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After browsing a few forums it seems that 39ohms is a commonly recommended load resistance for the outputs of the es9018, to keep it operating deep in current mode and keeping thd etc low.
Just wondering what the voltage amplitude of the output would be at 0dbfs? Just looking at options for an output stage, and need to know how much gain would be needed in a situation like this.
Does the output change much in amplitude when varying the load, say to 20ohm or 40ohm etc.
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Despite what you have been told, *ANY* resistance will have the DAC working in voltage mode. The fact is for all intents and purposes it is a voltage source unless it is working into a virtual ground. :) In order for the DAC to act like a current source it really has to operate into a virtual ground. (though this will usually not be 0V but closer to AVCC/2) Anyone who tells you differently is completely mistaken. :) The effect of adding a resistor to GND is the same as having a voltage divider of RZ/195R where RZ is your resistor to GND. So Vout = Vin * RZ/195R. The reason the DAC *appears* as a current source into a virtual ground is because the output impedance is many times higher(as close to infinity as possible) than the input impedance of a virtual ground. 20R/195R is no where near high enough a ratio. Now .1R/195R would be much better. :) Edited by user Sunday, May 2, 2010 8:09:17 AM(UTC)
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Thanks for that, its cleared up a lot. I think Im 90% there. Just one or two more simple questions to make sure Im understanding it all properly - apologies in advance for any and all stupidity on my part.
You said "The effect of adding a resistor to GND is the same as having a voltage divider of 195R/RZ where RZ is your resistor to GND." .. So in the voltage divider representation, I can presume 195R is connected to Vin? In this case wouldnt Vout = Vin * RZ/195R since we are measuring voltage across RZ?
Is Vin the same value as AVCC or is it something else, and I guess more importantly, what is its value in the buffalo II Boards?
Last question, Im not sure if there is any bias/offset voltage present on the output pins, but if there is, is there a way to calculate it if using a simple R I/V. Is it dependant on anything in particular (Vin, Rz anything else?)
Thanks again for your help.
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Ah yes, clearly a case of posting before having adequate supply of coffee.  Thanks for pointing out my silly mistake.  I have corrected my post. From the manual: "Each analog output at 0DBFS is equivalent to a voltage of approximately 92.4% of AVCC in series with 195R. So given 3.3VDC AVCC it will be about 3.05Vpp across 195R. The output will be DC biased at AVCC/2. This works out to about 16ma peak to peak at each output. The amount of bias current will depend of the voltage of the virtual ground." When I said Vin I just meant the voltage into the divider, which can be ascertained from the above. Edited by user Sunday, May 2, 2010 8:16:31 AM(UTC)
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Thanks Russ, thats brilliant. That covers it all clearly.
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I have been using the Welborne/Borbely Cato 6GM8- Mosfet unit with my BII. Increased the gain to 35dB (within specs) and am using .75 ohm drop down resistors. (Started at 39 ohms with 20dB of gain). Must say that as currently configured the sound is fantastic. Russ I don't think I can hit the .1 ohms  but it has been a great experiment trying. The Legato will have to be next. Best Bob
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