Laszlo's Valve Output Stage with Lundahl transformer - diyAudio
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Old 18th April 2007, 12:52 PM   #1
oshifis is online now oshifis  Hungary
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Default Laszlo's Valve Output Stage with Lundahl transformer

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.
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Old 18th April 2007, 07:41 PM   #2
oshifis is online now oshifis  Hungary
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Picture of the experimental unit
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Old 3rd May 2007, 04:13 PM   #3
oshifis is online now oshifis  Hungary
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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.
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Old 4th May 2007, 12:10 AM   #4
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Would you know what output impedence you have?
Cheers George
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Old 4th May 2007, 06:36 AM   #5
oshifis is online now oshifis  Hungary
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It is in the order of 1 k (700 ohms to 1300 ohms). I will measure it more exactly.
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Old 4th May 2007, 05:44 PM   #6
kevinkr is offline kevinkr  United States
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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.)
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Old 13th May 2007, 09:20 PM   #7
agent.5 is offline agent.5  United States
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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..._2/page14.html

Iout is applied to the source instead of the gate of the FET.
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Old 14th May 2007, 01:11 PM   #8
oshifis is online now oshifis  Hungary
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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...
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Old 14th May 2007, 06:57 PM   #9
oshifis is online now oshifis  Hungary
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I made some more accurate measurements from a test CD with 1 kHz 0 dB sine tone (L/R ch):

Vout = 2.06/2.03 V
Rout = 1.93/1.88 kohm
S/N = 72.0/66.6 dB lin
S/N = 88.2/82.8 dB (A weighted)
S/N = 90.1/85.6 dB (A weighted + 400 Hz/30 kHz filter)
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Old 14th May 2007, 08:18 PM   #10
agent.5 is offline agent.5  United States
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Quote:
Originally posted by oshifis

If noise is your concern, my circuit is _very_ quiet...
Thanks. That is good to know. Noise is my concern.
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