# Measure ac voltage and current

Hi all,

I have a spot welder design for which I bend to measure the ac voltage of 2-3v. Since the voltage is ac should it be rectified and filtered before sending it via the ADC ? Amp is several 1000's. To measure the amps put into spot by the electrode a formula came across is menationed on another post in a forum (http://www.eevblog.com/forum/projects/measuring-3vac-amps-and-resistance/msg1301104/?PHPSESSID=f3a4mu5d3fnngb2ag5r4usfds3#msg1301104)

Here the measurement is done by tapping into two points on a copper cable in the secondary with known points apart from each other.

For the current, you can use a hall-effect current sensor, or a current transformer. But, that's a lot of current and I don't know how common high-current sensors/transformers are. (Higher currents are easier to measure than lower currents, but the components may not be as common.)

The most common way to measure current is to use a low-value resistor and measure the voltage drop across it (that's how digital multi-meters work), but I don't think that's practical in this situation.

For the AC, yes that needs to be dealt with because negative voltages (or voltages greater than +5V) can damage the Arduino. Rectification is one solution, but standard silicon diodes have about a 0.7V drop so a simple diode or bridge rectifier isn't good for low voltages. You can make a precision rectifier with an op-amp, or a full-wave precision rectifier with a couple-more op-amps. (Full-wave rectification may not be necessary, since the waveforms are approximately symmetrical.)

Another solution (common in audio applications) is to [u]bias[/u] at 2.5V. With the bias, it can handle a 5V P-P AC voltage (~ 1.75V RMS). If you need to handle a higher voltage, you can use a voltage divider (2 more resistors).

And, since you're deaing with high-power connections to the "uncontrolled outside world", it's a good idea to have an [u]over-voltage protection circuit[/u]. (And you may need that for both the voltage & current inputs, depending on what you use as a current sensor.)

well that is log of more work and expense on additional components. I'll just leave it.