B&B Electronics 232OPSDA manual To 10VDC A/D Input, 20mA Current Loop A/D Input

inverting amplifier, remove R15 and calculate values for R13 and R14 using the equation below.

Gain = VinV0 = 1+ R13R14

NOTE: V0 is the voltage read by the A/D converter chip, and Vin is the voltage at TB1. R13 and R14 should be chosen so that V0 does not exceed 5.00VDC.

0 to 10VDC A/D Input

The 232OPSDA contains one A/D input than is capable of handling voltages between 0VDC and 10VDC. This channel is A/D 3 and is located on Terminal Block 3. The gain of the signal conditioning circuitry for this channel is 0.5. If 10VDC is applied to Terminal Block 3, the A/D Converter chip will read 5.00V. The input resistance of this channel is 200KΩ , so the driving source impedance should be less than 1KΩ to minimize voltage division error.

4-20mA Current Loop A/D Input

The 232OPSDA has one A/D channel capable of monitoring the loop current in a 4-20mA analog current loop. See figure B.1 in Appendix B for a circuit schematic. A 10Ω resistor is connected between TB 0 and GND inside the 232OPSDA. The voltage drop across this resistor is proportional to the current in the current loop. With the original configuration, the following equation can be used to convert the voltage read by the A/D converter chip to the actual current in the loop. The value 23.064 is the gain of the signal conditioning circuitry

( )		1000 ×	AD0
LoopCurrent mA	=
		23.064	× 10Ω

NOTE: AD0 is the voltage read by the A/D converter chip. This voltage is between 0 and 5.00VDC.

16Documentation Number 232OPSDA1397 Manual

B&B Electronics -- PO Box 1040 -- Ottawa, IL 61350

PH (815) 433-5100 -- FAX (815) 433-5105