMS with the DAC connected. The listening tests are very positive so far.
Attachments
When I calculated the ideal turns ratio of the transformer, I found an interesting thing: the output voltage is inversely proportional to the turns ratio. It is not directly proportional, as one would assume. How it comes?
Let's assume the peak-to-peak output current of the DAC is 4 mA, that is equivalent with 1.41 mA RMS. With 1:16 turns ratio the secondary AC current is transformed down to 88.4 uA RMS. I measured about 64 mV on the secondary. From these two values one can calculate the load represented by the cathode input impedance. It gives 724 ohms. Since I connected 1.5 k parallel with the cathodes, the true cathode input impedance is 1.4 kohm. The tube circuit in itself has a fixed 32x voltage gain, so we yield around 2.05 V RMS audio output voltage.
Now what happens with an 1:8 turns ratio? The secondary current will be double than before, 176.8 uA. This gives twice as much voltage at the fixed 724 ohms cathode impedance, i. e. 128 mV. Since the gain is fixed, we get double audio output voltage, that is 4.1 V.
I concluded to this only in theory. I will try the 1:8 turns ratio just to see if the above is valid.
Let's assume the peak-to-peak output current of the DAC is 4 mA, that is equivalent with 1.41 mA RMS. With 1:16 turns ratio the secondary AC current is transformed down to 88.4 uA RMS. I measured about 64 mV on the secondary. From these two values one can calculate the load represented by the cathode input impedance. It gives 724 ohms. Since I connected 1.5 k parallel with the cathodes, the true cathode input impedance is 1.4 kohm. The tube circuit in itself has a fixed 32x voltage gain, so we yield around 2.05 V RMS audio output voltage.
Now what happens with an 1:8 turns ratio? The secondary current will be double than before, 176.8 uA. This gives twice as much voltage at the fixed 724 ohms cathode impedance, i. e. 128 mV. Since the gain is fixed, we get double audio output voltage, that is 4.1 V.
I concluded to this only in theory. I will try the 1:8 turns ratio just to see if the above is valid.
One thing to think about is the impedance reflected back to the input of your dac chip, given the change in ratio this will now be about 4X higher than it was before. This should result in an appreciable voltage on the output of your dac and possibly adverse effects on INL/DNL performance in the dac. (Linearity basically.)
Very interesting circuit btw. Operating in the current domain as it apparently does makes casual analysis rather difficult, but as I think about it the counter-intuitive results you are getting probably make sense. If you were to make the impedance reflect back to the primary identical for both 1:8 and 1:16 then I think you would see the normal expected voltage relationship restored. (I'm not saying you want to do this, but it would be an interesting experiment.)
Very interesting circuit btw. Operating in the current domain as it apparently does makes casual analysis rather difficult, but as I think about it the counter-intuitive results you are getting probably make sense. If you were to make the impedance reflect back to the primary identical for both 1:8 and 1:16 then I think you would see the normal expected voltage relationship restored. (I'm not saying you want to do this, but it would be an interesting experiment.)
If I use a 1:1 input transformer and a JFET instead of tube, do you guys think that a 10 ohm resistor can be used for I/V, instead of the over 1K resistor?
A hybrid cascode with FET as stage one is my thinking, similar to the left drawing at
http://www.tubecad.com/articles_2002/RIAA_Preamps_Part_2/page14.html
Iout is applied to the source instead of the gate of the FET.
A hybrid cascode with FET as stage one is my thinking, similar to the left drawing at
http://www.tubecad.com/articles_2002/RIAA_Preamps_Part_2/page14.html
Iout is applied to the source instead of the gate of the FET.
Hi agent.5,
A 10 ohm I/V resistor is probably too high for the TDA1541A, because its voltage compliance is 25 mV peak-to-peak, and 10 ohm will result in 40 mV. I am interested in your result with the FET/Tube Loftin-White stage. If you build it, please compare the circuit with a simple version where the tube is replaced with a single 1.5 k resistor in the drain of the FET (anode voltage adjusted accordingly lower).
If noise is your concern, my circuit is _very_ quiet...
A 10 ohm I/V resistor is probably too high for the TDA1541A, because its voltage compliance is 25 mV peak-to-peak, and 10 ohm will result in 40 mV. I am interested in your result with the FET/Tube Loftin-White stage. If you build it, please compare the circuit with a simple version where the tube is replaced with a single 1.5 k resistor in the drain of the FET (anode voltage adjusted accordingly lower).
If noise is your concern, my circuit is _very_ quiet...
single ended - balaced operation
Hi!
Great work!
The measurements you made (IMD, S/N, etc.) represent the single ended operation (2 mA standing DC current), or the projected balanced mode?
