 # How to measure Instantaneous Current and Voltage in AC system

Hi frndz..!

I am working on a project in which i need to calculate instantaneous values of current and voltage.

For Current: If i use the CT, its secondary will provide AC current, and due to the negative cycle my ADC would not detect that negative voltage. If after CT i use bridge rectifier, then i will get DC value from which i can only calculate RMS value, but i need instantaneous values.

For Voltage: If i use step down transformer, nd then use bridge rectifer again i would get DC and i can calculate instantaneous value again.

If i clamp the voltage to bring all the voltages to positive level, then ADC would work but i would need another small voltage source, which voltage may not remain constant when talking about months nd so ..

Plz provide me a circuit diagram for this. THanks

They make High Voltage monitors for uCs.

Industry applications Isolated current and voltage measurement with instantaneous current or voltage output. Non contact or with integrated conductor or clip-on current measurement from 0.1 A to 20,000 A. High voltage measurement from 10 V up to 6,400 V. Different technologies used to cover required performances, such as accuracy from some ppm to several percents, or response time, bandwidth.

I drew this up for use with a DC motor being used as a generator, where he also needed to measure voltage when the motor is turning backwards and therefore generating a negative voltage.

Insert the current transformer in place of the motor. You may want to change that 0.1uF cap across it into a 1nF instead. Select RL so that the maximum peak voltage is never greater than 2.5V.

Oh, and connect A1 to the junction of the 10k resistors so you can measure that as a reference point.

I don't see anything powering the motor. How does that work ?

I drew this up for use with a DC motor being used as a generator...

In another thread, a gentleman's son is doing a science fair project, I think it was, using a DC motor as a wind generator.

Rather than rectify (which is distinctly non-linear in response and could lead to false readings), just offset the output of the transformer to mid-rail. The simplest circuit is string two 10k resistors as a divider between gnd/Vcc and take the midpoint of the divider to one side of the transformer output and the other to an analog pin.

You can then directly read a signed value (once you've calibrated the mid-point reading) for each of voltage and current, giving you the ability to measure instaneously power and direction too.

Of course I haven't addressed issue of scaling the voltage, but that may just be another divider on the output of the relevant transformer.

Obligatory warning - never operate a current transformer without its burden resistor, they could generate extremely high voltages. Many CT units have built-in burden resistors so they are nice and safe, but check...

Um... MarkT, that is precisely what my circuit does.

Assuming that motor is being used as a generator and you are measuring the generator output, then the voltage measured at A0
will have to take into account the voltage (2.5V) at the midpoint of the voltage divider and the voltage measured should subtract the 2.5V . So if it generates 0V , (motionless), then you would read 2.5V with a series 1 k ohm resistor.If it generated 1V, it would either ADD or SUBTRACT from 2.5V , depending on the direction of rotation. At any point in time , there is at least 10 k + 1 k ohm series resistance , limiting the current. The title of the post is measuring voltage and current. Any attempt to measure the current will need to take into account those resistances. For example , your initial measurement when it is stopped, is your reference voltage. If it generates -2.5V then (2.5V-2.5V) = 0V.

He is using a current transformer. The voltage generated is an AC 50 or 60Hz sine wave directly proportional to the current being drawn. That is what that circuit is for. It simply sets the midpoint at 2.5V, so the positive peaks from the CT (current transformer) are added to 2.5V, and negative peaks added (adding a negative) to 2.5V. RL selected so the peaks never go over ±2.5V. The 1k resistor is there to protect the internal diodes in case the sum of voltage goes higher than 5V or below 0V.

http://sound.westhost.com/xfmr2.htm#s17

The input resistance of the Arduino is on the order of 100M:

http://forum.arduino.cc/index.php?topic=65134.0

So this is in series with the 5k thevenin resistance of 10k//10k and the 1k series. So those resistances will have infinitesimal effect on the signal voltage.

I am addressing the need to measure current.

The same circuit can be used to measure voltage. Merely insert the secondary of an AC stepdown transformer in there, one with an output voltage of no more than 2.5V peak at the maximum AC primary voltage.

This will also give you pretty accurate phase information, and everything is isolated.

polymorph: Um... MarkT, that is precisely what my circuit does.

Ah, yes!

Treat it as a vote of confidence.

Useful input.

Sometimes it seems like my answers get skipped over, and someone who says the same thing in a later reply gets a "Thanks!" and I get ... nothing. Value for dollar, I suppose.

It's a conspiracy.

I -knew- it!