AutomationDirect DL205 - Module 12 Bit Input Resolution; Module 14 Bit Input Resolution; Module 16 Bit Input Resolution

DL205 Analog Manual, 7th Edition Rev. D
15-20
Chapter 15: F2-8AD4DA-1, 8-Ch. In / 4-Ch. Out Analog Current Combination
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Module 12 Bit Input Resolution
When the 0–20 mA module inputs are
configured for 12 bit resolution, the analog
signal is converted into 4096 (2
12
) counts
ranging from 0–4095. For example, a 0mA
signal would be 0, and a 20mA signal would
be 4095. This is equivalent to a binary value
of 0000 0000 0000 to 1111 1111 1111, or
000 to FFF hexadecimal. The diagram shows
how this relates to the signal range.
Each count can also be expressed in terms of
the signal level by using the equation shown.
Module 14 Bit Input Resolution
When the 0–20 mA module inputs are
configured for 14 bit resolution, the analog
signal is converted into 16384 (2
14
) counts
ranging from 0–16383. For example, a 0mA
signal would be 0, and a 20mA signal would
be 16383. This is equivalent to a binary
value of 00 0000 0000 0000 to 11 1111
1111 1111, or 0000 to 3FFF hexadecimal.
The diagram shows how this relates to the
signal range.
Each count can also be expressed in terms of
the signal level by using the equation shown.
Module 16 Bit Input Resolution
When the 0–20 mA module inputs are
configured for 16 bit resolution, the analog
signal is converted into 65536 (2
16
) counts
ranging from 0–65535. For example, a 0mA
signal would be 0, and a 20mA signal would
be 65535. This is equivalent to a binary value
of 0000 0000 0000 0000 to 1111 1111 1111
1111, or 0000 to FFFF hexadecimal. The
diagram shows how this relates to the signal
range.
Each count can also be expressed in terms of
the signal level by using the equation shown.
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20mA
0mA
0 4095
12 BitResolution =
H − L
4095
H=high limitofthe signal range
L=low limit of thesignal range
20mA/4095 =4.88Apercount
12 Bit Resolution
μ
0–20 mA
20mA
0mA
0-- 20 mA
14 BitResolution
=
H − L
16383
H=high limitofthe signal range
L=low limit of thesignal range
Aper count
14 Bit Resolution
μ
0 16383
20mA / 16383 = 1.22
20mA
0mA
0-- 20 mA
16 BitResolution
=
H − L
65535
H=high limitofthe signal range
L=low limit of thesignal range
Aper count
16 Bit Resolution
μ
0 65535
20mA / 65535 = 0.305