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Digital Circuits and Logic Design



                   Notes
                                    3.  The full-scale output voltage is equal to the positive voltage of the digital input signal. (The
                                      divider would work equally well with input voltages of 0 and  V.)
                                    4.  The LSB has a weight of 1/(2  – 1), where n is the number of input bits.
                                                             n
                                                                                                n
                                    5.  The change in output voltage due to a change in the LSB is equal to V/(2  – 1), where V is
                                      the digital input voltage level.
                                    6.  The output voltage V  can be found for any digital input signal by using the following
                                                        A
                                      modified form of Millman’s theorem:
                                                                                       3
                                                                     0
                                                                                 2
                                                                           1
                                                                  V 2 +  V 2 +  V 2 + V 2 + ... + V  2 n − 1
                                                            V   =   0    1     2      3        n − 1     ...(12.1)
                                                             A                     n
                                                                                  2 −  1
                                 Where V , V , V , V ,………, V n – 1  are the digital input voltage levels (0 or V) and n is the number
                                        0
                                                 3
                                              2
                                           1
                                 of input bits.
                                               12.3:
                                 For a 5-bit resistive divider, determine the following: (a) the weight assigned to the LSB; (b) the
                                 weight assigned to the second and third LSB; (c) the change in output voltage due to a change
                                 in the LSB, the second LSB, and the third LSB; (d) the output voltage for a digital input of 10101.
                                 Assume 0 = 0 V and 1 = + 10 V.
                                 Solution:
                                   (a)   The LSB weight is 1/(2  – 1) = 1/31.
                                                         5
                                   (b)  The second LSB weight is 2/31, and the third LSB weight is 4/31.
                                   (c)  The LSB causes a change in the output voltage of 10/31 V. The second LSB causes an output
                                      voltage change of 20/31 V, and the third LSB causes an output voltage change of 40/31 V.
                                   (d)  The output voltage for a digital input of 10101 is
                                                                      0
                                                                              1
                                                                                     2
                                                                                            3
                                                                    *
                                                                            *
                                                                                                 *
                                                                                         *
                                                                                   *
                                                            V   =   10 2 +  102 +  10 2 +  0 2 + 102 4
                                                             A                    5
                                                                                 2 − 1
                                                                    (
                                                                        4
                                                                  10 1 ++  16)  210
                                                                =             =     = +6.77 V
                                                                     32 − 1     31
                                 This resistive divider has two serious drawbacks. The first is the fact that each resistor in the
                                 network has a different value. Since these dividers are usually constructed by using precision
                                 resistors, the added expense becomes unattractive. Moreover, the resistor used for the MSB is
                                 required to handle a much greater current than that used for the LSB resistor. For example, in
                                 a 10-bit system, the current through the MSB resistor is approximately 500 times as large as the
                                 current through the LSB resistor. For these reasons, a second type of resistive network, called a
                                 ladder, has been developed.
                                 12.2 Binary Ladders
                                 The binary ladder is a resistive network whose output voltage is a properly weighted sum of
                                 the digital inputs. Such a ladder, designed for 4 bits, is shown in Figure 12.6. It is constructed
                                 of resistors that have only two values and thus overcomes one of the objections to the resistive
                                 divider previously discussed. The left end of the ladder is terminated in a resistance of 2R, and
                                 we shall assume for the moment that the right end of the ladder (the output) is open-circuited.
                                 Let us now examine the resistive properties of the network, assuming that all the digital inputs
                                 are at ground. Beginning at node A, the total resistance looking into the terminating resister is 2R.



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