50A Current Sensor(SKU:SEN0098) Contents 1 Introduction 2 Specifications 3 PinOut 4 Tutorial 4.1 Connection Diagram 4.2 DC load 4.2.1 Sketch 4.2.2 Result 4.3 AC Load 4.3.1 Sketch 4.4 Result of AC current 5 Trouble Shooting...
Introduction This is a breakout board for the fully integrated Hall Effect based linear ACS758 current sensor. The sensor gives precise current measurement for both AC and DC signals.The thickness of the copper conductor allows survival of the device at high overcurrent conditions. The ACS758 outputs an analog voltage output signal that varies linearly with sensed current.
Tutorial This tutorial is going to test DC and AC current. The result of AC current is its effective value, i.e. the same as an Ammeter reading. Connection Diagram DANGER: This diagram can also be applied to AC current connection. But notice that the module should be cascaded on one electric wire ( +/- ).
Smoothing algorithm (http://www.arduino.cc/en/Tutorial/Smoothing) is used to make the outputting current value more reliable; Created 27 December 2011 @Barry Machine @www.dfrobot.com const int numReadings = 30; float readings[numReadings];...
readings[index] = (readings[index]-512)*5/1024/0.04-0.04; total= total + readings[index]; index = index + 1; if (index >= numReadings) index = 0; average = total/numReadings; //Smoothing algorithm (http://www.arduino. cc/en/Tutorial/Smoothing) currentValue= average; Serial.println(currentValue); delay(10); Result After uploading the sample sketch, open the serial monitor, when the current is 0A, you may found the reading is not 0A, then if you need amore ccurate readings, you have to revise the code with the reading when the input current is 0A.These are the steps: 1 Check the initial current value when the input is 0A;...
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2 Caculate the initial current value: -0.34 = -0.04 - ( +0.30 ); 3 Revise the sample sketch: readings[index] = (readings[index]-510)*5/1024/0.04-0.34;//...
4 Upload your new sketch again, then you will found the readinga are around 0A, if not, please do the steps above several times to make it. 5 As shown above, it can be applied to detect the current value accurately, and output analog voltage signals.
4 void setup(){ Serial.begin(115200); 7 void loop() { reading = analogRead(*); //Raw data reading currentValue = (reading - 510) * 5 / 1024 / 0.04 - 0.34; Serial.println(currentValue); delay(2); 12 } Result of AC current I tested with a lamp @220V~, and got the result shown below: Read more about the experiment.
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