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M16C/62P Group (M16C/62P, M16C/62PT)
18.2.5 Current Consumption Reducing Function
When not using the A/D converter, its resistor ladder and reference voltage input pin (VREF) can be
separated using the VCUT bit in the ADCON1 register. When separated, no current will flow from the
VREF pin into the resistor ladder, helping to reduce the power consumption of the chip.
To use the A/D converter, set the VCUT bit to "1" (Vref connected) and then set the ADST bit in the
ADCON0 register to "1" (A/D conversion start). The VCUT and ADST bits cannot be set to "1" at the same
time.
Nor can the VCUT bit be set to "0" (Vref unconnected) during A/D conversion.
Note that this does not affect VREF for the D/A converter (irrelevant).
18.2.6 Output Impedance of Sensor under A/D Conversion
To carry out A/D conversion properly, charging the internal capacitor C shown in Figure 18.10 has to
be completed within a specified period of time. T (sampling time) as the specified time. Let output
impedance of sensor equivalent circuit be R0, microcomputer's internal resistance be R, precision
(error) of the A/D converter be X, and the A/D converter's resolution be Y (Y is 1024 in the 10-bit
mode, and 256 in the 8-bit mode).
VC is generally VC = VIN {1 – e
And when t = T,
Hence, R0 = –
Figure 18.10 shows analog input pin and external sensor equivalent circuit. When the difference
between VIN and VC becomes 0.1LSB, we find impedance R0 when voltage between pins VC
changes from 0 to VIN-(0.1/1024) VIN in time T. (0.1/1024) means that A/D precision drop due to
insufficient capacitor charge is held to 0.1LSB at time of A/D conversion in the 10-bit mode. Actual
error however is the value of absolute precision added to 0.1LSB. When f(XIN) = 10 MHz, T = 0.3 µs
in the A/D conversion mode with sample & hold. Output impedance R0 for sufficiently charging ca-
pacitor C within time T is determined as follows.
T = 0.3 µs, R = 7.8 kΩ, C = 1.5 pF, X = 0.1, and Y = 1024 . Hence,
R0 = –
1.5 X 10
Thus, the allowable output impedance of the sensor circuit capable of thoroughly driving the A/D con-
verter turns out to be approximately 13.9 kΩ.
R
e
. v
2
3 .
0
S
e
p
0
, 1
2
0
0
4
R
E
J
0
9
B
0
1
8
5
0 -
2
3
0
Z
–
X
VC=VIN –
Y
1
–
C (R0 + R)
e
1
–
T= ln
C (R0 +R)
T
– R
X
C • ln
Y
0.3 X 10
-6
0.1
–12
• ln
1024
page 222
f o
3
6
4
1
t
C (R0 + R)
}
X
VIN=VIN(1 –
)
Y
T
X
=
Y
X
Y
3
3
–7.8 X10
13.9 X 10
18. A/D Converter
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