There is an integrated DAC word length N=8 bits and a range of reference voltage 0V....+10v. Determine the value of the output voltage when the input code 170.
Correct answer depends on the DAC reference voltage.There is an integrated DAC word length N=8 bits and a range of reference voltage 0V....+10v. Determine the value of the output voltage when the input code 170.
TS - "0V....+10v"Correct answer depends on the DAC reference voltage.
Are you certain?\( Vout = \frac{170\times Vref}{2^n-1}\)
TS - "0V....+10v"
Whatever that means.
And the numbering notation. Is 170 decimal, hexadecimal or octal?
Pretty certain, since the reference is in the numerator .Are you certain?
Even in that data sheet, the term in question is in the numerator.Not according to this MCP4901 DAC datasheet from Microchip.
Your first result is the correct one. An 8-bit DAC has 256 possible output codes. If it's unipolar with a 10V reference, an LSB is worth 10/256. So the minimum output voltage is 0 * LSB = 0V, and the maximum output voltage is 255 * LSB = 9.96 V.Even in that data sheet, the term in question is in the numerator.
But, there appears to be an inherent error in the equation given in the data sheet in that the denominator is given as decimal 256 for an 8 bit DAC.
Given a reference of 10 volts and a gain of 1, with the input code to the DAC being decimal 255 and the denominator being 256, Vout would be:
\( V_{out} = \frac{10V\times255}{256} \approx 9.96 volts\), which is wrong.
With the denominator at 255, however, we have:
\( V_{out} = \frac{10V\times255}{255} = 10.00 volts\), which is right.
yeah, from 0 to 255.An 8-bit DAC has 256 possible output codes.
The datasheet is definitely not wrong. If you're splitting up 10V into equal chunks numbered 0 through 255, the smallest chunk size is 10/256.yeah, from 0 to 255.
the microchip datasheet is wrong.
Oh, good grief. If the Microchip data sheet is wrong, then the data sheets for ALL of the DACs I've ever seen and used in the last 40+ years, from EVERY manufacturer, are also wrong-- because they all say the same thing: the output of an N-bit DAC is the reference voltage (or current, whichever the case) times the input code divided by 2^N, NOT divided by (2^N-1).yeah, from 0 to 255.
the microchip datasheet is wrong.
Thanks for the reality check You're right, my first result was the correct one. I got chunks and levels confused. Here's the proof:Your first result is the correct one. An 8-bit DAC has 256 possible output codes. If it's unipolar with a 10V reference, an LSB is worth 10/256. So the minimum output voltage is 0 * LSB = 0V, and the maximum output voltage is 255 * LSB = 9.96 V.
by Jake Hertz
by Aaron Carman
by Aaron Carman