Problem 16.4 (a) In the successive-approximation ADO examples from Section 16.2.

Question

Problem 16.4 (a) In the successive-approximation ADO examples from Section 16.2.2, the convention is that the SAR starts with the 01111111 8-bit (as in the 74LS502 and 74LS503 SAR's in Section 16.4). Another (and reasonable) convention is to start off with the the altermate possible end result of the two schemes for an midpoint word 10000000. Briefly describe the difference in the arbitrary input. conversion levels are 0, 0.01, 0.02. (b) Consider the ADC circuit shown below, where the DAC 01111111 2.55 V. Suppose the input analog voltage of 0.45 V, and as mentioned the SAR starts with on the first clock cycle (the "start" cycle). What is the SAR output after 3 more clock cycles?

Explanation / Answer

a, If the conversion starts at 01111111 after the 2^8 clock pulses that is the ending address is at1000000.

If the conversion starts at 1000000 after the 2^8 clock pulses that is the ending address is at 01111111.

here only difference is the ending address.

b. here resolution is 2.55/(2^8-1)= 2.55/255 = 0.01

what resolution says? actualy resolution is change in analog output for one bit change in digital input.

here the conversion starts at 01111111 - equivalent aanalog value 0.45

after one pulse it would be 10000000 - eq analog value 0.45+0.01 = 0.46

after second clock pulse 10000001 equivalent analog value 0.46+0.01 = 0.47

after third clock pulse 10000010 equivalent analog value is 0.47+0.01 = 0.48

Therefore the output of SAR after 3 pulses is 0.48

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