I want to implement a alert system using a comparator circuit. The circuit will alert the user when he is a certain distance away from an object of interest. The comparator circuit will comprise of an op-amp powered with only a +5V voltage supply. There will be a Infra-Red sensor to sense the distance. The idea is to turn on a vibration motor only when the user is 2 cm or closer to an object. The output voltage of the IR sensor is proportional to the distance of the object it is sensing. This output can, thus, simply be compared against a voltage of interest to produce an output voltage.
simulate this circuit – Schematic created using CircuitLab
As shown in the circuit above, a 5 V voltage supply will be used to power the IR sensor, vibration motor, and the operational amplifier. The voltage output of the IR sensor is proportional to the distance to the object it is sensing, and varies from 2.65 V to 0.4 V as the distance changes from 2 cm to 28 cm. This voltage is fed to the non-inverting output of the op-amp. The inverting output is connected to one terminal of the potentiometer. The knob of the potentiometer can be adjusted such that the input to the inverting input varies from 2.65 V to 0.4 V. If, suppose, the input to the inverting input is held at 2.65 V, then the output from the op-amp is 5 V (the positive voltage rail) only for distances the IR sensor detects to be less than 2 cm. If the distance is more than 2 cm, the input voltage at the non-inverting output will be less than 2.65 V, which means it will be lower than the 2.65 V voltage at the inverting output. This would normally force the output to saturate at the negative power supply voltage, which is simply not present for our circuit. This means the vibration motor will turn off for distances more than 2 cm.
Total Power consumption Estimation
- Op-Amp input ~0
- Op-Amp output/Vibration motor: 5 V x 75 mA
- IR sensor: 5 V x 40 mA
My question is whether the idea of using a single voltage source, and estimates of power correct?