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Old 14th October 2008, 09:38 PM   #1
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Default attenuation using I/V

Hi there.
I feel great writing in this wonderful forum.

I have builted a dac using the AD1865,2 chips per channel,and i am using only an 150R resistor as I/V,following a tube stage working in A2 class.

The dac sounds great!!but,i don't want to use any pot after the tube stage for attenuation.

Is it possible to make an variable I/V resistor,and use it as analog attenuation??Has anyone tried it before?Any opinions?

Thank you in advance
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Old 14th October 2008, 10:05 PM   #2
DougL is offline DougL  United States
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I am going to try it.

I have a RAKK DAC with a passive board.

I am going to switch in a shunt resistor.

I will let you know how it works.

Doug
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Old 14th October 2008, 10:42 PM   #3
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The RAKK dac Passive Output has transformers on it so it interfaces nicely with a potentiometer as shown here: http://www.raleighaudio.com/passive_output.htm

The concept of using a potentiometer as an I/V resistor (without a transformer) is a good one, but runs into trouble in implementation. There aren’t any sub-100 Ohm potentiometers available that I know of. I would like to know of one if any of you have that information.

Dave
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Old 15th October 2008, 02:56 AM   #4
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Thanks Doug.I am using an 150R from B.I and i was thinking of making a ladder attenuator down to 15R,that is 20db
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Old 15th October 2008, 02:56 AM   #5
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Hi Dave.
I know there isn't anything like this on the market,so i am going to use some B.I resistors and make a discrete pot,like a R2R ladder,from 150R to abut 15~10R.What do you think?
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Old 15th October 2008, 03:24 AM   #6
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I have tried a discrete stepped attenuator for the I/V conversion for a current-output DAC and found that each step produced a loud “pop” and thus was unacceptable.

Dave
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Old 15th October 2008, 12:51 PM   #7
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David,
you could try and use the stepper attenuator parallel to a 150R resistror,so there is always a resistor in the path of the signal,and you won't hear the pop.

Was the sound better than the classic pot attenuation?
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Old 15th October 2008, 03:18 PM   #8
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You need this 75 ohm rotary attenuator!

Click the image to open in full size.
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Thanx!
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Old 15th October 2008, 03:56 PM   #9
DougL is offline DougL  United States
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Quote:
I have tried a discrete stepped attenuator for the I/V conversion for a current-output DAC and found that each step produced a loud “pop” and thus was unacceptable.
Never thought of that.
I have the parts now, may as well see.

Thanks Dave.

Doug
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Old 15th October 2008, 05:24 PM   #10
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I did not try a stepped attenuator in parallel with a fixed resistor, however I suspect that it will still pop because of the sharp, discrete change in resistance with each step rather than a smooth transition that a potentiometer will give. I have not tried it so I do not know for sure and I could be wrong about this, so report back your findings if you try it. You could breadboard a simple single-step test with two or three resistors and a switch to see what happens.

The advantage of this attenuator is that the I/V resistor is always as small as possible, which is a good thing for current output DACs. Of course, it depends on the implementation but a smaller variable resistor may sound better than a fixed larger resistor there, and the character of the sound will depend on the type of resistor used. I have found that the type of resistor makes more difference in the character of the sound than does the size of the resistor. The size is subtle and the type is noticeable. Before you ask, there is no “best” resistor to use there, only differences and preferences.

Does it sound better than the classic pot attenuation? Again, it depends on the implementation. To me, a properly designed shunt attenuator sounds better than the classic potentiometer. Use a high-quality resistor for the series leg and I prefer a stepped attenuator, like the DACT or the K&K Audio, for the shunt leg. Keep the series leg as small as possible – equal to or less than the nominal resistance of the shunt leg. Assuming this, I have not found a consistent difference to where in the attenuator is placed in the signal path. I have placed it just before the tube in the active stage, and now, for convenience, have moved it to the output of the active stage. The important thing here is that my series leg is 5K.

Dave
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