Using a Optocoupler - Determining Parameters

Question

I'm looking at possibly using an optocoupler to drive a higher powered circuit from a lower powered one.

This is the device I am looking at: http://www.vishay.com/docs/83641/il755.pdf

The circuit I am switching is a 30V @ .5A circuit. Am I reading this correctly?

• V_CEO = 60V, meaning the circuit on the output side can handle a voltage of up to 60 volts

• V_F = 1.2, meaning, if I apply 1.2 volts, the output should be fully on.

• How much current will it draw? Under the forward voltage condition of 1.2 V it pulls 10mA.

• Where in the data sheet does it state the max current the switching side of the coupler can handle?

So then, if I supply 5V to this thing, how much resistance to I need to put in front of it to get the required 1.2 V? (5 - 1.2)/.01 = 380 ohms.

Am I reading this all correctly? Especially with regard to the last point. Is it suffient to place a resistor in front of it to induce a voltage drop of 3.8V (leaving the 1.2 to supply to the transistor)? Or do I need a dedicated supply like a LDO?

Answer 1

380 ohms in series with the input, driven from 5v, will be just fine for getting approximately 10mA into the input LED.

The min CTR (current transfer ratio) is 750%, so with 10mA in the LED, you could expect to sink at least 75mA in the output. With a VCEsat of 1v, that's an output dissipation of around 75mW, which with 10(ish)mW on the input LED is within the package limit of 200mW. However, that limit is at 25C, it will drop with temperature.

The data sheet doesn't give a maximum output current anywhere. In figure 4, the curves only go to 40(ish)mA and are curving quite quickly, so that is a hint at what the manufacturer's expect. The package dissipation limit will give you a bound on the output current, much above 75mA as calculated above will be getting close to that at an elevated package temperature.

Answer 2

Simply treat this opto-Darlington as a transistor switch with isolated returns.

The current gain or CTR is ~10 @ 1mA = 10mA out
and 7.5min ~2mA or 15mA out.
So you need another transistor switch on the output = "open collector PNP switch" Inverting switches at saturation (Vce=Vce(sat) are rated a current gain of 10 (min).

Thus each stage is 1 mA (hi) -- 10mA (low) -- 500mA (hi) Since last ratio =50 , you must choose either a PNP with hFE >500 or better use Pch Enh FET with Ron << 1Ohm.

Using Ohm's Law you correctly compute the Voltage R series drop.

I'm sure you can find many examples.