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AMT21 数据表(PDF) 2 Page - CUI INC

部件名 AMT21
功能描述  MODULAR ABSOLUTE ENCODER
PDF  10 Pages
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制造商  CUI [CUI INC]
网页  http://www.cui.com
标志 CUI - CUI INC

AMT21 数据表(HTML) 2 Page - CUI INC

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cui.com
date 06/29/2018
│ page 2 of 10
CUI Inc
│ SERIES: AMT21 │ DESCRIPTION: MODULAR ABSOLUTE ENCODER
0x54 ReadSingle Turn Position
e
s
n
o
p
s
e
r
e
t
y
b
h
g
i
h
1
6
x
0
e
s
n
o
p
s
e
r
e
t
y
b
w
o
l
B
A
x
0
STOP
K1
K0
H5
5
STARTH2H3H4
H1
L1 L2 L3 L4 L5 L6 L7 STOP
H0
Command
67
Response
L0
0123
P
O
T
S
T
R
A
T
S
START
4
Read Single-Turn Configuration
Figure 1
AMT21 RS485 Interface
The AMT21 utilizes a half-duplex RS485 interface where the host sends a command over the differential bus at the proper
baud rate (default 2 Mbps), then the encoder takes control and responds. The host sends a single character using 8 bits, 1
stop bit, and no parity. The encoder responds with 14-bit position, and 2 check bits. It sends this as two bytes also using
8 bits, 1 stop bit, and no parity. The low byte comes first. 12-bit encoders use the same format but the lowest two bits are
zero.
Commands:
Read Single-Turn Position
Command char: T
Hex value: 0x54
RS485 INTERFACE
parameter
conditions/description
min
typ
max
units
protocol
RS485 balanced digital multi-port interface
data rate
8 data bits, no parity, 1 stop bit, asynchronous
2
4
Mbps
transceiver
Texas Instruments SN65HVD75
latency
time before encoder responds with position
18
29
µs
driver differential output
voltage magnitude
load resistance = 54 Ω
1.5
2
3.3
V
Values K1 and K0 in the response are checkbits. The checkbits are odd parity over the odd and even bits in the position
response shown in the equation below. The checkbits are not part of the position, but are used to verify its validity. The
lower 14 bits are the encoder position.
Full response: 0x61AB
14-bit position: 0x21AB (8619 decimal)
Checkbit Formula
Odd: K1 = !(H5^H3^H1^L7^L5^L3^L1)
Even: K0 = !(H4^H2^H0^L6^L4^L2^L0)
From the above response 0x61AB:
Odd: 0 = !(1^0^0^1^1^1^1) = correct
Even: 1 = !(0^0^1^0^0^0^1) = correct
For 12-bit applications L0 and L1 are always 0. Because the low two bits are 0, 12-bit data will need right-shifted two bits.
The checkbit calculation remains the same and unaffected.
For more information, please visit the product page.



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