* it seems to be safe in the meaning of safety. It provides galvanuc isolation between mains and PIC supply.My first question is, is that circuit (img1) safe to use? I didn't see any bridge rectifiers, and signals are assumed to still be AC, so how come they can be fed into the PIC? Plus only in AC zero crossing takes place, how come in DC (after supposed rectification that must be done) we can know that the AC version of the signal crossed zero?
As said before the circuit is not good. CT's need a load resistor to calculate the output voltage.Also if the circuit is safe, and ideal to be practiced, what is the current transformer ratings?
This is a typical question that is answered in the PIC datasheet. Read it.what is the maximum input current to the PIC analogue pin?
* it seems to be safe in the meaning of safety. It provides galvanuc isolation between mains and PIC supply.
* from function side, i don't like the circuit. Missing biasing resistors, i'd use a comparator instead an opamp.
* a bridge rectifier is not necessary.
* the presence of AC is no problem, as long as the voltage fed to the PIC is within the PIC's io specification.
As said before the circuit is not good. CT's need a load resistor to calculate the output voltage.
This is a typical question that is answered in the PIC datasheet. Read it.
Klaus
Yes.Do you have a better circuit diagram? The burden resistor is branched along the CT two terminals?
You don't give current to the PIC, but voltage. Intput current to PIC pin is near zero.PIC's maximum input current is of 20mA
Some problems of the circuit have been already mentioned like missing CT termination. The circuit misses also a correct OP input bias, and the LM358 supply voltage must be either limited to 5V or a clamp circuit is needed to protect the OP inputs.
Yes OP means operational amplifier.
The circuit has at least these problems:
- Missing current transformer termination
- OP inputs are floating
- The OP output voltage with about 35 mA short circuit current will damage the microcontroller inputs
Of course processing voltage and current polarity is much simpler than multiplying and integrating voltage and current adc samples as a full featured power meter does.
The thing to remember:Could you please provide resistors value? Cannot know how, and according to what KlausST has earlier said, that we cannot input a current to PIC, so I assume that the LM358, makes a source transformation of current, to input a voltage to PIC?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?