# Digital Volt and Amp Meter

Posted on Oct 8, 2012

The PIC Microchip Processor must be programmed before it will function as a Volt & Amp meter. There are many internet sites and PIC programmers that you can use. I used a Microchip MPLAB ICD 2 during the project. You might need to made changes to the circuit to accommodate a different type of programmer, do read the programmers instructions carefully. The circuit relies on the internal analogue to digital converter (ADC) of the PIC Microchip Processor. The accuracy is dependant on scaling the input voltage for the ADC for all three measurements. The good news is that both the PIC's which can be used for this project have 10-Bit resolution ADC units which should work adequately in most circumstances.

In order to determine the resolution, simple to advanced mathematics can be used - I will use simple mathematics and present a basic explanation in order for you to get going on the project. R17 and VR1 form a voltage divider as input to the ADC Input of the PIC. As the input voltage changes, so will the output of the Voltage divider on PIN 3 of the PIC. When calibrating the project for use it is important to remove the PIC from the circuit and adjust the Value of VR1 in such a manner that the output of the Voltage divider is never greater than 5.0V. Failing to do so might damage the PIC. The current measurement is more complex and involves an Op-Amp configured as an inverting amplifier to provide the input for the PIC. The resistor R7, in the Power Supply circuit is used as a shunt resistor. The small voltage drop across the resistor varies according to the amount of current a given load will draw from the PSU. In order to measure with greater accuracy the small voltage drop is amplified using an Op-Amp circuit. Using the formula for an inverting amplifier the output voltage of the Op-Amp can be calculated as follows: 1. The maximum current through R7 maybe 3.0A 2. The voltage drop over R7 = V = I * R = 3.0A * 0.47R = 1.41V 3. Op-Amp output = Vout = -(R12/R16)(Vin) = -(33K/10K)/1.41V = -4.653V. However, the input is a negative voltage and if we negate the answer we should measure close to 4.653V for the maximum...

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