LTC1446/LTC1446L
U U
DEFI ITIO S
Resolution (n)
Nominal LSBs:
Resolution is defined as the number of digital input bits,
n. It defines the number of DAC output states (2n) that
divide the full-scale range. The resolution does not imply
linearity.
LTC1446 LSB = 4.095V/4095 = 1mV
LTC1446L LSB = 2.5V/4095 = 0.610mV
Zero Scale Error (ZSE)
The output voltage when the DAC is loaded with all zeros.
Since this is a single supply part this value cannot be less
than 0V.
Full-Scale Voltage (VFS)
This is the output of the DAC when all bits are set to one.
Voltage Offset Error (VOS)
Integral Nonlinearity (INL)
The theoretical voltage at the output when the DAC is
loaded with all zeros. The output amplifier can have a true
negative offset, but because the part is operated from a
single supply, the output cannot go below zero. If the
offset is negative, the output will remain near 0V resulting
in the transfer curve shown in Figure 1.
End-point INL is the maximum deviation from a straight
line passing through the end points of the DAC transfer
curve. Because the part operates from a single supply and
the output cannot go below 0, the linearity is measured
between full scale and the code corresponding to the
maximum offset specification. The INL error at a given
input code is calculated as follows :
OUTPUT
VOLTAGE
INL = [VOUT – VOS – (VFS – VOS)(Code/4095)]/LSB
VOUT=theoutputvoltageoftheDACmeasuredatthegiven
input code
0V
NEGATIVE
OFFSET
DAC CODE
1446/46L F01
Differential Nonlinearity (DNL)
Figure 1. Effect of Negative Offset
DNL is the difference between the measured change and
the ideal 1LSB change between any two adjacent codes.
The DNL error between any two codes is calculated as
follows:
The offset of the part is measured at the code that corre-
sponds to the maximum offset specification:
VOS = VOUT – [(Code)(VFS)/(2n – 1)]
DNL = (∆VOUT – LSB)/LSB
Least Significant Bit (LSB)
∆VOUT = The measured voltage difference between two
adjacent codes
One LSB is the ideal voltage difference between two
successive codes.
LSB = (VFS – VOS)/(2n – 1) = (VFS – VOS)/4095
7