Hi All,

Just got a quick question regarding the attached circuit.
(Source:http://picmania.garcia-cuervo.net/r..._high-level_loads_with_optocouplers_83704.pdf)

It says;
"

The voltage
rating of the phototransistor is irrelevant since its
maximum collector-emitter voltage is the base emitter​

voltage of Q1 (about 0.7 V).
"
I don't understand how. If Q1 is conducting, would the voltage between the collector and emitter of the optocoupler not be whatever the source voltage is? I bet it is just my lack of understanding of something, but I just can't get my head around it.

Thanks

 

hi mark,
Are you mixing up fig #7 and fig #8

Your drawing is fig 8 and your text clip for fig 7.???

E

EDIT:
The voltage across the opto collector/emitter is never higher than the Vbe of the driven transistor. [0.7v]

 

If the transistor is conducting its emitter is at +V and its base is at +V -0.7. Therefore the voltage between the collector and emitter of the photo-transistor is +V - (+V -0.7) = 0.7

 

Thanks for the feedback guys,

Eric, No, I am not mixing them up. I just need the PNP version. I think the text is valid for both.

I still, however, don't understand why the Vce is not at +V. What drops the voltage?

Thanks again.

 

Vce indeed sees V+. But the Vbe drop is only about 0.7V so that is the voltage the opto sees. Remember the Vbe is relative, not absolute to ground.

 

Remember the Vbe is relative, not absolute to ground.

That made me click. Now I get what is going on

Thanks all.