Very high I/V resistor in Sony dataprocessing IC.

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I am considering building an experimental CD player using a salvaged transport from a portable CD player. It uses an obscure Sony Data processing chip with a built in 1bit DAC. On the chip it has a current output and then an internal OPamp for generating a voltage output. However the sttange part is that on the suggested implementation it has a network of I/V and feedback resistors with values of 12K. The current output seems to suggest that its capable of 4mA signal output - but this is difficult to interpret from the datasheet and doesn't seem to fit with a 12K I/V resistor.
I want to put it into a tube output stage and use just the current output with a resistor load - but This would suggest a value of I/V more in the range of 100's of ohms.

I am new to this, but it doesn't make much sense so far.

Shoog
 
Tricky that since I only have a sister datasheet from the same period, and even that was a bitch to find. The datasheet i have is for the CXD2548R

http://pdf.datasheetcatalog.com/datasheet/sony/a6800323.pdf

Page 8 has the output data and page 12 has the suggested output network.

I know your going to tell me this is a pile and not to bother - but this is a no cost experiment to see if its worth seeking out a decent transport to do the same.

Shoog
 
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I can see 1V15 and 8K on p.13 , but I can't see 4mA anywhere on page 8

Output voltage (3) sez (-0,28/+0,36)mA

I saw the output voltage 3, but wonder why there is an output voltage 1,2,3. As I said this is new to me and it really doesn't make a lot of sense yet.

8k is the min. load resistance, which is about what you would get when all of the output network were summed together.

Still if that is correct then the calculated output voltage would be -2.24V to +2.88V which would be a reasonable output voltage.
Its a workable system - but certainly far from typical for a current output DAC.

Shoog
 
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see - number in brackets is referring to applicable pins

so , (3) is current at those AIN pins , where 12K resistors are connected and that's , in this moment , still pretty much black box to me

however , -2V24/+2V88 is pretty decent output voltage

even if figure of 8K as min load is suggesting that LOUT pins aren't current output pins , but voltage ones
 
Looking at your data sheet, page 12, the circuit with the 12k resistors etc is a low pass filter circuit for measuring distortion and noise figures at 1 kHz. It is not the suggested circuit for normal operation. Normal operation seems to be the page 12 bottom diagram, which seems to be a direct feed with levels given on page 13, top.
 
Looking at your data sheet, page 12, the circuit with the 12k resistors etc is a low pass filter circuit for measuring distortion and noise figures at 1 kHz. It is not the suggested circuit for normal operation. Normal operation seems to be the page 12 bottom diagram, which seems to be a direct feed with levels given on page 13, top.
The same circuit is suggested as the output configuration into the final load later in the datasheet, so those figures are the recommended one's.

Shoog
 
Hi,

the Datasheet specifies Lout1 and Lout2 as voltage outputs of the internal OPamps.
Chapter 3-11, Diagram 3-12 on page73 clears up things.
It reads LPF-Block and that it is.
The Opamp between the Lout and Ain pins is working as a unity gain filter block.
As such a variation in the feedback resistors varies the filter´s response but not the output amplitude.
The second OPamp feeding into Aout generates a DC Reference voltage.
It´ll probabely work as DC-Servo here.
So no hint of a current output at all.
I/V conversion is fixed inside the Chip and fixed is the output volage level.

jauu
Calvin
 
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