not exactly, but..... not far
even if they are in the same package, the optocoupler contains 2 devices : an LED and a phototransistor .
When used in switching applications, a transistor may be compared with a....switch :grin: . 2 states : open/closed
What you want is : no current when in open state, and the less "resistance" when in closed state. Then the lower Vce(SAT) the best.
with a "normal" transistor, the saturation state is obtained by setting Ib >> Ic/hfe . When it is fully saturated, the voltage between collector and emitter is at its lowest value. The Ic and Ib values needed are given in the transistor datasheet.
To make it simple, in an optocoupler, Ib depends on the light it receives, the more light it receives, the higher Ib is . Thus the 4 different curves in the figure.
But when it is not clearly listed (as in the Fairchild document) I have no idea how to find it.
When the datasheet doesn't indicates it , and if you need to know it, you may calculate it with the maximum output power (Pmax) :
let's say you know what Ic you need -> look at the curves and note the corresponding Vce(SAT) . If Vce(SAT) x Ic is less than Pmax, then you're OK - But ...... being within the limits only means you won't fry it , it doesn't mean you are using it well.
Example : the vishay datasheet says IcMax = 50mA - If I look at the curves, I cannot even see the Vce(SAT) value for Ic=50mA .
If I set Ic=10mA (which is far below 50mA) , I can see tha Vce(SAT) (with If=10mA too) is more than 2V !
OK 2V x 0,010 = 0,02 W , below 150mW, but 2V are lost in the transistor .
Edit : while I was writing this, Jremington posted a shorter an actually, maybe better explanation 8) - I post mine anyway, but... ;)