Measuring AC Mains Property

[parthiban]

Before some days i planned to design a energy meter to monitor a ac main P=IV. After googled i got a circuit, the circuit works fine. But i am unable understand how the circuit woks? can anyone help to get a concept?

Vac = 250v
In that circuit after R1(100K/2W) Vdrop = 26Vac,
After D1(1n4007) Vdrop = 52dc,
after R2(1K) Vdrop = 52dc,
after R3(22K) Vdrop = 2.1dc, this voltage goes to adc of microcontroller,

How do i calculate this values by formula?

 

[hamid159]

D1 is here to rectify the ac signal...microcontroller will read the voltage across R4 w.r.t some reference ( usually VDD )..then you can multiply the ADC reading ( upto 255 for 8 bit and 1023 for 10 bit ) with some constant to get exact voltage reading...
For current measurement you can use current Transformer or linear hall sensor..then the take its output to microcontroller.Multiply these two values ( current and voltage ) to get power..simple

 

[crutschow]

That is a very dangerous circuit since it's not isolated form the AC line. It's okay if the final circuit is in an insulated case but you should use an isolation transformer when testing the circuit. Otherwise if the neutral and hot leads are accidentally reversed there will be the mains voltage on the circuit common. At a minimum you should plug the circuit only into an outlet that has a GFI (Ground Fault Interrupter).

 

[schmitt trigger]

Designing an AC energy meter without taking into consideration the power factor and waveform distortion is pointless.
You are losing both with such a simple circuit, in addition to the risks of having a non-isolated connection.

 

[crutschow]

.................................
For current measurement you can use current Transformer or linear hall sensor..then the take its output to microcontroller.Multiply these two values ( current and voltage ) to get power..simple

To get true power you need to multiply the instantaneous voltage by the instantaneous current. (Don't rectify the voltage or current measurements). This calculation automatically takes into account the power factor (phase-shift) between current and voltage. To do this you take multiple simultaneous samples per cycle of the mains frequency for both voltage and current and multiple each sample-pair together. You add up these multiplied values for one cycle and then divide by the number of samples to get true average power. The number of samples per cycle determines the highest harmonic frequency that will be included in the calculation (equal to 1/2 the sample frequency).

 

[parthiban]

ya my microcontroller reads voltage by ADC and my doubt is how 150k reduces 250Vac to 16Vac, please could you tell me?

 

[crutschow]

ya my microcontroller reads voltage by ADC and my doubt is how 150k reduces 250Vac to 16Vac, please could you tell me?

You use a voltage divider and Ohm's law. 150kΩ in series with 10.2kΩ to ground will give ≈16V (15.92V±1% with 1% resistors) at the junction of the two resistors to ground. For a value closer to 16V you could use 133kΩ in series with 9.09kΩ which gives an output of 15.99V±1% for 250V input using 1% resistors.

 

[schmitt trigger]

You use a voltage divider and Ohm's law. .

crutschow;

I had been thinking about replying a similar thing; that if one ignores basic electrical knowledge like ohm's law and its effects on dividing voltages, then one should not be attempting a non-isolated mains circuit.

 

[parthiban]

sorry for all till now i did not get answer, i asked how voltage divided across all parts and how to simplify circuit like thevenin's circuit?

 

[parthiban]

and finally i got a answer from Vrms, Vmax, Vave calculation formula as well as basic ohms law, thanks to all.