Opto isolator question

[Doddy]

Your thinking is broadly correct, but you don't understand Opto-Isolators. Your chance of success lies with the type of illumination provided by the switches.

In operation, the internal LED translates current to light, and the internal transistor translates light to (essentially) current. The device has a characteristic called Current Transfer Ratio - which for an EL817 (cheap) device (and given the markings on the board a likely candidate for the devices you have) is specified between 50% to 600% (the variance is depending on manufacturing deviations, temperature, etc). Let's imagine a CTR of 200% - not unreasonable. Now, examine the resistor on the input to the Opto Isolator on the driver board, and what looks like a series-connected LED for status-indication. The resistor is likely to be rated to provide a forward current through the Opto Isolator between 1mA and 5mA (once upon a time likely higher, but modern LEDs have better efficiency and the trend is to reduce the forward current) - Let's say 2mA. With a CTR of 200% that means you can only expect 2mA x 200%. = 4mA through the collector/emitter junction of the transistor in the opto-isolator. Now, if your 24V panel indicator/switch is an incandescent lamp I'd expect the operating current to be around 50mA or so - so you're not going to get the required current to illuminate the switch. Additionally, there is a series resistor on the collector to the opto-isolator which is going to limit the maximum current available through the device.

You've a chance of this working if your indicators are, themselves, LED indicators, as these are likely to operate with the relatively low current provided by the device.

A better solution is to wire up any number of NPN transistors, rather than use the opto-isolator board, or even use these after the opto-isolator board if you actually require galvanic isolation (unlikely) - happy to explain further if you want to pursue this option.


***FURTHER INFO***

An image of this board from eBay shows the LED current-limiting resistor is 1k, in series with the on-board status LED, so you have the forward voltage drop across both devices (call this 3.6V), so if driving from 5V (I'll be generous and ignore the drop across the driving device) that gives 1.4V through 1k resistor, or 1.4mA forward current. Now, on the collector there's a 3k3 resistor - which means that at 24V across the C-E on the isolator's transistor, you're limited to 24/3300 = 7mA, before considering the CTR of the device. Realistically, given the LED forward current limiting I'd work on switching no more than 1mA reliably.

*** FURTHER FURTHER INFO ***

I'm concerned of your wiring in the image, and I'd agree with VoiceCoil's post - If the link-options (red jumpers) provide a shorting link from the opto isolator's emitters to the common ground from the input-side, then you need to be wiring your indictors to the upper terminal of each channel - the otto-isolators will be switching the ground level - so the other side of the indicators would be wired to +24V. But for the reasons above this is still unlikely to work.

Mike