Im thinking of possibly using 2 or 4 paralled LL1678 in single ended 1:32 mode for reducing core magnetization by standing dc currrent.
Hi!
Great work!
The measurements you made (IMD, S/N, etc.) represent the single ended operation (2 mA standing DC current), or the projected balanced mode?
Im thinking of possibly using 2 or 4 paralled LL1678 in single ended 1:32 mode for reducing core magnetization by standing dc currrent.
Hi,
I measured with single-ended drive, one end of the input coil was unconnected. So there was 2 mA DC biasing the transformer primary. This could have been easily cancelled by a biasing resistor or CCS on the unconnected winding, but I did not bother. In the final build I intend to use both DACs of a TDA1541A per channel, one side driven with inverted binary signal train, that will facilitate complete bias cancellation, harmonic cancellation and double output level.
Laszlo
I measured with single-ended drive, one end of the input coil was unconnected. So there was 2 mA DC biasing the transformer primary. This could have been easily cancelled by a biasing resistor or CCS on the unconnected winding, but I did not bother. In the final build I intend to use both DACs of a TDA1541A per channel, one side driven with inverted binary signal train, that will facilitate complete bias cancellation, harmonic cancellation and double output level.
Laszlo
Thats why i'd like to parallel some transformers and drive them single-ended.
With 2 LL1678 paralleled the DCR of the primaries and the reflected DCR of the secondaries in 1:32 mode together are only about 1 Ohm.
With 4 paralleled, only about 0.5 ohm's.
This leaves more headroom regarding the voltage compliance of the dac for the signal swing.
The THD and IMD figures of one LL1678 look to me as if the core has not run into saturation. So with 2 or 4 one would be on the safe side.
Or where do I take a wrong turn in thinking?
With 2 LL1678 paralleled the DCR of the primaries and the reflected DCR of the secondaries in 1:32 mode together are only about 1 Ohm.
With 4 paralleled, only about 0.5 ohm's.
This leaves more headroom regarding the voltage compliance of the dac for the signal swing.
The THD and IMD figures of one LL1678 look to me as if the core has not run into saturation. So with 2 or 4 one would be on the safe side.
Or where do I take a wrong turn in thinking?
Hi Laszlo
Sorry for the noob question. In your design, is the reflected resistance of 5 Ohms on the primary, in series with Iout from the TDA1541? Or is it equivalent to a 5R resistor connectd to Iout on the primary & its other end beng connected to the analog ground? Many thanks
Regards
Fib
Sorry for the noob question. In your design, is the reflected resistance of 5 Ohms on the primary, in series with Iout from the TDA1541? Or is it equivalent to a 5R resistor connectd to Iout on the primary & its other end beng connected to the analog ground? Many thanks
Regards
Fib
oshifis said:
Thanks Laszlo
In my present tube output stage, I am planning to reduce the size of the i/v resistor after TDA1541S1 & make up the gain with Lundahl 1678 in 1:16.
However, as I have a single chip feeding the trannys, I may have to put the 5R resistor on the primary as my tube output design is different from yours. It was such a leap from discreet output stage to tubes. Now I am hoping to extract the best performance from this magical dac chip. Thanks again
Best regards
Fib
This is due to the constant input current. I have done another calculation:oshifis said:
Let's take a 1:10 transformer and 1 mA current at the primary. Let's have 1 kohm secondary load. The load will be transformed to the primary as 10 ohm. 1 mA will drop 10 mV on this virtual resistance, which in turn will be transformed to the secondary as 100 mV.
Now let's have a 1:20 transformer. The same 1 kohm load will be transformed to the primary as 1000/20^2 = 2.5 ohm. The same 1 mA current will drop 2.5 mV on it. This will be transformed to the secondary as 50 mV.
The impedance transformed to the primary and the output voltage is inversely proportional with the turns ratio, indeed.
In this sense my push-pull TDA1541s give 8 mA peak-to-peak, 2.82 mA RMS current. The gain of the push-pull driven ECC88/6DJ8 is 32. For 2 V analog output the cathodes should see 62.5 mV each. The cathode resistance is about 1.35 k, so the secondary current will be 0.0463 mA in each cathode, 0.0926 mA push-pull. The turns ratio will be then 1:30 (1+1:30+30). The transformed impedance (what each DAC output will see) is 1350/30^2 = 1.5 ohm.
I also calculated the necessary inductance of the transformer with the T = L/R formula. T = 16 ms for 10 Hz cutoff. L will be then 0.016 * 1350 = 21.6 H, rounded 20 H at the half secondary. This is equivalent to 24 mH at the half primary (between end and center tap), whichever is easier to measure.
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