Page 231 - DCAP108_DIGITAL_CIRCUITS_AND_LOGIC_DESIGNS
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Digital Circuits and Logic Design



                   Notes         Solution:
                                 There is obviously some malfunction since the actual output waveform is not continuously
                                 increasing as it should be. The actual digital inputs are shown directly below the waveform. Notice
                                 that the converter functions correctly up to count 3. At count 4, however, the output should be 4
                                 units in amplitude. Instead, it drops to 0. It remains 4 units below the correct level until it reaches
                                 count 8. Then, from count 8 to 11, the output level is correct. But again at count 12 the output
                                 falls 4 units below the correct level and remains there for the next four levels. If you examine the
                                 waveform carefully, you will note that the output is 4 units below normal during the time when
                                 the 2  bit is supposed to be high. This then suggests that the 2  bit is being dropped (i.e. the 2
                                     2
                                                                                   2
                                                                                                             2
                                 input to the ladder is not being held high). This means that the 2 -level amplifier is malfunctioning
                                                                                   2
                                 or the 2  AND gate is not operating properly. In any case, the monotonicity check has clearly
                                        2
                                 shown that the second MSB is not being used and that the converter is not operating properly.
                                 12.3.3 Available D/A Converters
                                 D/A converters, as well as sample-and-hold amplifiers, are readily obtainable commercial
                                 products. Each unit is constructed in a single package; general-purpose economy units are available
                                 with 6-, 8-, 10-, and 12-bit resolution, and high-resolution units with up to 16-bit resolution are
                                 available.
                                 An inexpensive and very popular D/A converter is the DAC0808, an 8-bit D/A converter available
                                 from National Semiconductor. Motorola manufactures an 8-bit D/A converter, the MC1508/1408.
                                 In Figure 12.18, a DAC0808 is connected to provide a full-scale output voltage of V  = +10 Vdc
                                                                                                    0
                                 when all 8 digital inputs are 1s (high). If the 8 digital inputs are all 0s (low), the output voltage
                                 will be V  = 0 Vdc. Let us look at this circuit in detail.
                                         0
                                 First of all, two dc power-supply voltages are required for the DAC0808: Vcc = +5 Vdc and Vee
                                 = -15 Vdc. The 0.1-µF capacitor is to prevent unwanted circuit oscillations, and to isolate any
                                 variations in Vee. Pin 2 is ground (GND), and pin 15 is also referenced to ground through a resistor.
                                                            Figure 12.18: D/A Converters
























                                 The output of the D/A converter on pin 4 has a very limited voltage range (+0.5 to -0.6 V). Rather,
                                 it is designed to provide an output current I . The minimum current (all digital inputs low) is 0.0
                                                                    0
                                 mA, and the maximum current (all digital inputs high) is Iref. This reference current is established
                                 with the resistor at pin 14 and the reference voltage as

                                                            I   =  V /R ref                            ...(12.4)
                                                            ref
                                                                  ref



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