Hi the EPE mag isssue is Feb 2005. But i found something better(dont know its authenticity) its supposed to have all projects published in EPE since 1995.
thanks for you all
i am working on graduation project and calculating power is a part of it .
i need to connect isolator to protect PIC from high voltage.
i have question about how to connect isolator to protect PIC from high voltage.
i used 4n26 IC but it doesn't work ...
i used a voltage divider to minimize voltage and shunt resistor to sense current
help please
can i use a ordinary resistor instead of a shunt resistor in measuring current?? there's no available shunt resistor at our place so instead of using shunt any other way or do you know other circuit regarding my problem?
Well, a shunt resistor typically MEANS it's a low ohms resistor, also usually means high precision and high power rating. So NO, you cannot replace it with a non-shunt resistor unless you put a lot of them in parallel.
A "normal" 100ohm resistor with 1 amp pushed through it drops 100V and 100W. Nowhere near usable.
If it's 30ma you don't need a 'shunt' Many electronics devices I've pulled apart have 1ohm or 10ohm resistors used as shunts, as the current is so low modest power loss isn't an issue as long as it's a plug in device. High precision is generally speaking not an issue as it'll have to be calibrated regardless. That being said lower values of resistance mean lower power dissipation in it which means better stability. You could if you wanted and had a decent low noise opamp simply tap any two slightly different points of the power leading into a circuit and use that has a shunt, the current is already going through that trace, if you calibrate it there's no reason to use anything special.
I need to measure 220 V and currents around 13 Amperes using PIC micro-controller. The circuit I have designed on Proteus using differential amplifier that converts 220 Volts to 2.5 volts (0-5 volts range) taking 40 samples of the peak voltage is given below. The ADC of PIC is used for converting analog voltages into digital. The voltage divider divides the voltages into half so that it comes in the range of (0-5 volts) which can be read by PIC. It's showing the values of voltage and current on the input side accurately on LCD but I have some confusions regarding this circuit.
1. First of all is this circuit okay as I have some doubts about it?
2. Should I use a current transformer and voltage transformer on the input side?
3. The 1 ohm bypass resistor should be used if I use a current transformer or should I use only high power wire wound resistor only instead of CT for current measurement?
(I'm a student so I don't have much experience in this area. So kindly explain everything)