A 10-bit DAC has 1.023 V full-scale output. Its address inputs $A_9\ldots A_2$ connect to an 8-bit microprocessor’s $D_7\ldots D_0$, while $A_1,A_0$ are grounded. If the processor byte changes from 1010 1010 to 1010 1011, find output change in mV, to one decimal place.
The processor byte increases by one, but its least significant bit drives DAC bit A₂, worth four DAC counts. The DAC step is $1.023/(2^{10}-1)=0.001$ V = 1 mV. Therefore the output changes by $4(1)=4.0$ mV. Grounding the two lowest bits prevents ordinary one-count DAC steps.
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