Explain how a voltage divider scales a sensor signal for a PLC input.

Prepare for the OCC SACA Sensor Logic Systems 1 (C-205) Exam. Study with detailed questions and insightful explanations. Get ready for your certification!

Multiple Choice

Explain how a voltage divider scales a sensor signal for a PLC input.

Explanation:
The idea behind using a voltage divider is to attenuate a sensor’s voltage to a level the PLC can safely read. It uses two resistors in series between the sensor output and ground, and the junction between them provides a fraction of the sensor voltage. That fraction is set by the resistor ratio, giving Vout = Vin × (R2/(R1+R2)). This lets a higher sensor voltage be scaled down to the PLC’s input range (for example, 0–5 V or 0–10 V) without amplifying the signal. It’s a passive attenuation, so it won’t store energy or actively filter EMI by itself; if EMI filtering or energy storage is needed, separate components would be added. The PLC input’s impedance matters: if the input draws significant current or loads the divider, the ratio can shift, so you choose resistor values that keep the divider’s loading negligible or add buffering if needed.

The idea behind using a voltage divider is to attenuate a sensor’s voltage to a level the PLC can safely read. It uses two resistors in series between the sensor output and ground, and the junction between them provides a fraction of the sensor voltage. That fraction is set by the resistor ratio, giving Vout = Vin × (R2/(R1+R2)). This lets a higher sensor voltage be scaled down to the PLC’s input range (for example, 0–5 V or 0–10 V) without amplifying the signal. It’s a passive attenuation, so it won’t store energy or actively filter EMI by itself; if EMI filtering or energy storage is needed, separate components would be added. The PLC input’s impedance matters: if the input draws significant current or loads the divider, the ratio can shift, so you choose resistor values that keep the divider’s loading negligible or add buffering if needed.

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