Marek,
I tested it with a TDA1541A but using my active arrangement istead of using an I/V resistor:
http://www.diyaudio.com/forums/digi...s-valve-output-stage-lundahl-transformer.html
In either case, the low frequency cutoff is determined by the source resistance and the primary inductance according to the formula T = L / R, where T is 1 / (2 * PI * f). For example, if we want f = 10 Hz, T = 0,016 s, and R = 10 ohms. The primary resistance must be sustracted from this value. This is the maximum value of the I/V resistor across the primary, directly at the DAC output. Using an I/V resistor across the secondary, R will be n^2 times this value, i.e. 2.56 kohm (minus the secondary resistance 375 ohms, which gives round 2.2 kohm).
I tested it with a TDA1541A but using my active arrangement istead of using an I/V resistor:
http://www.diyaudio.com/forums/digi...s-valve-output-stage-lundahl-transformer.html
In either case, the low frequency cutoff is determined by the source resistance and the primary inductance according to the formula T = L / R, where T is 1 / (2 * PI * f). For example, if we want f = 10 Hz, T = 0,016 s, and R = 10 ohms. The primary resistance must be sustracted from this value. This is the maximum value of the I/V resistor across the primary, directly at the DAC output. Using an I/V resistor across the secondary, R will be n^2 times this value, i.e. 2.56 kohm (minus the secondary resistance 375 ohms, which gives round 2.2 kohm).
Sorry if I was not quite clear. Source resistance is what the primary sees, secondary unloaded (or the load is negligible). For example, using a current generator as source and a resistor parallel with the primary. Alternatively, using a voltage generator as a source, and a resistor in series with the primary. This is the primary source resistor I calculated before (10 ohms).
We can get the same result, if the secondary is loaded instead of the primary. The necessary load resistor is n^2 times the resistor we calculated for the primary (2.56 kohms). In this case we use infinite resistor across the primary (for current generator source) or zero resistance in series with the primary (for voltage generator source).
We can get the same result, if the secondary is loaded instead of the primary. The necessary load resistor is n^2 times the resistor we calculated for the primary (2.56 kohms). In this case we use infinite resistor across the primary (for current generator source) or zero resistance in series with the primary (for voltage generator source).
Ok I though that you mean internal source resistance of DAC (as it is real source, not ideal so it has finite internal resistance). BTW DAC source resistance has something to do with freq rersponse?
Yes, I know that secondary load resistor is equal to primary*N^2. BUT which configuration will be better? For now I used only secondary resistor - I think in this case transformer carry higher signals.
Analog Device document http://www.analog.com/static/imported-files/application_notes/AN_912.pdf even use both - secondary load as I/V and primary load as "DAC termination resistor". Should I use both resistors? it seems that primary load can decrease output impedance of transformer.
Yes, I know that secondary load resistor is equal to primary*N^2. BUT which configuration will be better? For now I used only secondary resistor - I think in this case transformer carry higher signals.
Analog Device document http://www.analog.com/static/imported-files/application_notes/AN_912.pdf even use both - secondary load as I/V and primary load as "DAC termination resistor". Should I use both resistors? it seems that primary load can decrease output impedance of transformer.
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I think the DAC internal source resistance is several order of magnitude higher than the I/V resistor, so its effect is negligible.
I really don't know which configuration is better: a low I/V resistor at the primary or an n^2 higher I/V resistor at the secondary. Perhaps it is true that in the latter case the transformer carries higher signal - I have to think a bit more about it.
I have another doubt. If I use single ended output DAC and trafo with N turns ratio, secondary load will be seen by DAC as Rload/n^2 BUT with balanced output DAC load seen by every output will be Rload/2*N^2 ?
example:
trafo 1:10, secondaries load 10000 -> 10000/100=100ohm so every complementary output see 100ohm or 50ohm?
example:
trafo 1:10, secondaries load 10000 -> 10000/100=100ohm so every complementary output see 100ohm or 50ohm?
You mean a transformer with 1+1:10 turns ratio and only the secondary loaded with 10000 ohms? In this case each primary will represent 100 ohms to the DAC output.
The AD application note mentions a different configuration, where there are load resistors directly at each complementary DAC output at the primaries of the transformer. Such load resistors are not necessary IMHO if the transformer is center-tapped and the secondary is loaded. The secondary load is reflected to each primary as (1 / n^2) * Rload.
The AD application note mentions a different configuration, where there are load resistors directly at each complementary DAC output at the primaries of the transformer. Such load resistors are not necessary IMHO if the transformer is center-tapped and the secondary is loaded. The secondary load is reflected to each primary as (1 / n^2) * Rload.
No, I mean primaries connected paralell (no center tap so no ground connected) and conected directly to complementary outputs of diffirental DAC, secondary loaded with 10kohm.
About "DAC termination resistors" from AD. There must be something why thay use load on both - primaries and secondaries..
I think I will make some measurments (BW, THD) using only X secondary load, then only primary load Y=X/N^2 and then simultaneously both 2*X on secondary and 2*Y on primary (so DAC will see the same load every time).
About "DAC termination resistors" from AD. There must be something why thay use load on both - primaries and secondaries..
I think I will make some measurments (BW, THD) using only X secondary load, then only primary load Y=X/N^2 and then simultaneously both 2*X on secondary and 2*Y on primary (so DAC will see the same load every time).
The AD Application Note is about current source or sink converters with complementary output. The circuit is closed through the center tap or through the external Ro resistors. The PCM1704 has +/- 1.2 mA differential output, so no center tap is necessary.
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