Page 238 - DCAP108_DIGITAL_CIRCUITS_AND_LOGIC_DESIGNS

Unit 12: A/D and D/A Converters

time required is longer. Since the counter always begins at zero and counts through its normal Notes
binary sequence, as many as 2 counts may be necessary before conversion is complete. The
n
average conversion time is, of course, 2 /2 or 2 n – 1 counts.
n
The counter advances one count for each cycle of the clock, and the clock therefore determines
the conversion rate. Suppose, for example, that we have a 10-bit converter. It requires 1024 clock
cycles for a full-scale count. If we are using a 1-MHz clock, the counter advances 1 count every
microsecond. Thus, to count full scale requires 1024 * 10 = 1.024 ms. The converter reaches one-
–6
half full scale in half this time, or in 0.512 ins. The time required to reach one-half full scale can
be considered the average conversion time for a large number of conversions.
12.12: Suppose that the converter shown in Figure 12.23 is an 8-bit converter

driven by a 500-kHz clock. Find (a) the maximum conversion time; (b) the average conversion
time; (c) the maximum conversion rate.

Solution:
(a) An 8-bit converter has a maximum of 2 = 256 counts. With a 500-kHz clock, the counter
8
advances at the rate of 1 count each 2 µs. To advance 256 counts requires 256 × 2 × 10 =
–6
512 × 10 = 512 µs.
–6
(b) The average conversion time is one-half the maximum conversion time. Thus it is 1/2 ×
0.512 × 10 = 0.256 ms.
–3
(c) The maximum conversion rate is determined by the longest conversion time. Since the
converter has a maximum conversion time of 0.512 ms, it is capable of making at least
1/(0.512 * 10 ) ≡ 1953 conversions per second.
–3
Figure 12.24: Control of the A/D Converter

To RESET
START
Counter
Clock
OS
Clock pulses
S Q to counter
R Q
Control
flip-flop
Comp.
Analog input Ref voltage
voltage
a
()
START
OS
Control flip-flop
Clock
Analog input
voltage level
Ret voltage
OV –
Comparator
output
()
b

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