D M D - 4 8 0 - S 2 S E R I E S A C P O W E R E D B R I D G E S E N S O R - R S 2 3 2 I N T E R F A C E
7
One Omega Drive, P.O. Box 4047
Stamford, CT 06907
http://www.omega.com
1. APPLY 3 mV TO THE INPUT AND READ THE 10%
DIGITAL COUNTS AS THE OUTPUT (K1).
2. APPLY 30 mV TO THE INPUT AND READ THE FULL SCALE
DIGITAL COUNTS AS THE OUTPUT (K2)
3. WRITE TWO EQUATIONS:
A)
(K2 x A) + C = 0.030 VOLTS
B)
(K1 x A) + C = 0.003 VOLTS
SUBTRACTING EQUATION B FROM A:
(K2 - K1) x A = 0.027 VOLTS
THEREFORE
A = 0.027 VOLTS
÷÷÷÷÷
(K2 - K1)
SOLVING FOR
C , C = -K1 x A
EXAMPLE 1:
WITH 3 mV APPLIED THE DMD-483-S2 OUTPUT
READS 4185 COUNTS.
WITH 30 mV APPLIED THE DMD-483-S2 OUTPUT
READS 41,745 COUNTS.
A)
(41,745 x A) + C = 0.030 VOLTS
B)
(4185 x A) + C = 0.003 VOLTS
SOLVING EQUATION A - EQUATION B
(37,560 x A) = 0.027 VOLTS. THEREFORE
A = (0.027
÷
37,560) = 7.1885 x 10
-7
= .00000071885
SOLVING FOR C, C = .003 - (K1 x A)
C = -8.38 x 10
-6
= .00000838
VERIFY YOUR RESULTS BY PLACING YOUR SOLUTIONS BACK INTO
EQUATIONS A AND B:
A)
(41,745 x .00000071885) - .00000838 = 0.030 VOLTS
B)
(4185 x .00000071885) - .00000838 = 0.003 VOLTS
IN A WEIGHING APPLICATION USING A LOAD CELL WITH A SENSITIVITY OF 3mV/V,
THE CONSTANT C WOULD THEN CORRESPOND TO THE TARE WEIGHT OF THE
SYSTEM.
EXAMPLE 2:
WITH THE TARE WEIGHT ONLY APPLIED, READ
1,500 COUNTS.
WITH A FULL LOAD OF 100 POUNDS APPLIED, READ ABOUT
41,745 COUNTS.
A)
(41,745 x A) + C = 100 POUNDS
B)
(1,500 x A) + C = 0 POUNDS
