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M
1
6
C
2 /
6
A
G
o r
u
p
(
M
1
6
14.5 Output Impedance of Sensor under A/D Conversion
To carry out A/D conversion properly, charging the internal capacitor C shown in Figure 14.5.1 has to be
completed within a specified period of time. T (sampling time) as the specified time. Let output imped-
ance 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,
Figure 14.5.1 shows analog input pin and externalsensor equivalent circuit. When the difference be-
tween 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 timer T. (0.1/1024) means that A/D precision drop due to insufficient
capacitor chage 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) = 10MHz, T=0.3µs in the A/D conversion
mode with sample & hold. Output inpedance R0 for sufficiently charging capacitor C within time T is
determined as follows.
T = 0.3µs, R = 7.8kΩ, C = 1.5pF, X = 0.1, and Y = 1024. Hence,
R0 = -
1.5X10
Thus, the allowable output impedance of the sensor circuit capable of thoroughly driving the A/D con-
verter turns out of be approximately 13.9kΩ.
Figure 14.5.1 Analog Input Pin and External Sensor Equivalent Circuit
R
e
. v
2
0 .
0
F
e
b
1 .
, 5
2
0
0
7
R
E
J
0
9
B
0
2
0
2
0 -
2
0
0
C
2 /
6
, A
M
1
6
C
2 /
6
, B
M
1
c(R0+R)
X
VC=VIN-
Y
1
T
c(R0+R)
e
=
1
-
T = ln
C(R0+R)
T
R0 = -
X
C•ln
Y
-6
0.3X10
- 7.8 X 10
0.1
-12
•ln
1024
Microcomputer
Sensor equivalent
circuit
R
0
R (7.8kΩ)
V
IN
C (1.5pF)
NOTES:
1. Reference value
page 212
f o
3
2
9
6
C
2 /
6
) T
1
t
}
X
VIN=VIN(1-
)
Y
X
Y
X
Y
- R
≅ 13.9 X 10
3
3
(1)
(1)
Sampling time
V
C
Sample-and-hold function enabled:
Sample-and-hold function disabled:
14. A/D Converter
3
φ
AD
2
φ
AD
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