How to calculate the resistor value.

Hi Every one.

I done Home Automation using Arduino. ATWINC1500 wifi and arduino promini used in this project.
TRIAC BTA12 used this project.( I planned snubberless TRIAC BTA12 800 BW or CW)

I done AC 230V bulb dimming using zero cross detection method(Interrupt 0 used (D2 pin))
So I just used 4N35 opto coupler to find the interrupts. So I need to reduce the voltage and current. So I used 2 watt 330Kresistor. But I get heat. what is the problem ?. I know the formula mistake(ohms law)

how to find the resistance and wattage.
Input Volt is DC 190V - 290V.(not constant this is the problem how to calculate ?).
Load 4N35.(2volt and 15mA).

I have attached the exact circuit. please find the attachment.
and how to find the resistance value ?.(If the input volt is not constant).

I think, you would read this article DIY - Isolated High Quality Mains Voltage Zero Crossing Detector. All is explained there. The circuit is little bit more complicated in compare with your but high quality in result. I am using this design for long time without any problem.

Simply calculate the resistance for 20mA with the maximum voltage which is root 2 * 290 V.

The wattage dissipated is this voltage times 0.02. The wattage rating of your resistor should be about twice this. It will get hot.

Why not use a 0.22 Y series in the 'phase' input to the bridge rectifier. With an Xc of around 14.5k at 50Hz or 12k at 60Hz, you will get voltage reduction with no heat generated. You can then use a much smaller dropper resistor to feed the opto unit.

jackrae:
Why not use a 0.22 Y series in the 'phase' input to the bridge rectifier. With an Xc of around 14.5k at 50Hz or 12k at 60Hz, you will get voltage reduction with no heat generated. You can then use a much smaller dropper resistor to feed the opto unit.

Hi, I think you will phase shift the current by 90Deg, so it will not correspond with zero voltage crossing.
You need to look here;
https://playground.arduino.cc/Main/ACPhaseControl
Tom... :slight_smile:

Hi,

You could use a small transformer to drop the mains to 5 or 10V. Its not as elegant as in post #1 but it will work.

290volts developed across 330k resistor dissipates just over 0.25 watts. So if your 2watt resistor is getting hot then it's either not a 330k or you have some incorrect wiring.

290volts developed across 330k resistor dissipates just over 0.25 watts. So if your 2watt resistor is getting hot then it's either not a 330k or you have some incorrect wiring.

Perhaps, but it depends on what "..But I get heat.." means.

I've seen resistors in that wattage range having a thermal rise of between 60 and 90 °C/Watt. Which would be a 15 to 20 °C rise. It also depends if the resistor is mounted to something or on leads in the air.

JohnRob:
Perhaps, but it depends on what "..But I get heat.." means.

I've seen resistors in that wattage range having a thermal rise of between 60 and 90 °C/Watt. Which would be a 15 to 20 °C rise. It also depends if the resistor is mounted to something or on leads in the air.

Suppose it goes to show how useless the remark "I get heat" is

Thank you for replying. I will try and test every one's idea and come back.

Thanks.