optobreakout board design help needed

[visky]

Hello!

I am designing optobreakout board .
On the outputs i will have 5Volts on the led diode.
For the inputs i will have 25.5Volts on the led diode.
I don't understand what exactly mean reverse voltage in the datasheet for quad optocouplers. But they are low,up to 5Volts max.

Can i use those optocouplers? and how to wire anode and catode with 25.5Votls?

For output i will have 5V on anode and catode to ground.

What about for inputs:

1option

25.5Volts/2.5mA on anode and catode to ground.

2option

5Volts on anode and catode 25.5Volts/2.5mA ->GND (on/off)

[irving2008]

Hmmm, where to start... not knowing how much electronics you know it is hard to know...

Lets start with basics... LEDs are just like other diodes. When there is sufficient voltage across the LED of more than 'Vf' volts, typically 1.3v or so, the LED comes on and forward current 'If' flows.. To operate an LED from a supply voltage Vs the current must be limited by a series resistor which can be calculated as (Vs - Vf)/If However the situation driving out from a parallel port isn't so simple as a parallel port doesn't put out 5v. In practice it puts out typically 3.3v at a relatively low current. It is usual therefore to use the parallel port output to drive the LED on when it in the LOW state and power the LED from an external 5v supply (taken from a USB port or game port, from a spare hard disc connector or use an external 5v wall-wart supply with an isolated output). The other advantage of doing this is to keep the logic polarity the same (LOW in = LOW out)

So what value to use for 'If'? well the recommended If is 16mA for best switching times as that guarantees to saturate the output transistor under all conditions but you can go lower if you know what collector current you plan to use. The transistor needs a collector load resistor to a supply voltage common to the system being driven (typically 5 -> 12v). The load resistor is calculated from the supply volts Vs and the saturated output transistor voltage (0.4v) and the collector current, say 5mA. For a 12v supply the resistor would be (12 - 0.4)/.005 = 2320ohm (use 2k2), for a 5v supply its (5 - 0.4)/.005 = 920ohm (use 910R or 1k). At 5mA collector current the LED forward current needs to be at least 8mA to saturate the transistor.

So we can calculate the LED resistor as (5 - 1.3)/.008 = 462ohm (use 390ohm).

For the input side the LED resistor is calculated in the same way (25.5 - 1.3)/.008 = 3025ohm (use 2k7). You can use the 1k collector resistor as before on the port side.

The LED reverse voltage is the maximum voltage the LED can stand before it breaks down and starts conducting and potentially destroying itself. Since your circuit is all DC and never going to put reverse voltages on the LEDs you don't need to worry about this.

The only other issue is that the devices you have selected are relatively low speed. At the values given above your maximum step rate will be around 15000steps/sec which is probably OK for most purposes. Most commercial Opto-breakout boards use HCPL2630 devices for the outputs and the lower speed devices for the inputs.

Hope this helps