I am struggling to understand datasheets for optocouplers to find one suitable for this project I have.
I am wanting to use an arduino to switch a signal line on an appliance I have. I have tried creating this circuit with a 5vDCcontrol/AC load relay but it wrecks havoc on my arduino causing it to freeze or reset itself and I learned an AC load relay is not suitable for DC loads.
After some reading I learned I should be using an optocoupler but i cant figure out what i should be looking for in terms of spec. the datasheets mention forward voltage, reverse voltage and other things but Im struggling to interpret this info and know what to do with my 5v Arduino circuit to successful operate the optocoupler.
in the circuit below I have used the VO617A-3X017T optoisolator because the seem highly available on JLCPCB.
To summarise- what should I be looking for in optocoupler datasheets that will allow me to know its suitable for a 5v arduino circuit so I can switch a 5vDC 1milliamp line on the appliance?
What does that mean, exactly? It is extremely important to know the detailed characteristics of the control input.
Can you post a link to or quote the relevant section of a document that describes the characteristics of this control input?
You MUST have a current limiting resistor in series with the input LED of an optocoupler to protect both the Arduino output and the LED. The maximum LED current is specified is specified in the data sheet. 220 Ohms for a 5V Arduino is usually the minimum.
You need to share a bit more information about the appliance. What is the voltage? What is the amperage? What is the wattage?
An optocoupler generally is 2 diodes. one wired to emit light and one wired to detect light...
I might recommend a solid-state relay, if I new what the power-handling requirement was, or a electromechanical relay with a 5-volt coil, again if I knew what the contacts should be rated for.
Your schematic does not have a resistor in series with the Arduino and the Opto input diode. You must have a diode else your Arduino will be very very unhappy.
Consider the input to be essentially and LED. Same rules apply.
So your appliance requires you to "close" the connection on a 5V system where when closed the current is 1 milliamp.
In your suggested PN the "3" after the dash matches the "bin" in the base part. The above table from the mfg data sheets puts the current transfer ratio at between 100% and 200%
The current transfer ratio is defined as the ratio of the current that can flow through the output transistor by the current going through the input diode.
For a 1 ma collector current you need a minimum of 1 ma in the input diode, assuming the min CTR of 100% for this device.
That's all well and good but the datasheet shows the "edges" or limits of the device. So you want to have a safety margin of at least 2. Think of a rope needing to hold 100 lbs, you would not use a rope specified for 100 lbs. Same goes here so you want the input current to be at least 2ma.
Now you might ask but won't this draw more than the appliance will generate? No the output transistor will be "Capable" of dealing with more than 1 ma but it will not require more than 1 ma.
