They are of little consequence, except that limit current in case of a latch-up. You can use any value from 1K to 10K (hence 4K is good value). The best accuracy is obtained when the input voltage is close to 1V. You can clamp the input voltage with a zener and work on the software.
They are acting as comparators rather than processing voltage and current magnitude.I cannot exactly understand the idea, about opamp functioning within the circuit.
Still, there aren't any rectifiers, and according to KlausST it is not a problem as long as Voutmax is withing margins.
Hi,
I`m not sure what you want to achieve.
When you use the circuit above you get two square wave signals. One for current, one for voltage.
But with this configuration you can not measure "how much" current or voltage.
But you are able to measure the timing.
Klaus
This statement is too general. And I don't think it's true here.If I`m not able to see a zero value on the multimeter, (this is the case of all digital multimeter, where when valuing an AC parameter, it cannot display its zero crossing, since its too fast to capture [50hz]), how come the ADC input of the PIC will detect a zero value of circuit's output.
No. It is likely, that this will kill your microcontroller.Shall I connect the output of the schematic, to PORTA.1 and make an if condition where when PORTA.1 is zero (low) then do whatever (like set a timer), to see if the functionality is good or not?
As described above. You may get a phase shift information, but not a true power factor information.Using this timer, I make software calculation to determine angle phi, and accordingly get the power factor.
Your schematic differs from my suggestion in post #14, it gives a large offset to the OP input which isn't useful. Please compare.
Hi,
...you don't give informations. Just new schematics and new questions. So it's hard to help.
I'll try anyway
*****
Yor new circuit has some issues:
* the output is purely on/off. No analog value. Still unclear if you want this.
* it introduces an offset, so it is impossible to switch at zero volts. Not 50% duty cycle.
* the output voltage is not compatible with your microcontroller.
*******
Doesn't the microcontroller have comparator inputs?
*******
This statement is too general. And I don't think it's true here.
I recommend you to use a scope instead of a DVM. It gives yo so much more information...
******
No. It is likely, that this will kill your microcontroller.
******
As described above. You may get a phase shift information, but not a true power factor information.
Klaus
This is possible with timing anlysis of zero cross of both U and I (wich is possible with a comparator circuit), AND you need sinuidal input waveform.all I want to achieve is a single idea, that is measuring power factor of an inductive load,
But power measuring is totally different than just power_factor measurement.because I have not well understand the concept of power measurement measuring
Seee your circuit. The OPAMP/comparator is supplied with +5V and -5V. Therefore the output voltage is +/-5V also. The -5V will kill your PIC.How come the output of the schematic is not safe to enter to PIC, but it is within the acceptable range (0;5V)?? Why would it damage the PIC?
From another part, I connected the + terminal of a dry battery cell, to both 2 and 3 inputs of LM358, and the other terminal - is branched to common GND of total circuit. This is to get a zero output since the opamp would subtract voltages, and since input voltages are the same (from same source), when i branched the two 10k resistors, I had a zero output. Isn't this a sign of correct circuit functioning?
The schematic is meaned exactly as shown. It's clear according to usual schematic drawing rules, I believe. 2x 10k is O.K.
View attachment 126341
The schematic is meaned exactly as shown. It's clear according to usual schematic drawing rules, I believe. 2x 10k is O.K.
View attachment 126341
Klaus, I apologize, it's my typing fault, I have never ment to measure Power.But the next time you say:
But power measuring is totally different than just power_factor measurement.
Power measurement is impossible with the comparator circuit.
For true power measurement you need the value of the current. I asked this, but I still don´t know the answer.
*****
Seee your circuit. The OPAMP/comparator is supplied with +5V and -5V. Therefore the output voltage is +/-5V also. The -5V will kill your PIC.
***
Klaus
As in AC voltage measurement: 3.5V
Hi,
I wonder how this is measured/calculated.
With 0V/5V and square wave I come to max 2.5V AC, rectified average as well as RMS.
Klaus
Should anything must be done??
Hi,
Maybe it is measuring AC + DC RMS.
Then it's true that a 0V/5V signal with 50% duty cycle gives 3.5V RMS.
Klaus
Hello
Is VT circuit diagram completed and ready to be entered to PIC and starr CT circuit?
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