Post by thread:Measuring AC volts

Andres Tarzia wrote: > I am interested in making an AC volt meter for monitoring the 115V power > line. > But I don't know how to use the A/D converter, which only accepts DC. Andres, all you need is DC input. make AC to DC and lower the voltage . maximum AC voltage should be your maximum DC 5v or your reference voltage. example 150 v - 5v 0xFF 75 v - 2.5 v 0x7D 0 v - 0.0 v 0x00 you can use pic16c711 or 71 with 4 segment LED display it will take 12 pins. you have one input available. Keep in your mind that if you need push buttons to add in this project you can add 12 buttons without changing the chip. Andre Abelian {Quote hidden}> > > Can someone suggest a technique to do just that? > > Thank you. > > Regards, > Andres Tarzia > Technology Consultant, SMART S.A. > e-mail: .....atarziaKILLspam@spam@smart.com.ar

{Quote hidden}>Andres Tarzia wrote: > >> I am interested in making an AC volt meter for monitoring the 115V power >> line. >> But I don't know how to use the A/D converter, which only accepts DC. > > >Andres, > >all you need is DC input. make AC to DC and lower the voltage . >maximum AC voltage should be your maximum DC 5v or your >reference voltage. > >example > >150 v - 5v 0xFF >75 v - 2.5 v 0x7D >0 v - 0.0 v 0x00 > >you can use pic16c711 or 71 with 4 segment LED display it will take 12 >pins. >you have one input available. Keep in your mind that if you need >push buttons to add in this project you can add 12 buttons without >changing the chip. > >Andre Abelian > Andre, I know that I have to get DC from AC. But I don't know whether to build a half-cycle rectifier, a full-cycle rectifier or not to rectify the AC altogether. Besides that, I am not sure about adding a capacitor to "smooth" the rectified AC or to repeatedly measure the rectified sine wave to get the "peak" (maximum) and then work on that... o-----/\/\/\/\---+-----------o R1 | \ 115V AC R2 / To PIC \ / | o--+-------------+-----------o | --- - . If R2/(R1+R2) = 1/25, then I can get the mains voltage within the 0-5V range. Using the PIC built-in input protection, I can forget reverse-voltage protection diodes, and so on (see AN521, available at ftp://ftp.microchip.com/Download/Appnote/Category/16c5x/00521C.PDF). But the problem is that I still have AC. ----------- o-----/\/\/\/\---+-------|~ | R1 | | +|------+---------o \ | Full | | 115V AC R2 / | Wave | --- C1 To PIC \ | Rectifier | --- / | | | | | -|------+---------o o----------------+-------|~ | ----------- Using the above circuit, with the same R1 and R2 as before I now have DC, but some issues remain. 1) I have to take into account the 0.7V drop from the rectifier. 2) What am I measuring? I mean 115V/25 = 4.6V, then 3.9V will be feed to the PIC. If the converter shows 3.9V, should I display 115V? I always confuse "peak" voltage and "mean" voltage and I always forget to which one "115V" refers... 3) Should I include C1 in the circuit above? Which capacitor is best suited to the task? A large electrolytic cap? Or a small one? Again, thank you for your help Regards, Andres Tarzia Technology Consultant, SMART S.A. e-mail: atarziaKILLspamsmart.com.ar

A few comments... 1. Watch out for line voltage isolation. Try not to electrocute people. 2. If you decide to rectify the AC, do it prior to a voltage divider so the diode knee voltage is a small portion of the AC, minimizing nonlinearity due to the knee. 3. You can rectify and drive a low pass filter (RC network) to get the average voltage, then correct to RMS assuming the voltage is a sine wave. 4. You can put an RMS to DC converter in front of the PIC. 5. You can have the PIC calculate the RMS. Of course, the PIC A/D only accepts 0..5 volts, so we're tempted to rectify the incoming AC prior to driving the PIC. However, we then have possible nonlinearity problems (which, as above, can be minimized by rectifying prior to voltage division). However, as someone cleverly pointed out on this thread, we could just apply the AC (voltage divided and current limited) to the PIC pin. The protection diode would remove the negative half cycle. We would then be able to A/D the positive half cycle. The RMS can be calculated for the positive half cycles, then doubled, assuming the AC is symmetrical. To calculate RMS, take a sample, square it, add to an accumulator. After some number of samples (like maybe 256), divide by the number of samples (256 is nice, since you can just throw out the low half of a 16 bit value), then take the square root. RMS is the square root of the mean (average) of the squares. Harold Harold Hallikainen .....haroldKILLspam.....hallikainen.com Hallikainen & Friends, Inc. See the FCC Rules at http://hallikainen.com/FccRules and comments filed in LPFM proceeding at http://hallikainen.com/lpfm ___________________________________________________________________ You don't need to buy Internet access to use free Internet e-mail. Get completely free e-mail from Juno at http://www.juno.com/getjuno.html or call Juno at (800) 654-JUNO [654-5866]

On Mon, 8 Feb 1999, Andres Tarzia wrote: {Quote hidden}> Andre, > > I know that I have to get DC from AC. > > But I don't know whether to build a half-cycle rectifier, a full-cycle > rectifier or not to rectify the AC altogether. Besides that, I am not sure > about adding a capacitor to "smooth" the rectified AC or to repeatedly > measure the rectified sine wave to get the "peak" (maximum) and then work on > that... > > o-----/\/\/\/\---+-----------o > R1 | > \ > 115V AC R2 / To PIC > \ > / > | > o--+-------------+-----------o > | > --- > - > . Since no one has suggested, I'll | in... If you add a "pull-up" resistor between the pic I/O pin and say +5V, then you can add a DC offset and keep the signal from swinging below 0V. This will allow you to sample the complete waveform to obtain a more accurate RMS calculation. However, if your voltage signal is fairly clean and you feel comfortable capturing the peak and approximating the RMS voltage, then I'd suggest building a circuit that would capture peaks in the range between low and high line (typically + and - 15% of 115Vrms) and scale that to a 0 to 5V range for the A/D converter. Might as well use the A/D resolution only where it's needed.

Harold, Following your suggestions, what about the following circuit: o---|>|-----/\/\/\/\-----*----------*----------*-------o Diode R1 | | | | | | \ | | / | + | 115V AC R2 \ C1 ===== C2 ===== To PIC / | | \ | | | | | Fuse R3 | | | o----~------/\/\/\/\-----*----------*----------*-------o GND Where the resistors are in the order (not EXACTLY equal to) of: R1=R3=5M and R2=400K. So we will get a voltage drop of aprox 4.5V on R2 with an input voltage of 115V (full 5V with 130V input). That way the circuit common ground will not be tied to the power line, but isolated by a 5M resistor. That will take care of the isolation. Shifting the diode in front of the voltage divider will cause the 0.7V drop to be subtracted from 115V instead of 5V, so impacting the measurements much less. Using a large electrolytic cap for C1 will get rectified DC. Adding a low value ceramic cap (C2) will filter any high freq noise that can be present on the power line. Any further suggestions? Regards, Andres Tarzia Technology Consultant, SMART S.A. e-mail: EraseMEatarziaspam_OUTTakeThisOuTsmart.com.ar {Original Message removed}