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My audio note dac kit has a pair 1:1 transformer at the output of the dac chip, with 330 Ohm resistor across the primary and 3.3k resistor across the secondary, which is connected to the grid of the output tube.
Im trying to adjust the output of the DAC to match the gain of the rest of the system. I understand I can play around with the values of the two 330 and 3.3K resistors.
Questions:
1) How do I calculate the AC resistance that the DAC chip see looking from the chip taking into account also dc resistance and inductance of the windings.
2) if I resistance load only one side of the transformers (either primary or secondary), does it matter which side?
Thanks
Follow Ups:
are you sure the tx is 1:1 ?1:10 would make more sense..
normally secondary is loaded so that the load impedance to the chip (AD1865?) is the reflected impedance across the winding as function of of turns ratio squared.
Shane
Edits: 10/11/14 10/11/14 10/11/14
The current is transformed into voltage across the 330 ohm resistor. If 6mA across a 330 ohm resistor results in 2 volts, a 33 ohm resistor with 6mA would give 0.2 volts. That is if the current source can handle the the 33 ohm load without trying to dump more current through it and overloading the chip.
I would leave the DAC as designed. You have a golden opportunity to decrease the gain in your amp and/or kill your preamp which will improve your sound. Muting the DAC will give you worse sound as you are going to be amplifying lower level signals which will give you a lower signal to noise ratio.
Vincin
The 330 and 3.3K resistors are in parallel.
The 330 is for I/V conversion.
The 3.3K is for the transformers performance.
DanL
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