Laszlo's Valve Output Stage with Lundahl transformer - Page 2 - diyAudio
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Old 17th September 2007, 11:17 AM   #11
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Question 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.

:-)
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Old 17th September 2007, 02:31 PM   #12
oshifis is offline oshifis  Hungary
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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
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Old 17th September 2007, 03:41 PM   #13
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Just bear in mind the dacs will superimpose so keep an eye on voltage compliance.

I.e the voltage as seen by the dac will be twice that indicated by the impedance it sees when it drives the transformer differentially.
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Old 17th September 2007, 07:52 PM   #14
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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?
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Old 18th September 2007, 12:40 PM   #15
oshifis is offline oshifis  Hungary
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Quote:
Originally posted by martinjufer
Thats why i'd like to parallel some transformers and drive them single-ended.
...
So with 2 or 4 one would be on the safe side.
Or where do I take a wrong turn in thinking?
In the price of the Lundahls
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Old 18th December 2007, 02:35 AM   #16
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Default Re: Laszlo's Valve Output Stage with Lundahl transformer

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
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Old 20th December 2007, 07:32 PM   #17
oshifis is offline oshifis  Hungary
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The 5 ohm is the input impedance of the circuit, that is the DAC is seeing 5 ohm load. It is equivalent with a 5 ohm resistor between the Iout and GND.
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Old 20th December 2007, 10:05 PM   #18
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Quote:
Originally posted by oshifis
The 5 ohm is the input impedance of the circuit, that is the DAC is seeing 5 ohm load. It is equivalent with a 5 ohm resistor between the Iout and GND.

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
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Old 7th July 2008, 03:55 PM   #19
oshifis is offline oshifis  Hungary
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Quote:
Originally posted by oshifis
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.
This is due to the constant input current. I have done another calculation:

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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Old 1st May 2010, 11:22 AM   #20
oshifis is offline oshifis  Hungary
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Inductance data of the Lundahl LL1678 (measured):

Lpri = 175 mH
Lsec = 44 H

Resistances (factory data):

Rpri = 4.5 ohms
Rsec = 375 ohms
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