# PNP with Optocouplers

Discussion in 'General Electronics Chat' started by markdem, Jun 24, 2014.

1. ### markdem Thread Starter Member

Jul 31, 2013
72
38
Hi All,

Just got a quick question regarding the attached circuit.

It says;
"
The voltage
rating of the phototransistor is irrelevant since its
maximum collector-emitter voltage is the base emitter​
voltage of Q
"
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

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2. ### ericgibbs Senior Member

Jan 29, 2010
2,413
369
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]

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Last edited: Jun 24, 2014
3. ### Alec_t AAC Fanatic!

Sep 17, 2013
5,590
1,070
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

4. ### markdem Thread Starter Member

Jul 31, 2013
72
38
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.

5. ### crutschow Expert

Mar 14, 2008
12,548
3,076
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.

Last edited: Jun 25, 2014
6. ### markdem Thread Starter Member

Jul 31, 2013
72
38
That made me click. Now I get what is going on

Thanks all.