Page 218 - DCAP108_DIGITAL_CIRCUITS_AND_LOGIC_DESIGNS

Unit 12: A/D and D/A Converters

signals in this Figure 12.1 into equivalent analog voltages. The smallest number represented is Notes
000; let us make this equal to 0 V. The largest number is 111; let us make this equal to +7 V. This
then establishes the range of the analog signal to be developed. (There is nothing special about
the voltage levels chosen; they were simply selected for convenience.)

Now, notice that between 000 and 111 there are seven discrete levels to be defined. Therefore, it
will be convenient to divide the analog signal into seven levels. The smallest incremental change
in the digital signal is represented by the least-significant bit (LSB), 2°. Thus we would like to have
this bit cause a change in the analog output that is equal to one-seventh of the full-scale analog
output voltage. The resistive divider will then be designed such that a 1 in the 2° position will
cause +7 *1/7 = + 1 V at the output.
Since 2 = 2 and 2° = 1, it can be clearly seen that the 2 bit represents a number that is twice the
1
1
size of the 2° bit. Therefore, a 1 in the 2 bit position must cause a change in the analog output
1
voltage that is twice the size of the LSB.
Figure 12.1: Binary Equivalent Truth Table
2 2 2 1 2°
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1

Figure 12.2: Binary Equivalent Weights

Bit Weight Bit Weight
2 0 1/7 2 0 1/15
2 1 2/7 2 1 2/15
2 2 4/7 2 2 4/15
2 3 8/15
Sum 7/7
Sum 15/15
(a) (b)
Similarly, 2 = 4 = 2 * 2 = 4 * 2 , and thus the 2 bit must cause a change in the output voltage equal
2
1
2
0
to four times that of the LSB. The 2 bit must then cause an output voltage change of +7 *4/7 = +4 V.
2
The process can be continued, and it will be seen that each successive bit must have a weight
twice that of the preceding bit. Thus the LSB is given a binary equivalent weight of 1/7 or 1 part
in 7. The next LSB is given a weight of 2/7, which is twice the LSB, or 2 parts in 7. The MSB (in the
case of this 3-bit system) is given a weight of 4/7, which is 4 times the LSB or 4 parts in 7. Notice
that the sum of the weights must equal 1. Thus 1/7 + 2/7 + 4/7 = 7/7 = 1. In general, the binary
equivalent weight assigned to the LSB is 1/(2n – 1), where n is the number of bits. The remaining
weights are found by multiplying by 2, 4, 8, and so on. Remember,
1
LSB weight =
2 ( n - 1)

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