Author Topic: Optocoupler Reverse Active mode  (Read 997 times)

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Offline csheldonTopic starter

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Optocoupler Reverse Active mode
« on: March 23, 2019, 07:27:25 am »
Hi.

Can someone explain how will the optocoupler behave in reverse active mode, if we swap the transistor side leads (swap C-E)?
Lets take for example PC817. This is what I found in datasheet for normal active mode if we set the current of the internal LED to 1.9mA:

PC817 Forward Current If = 1.9mA
PC817 Forward Voltage Vf =1.4V
PC817 transfer ratio @ 1.9mA = 670%
PC817 max collector current @ if->1.9mA Ic=12.73mA

Can someone now explain how will these parameters change if we reverse the C and E leads of the optotransistor side?
 

Online T3sl4co1l

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Re: Optocoupler Reverse Active mode
« Reply #1 on: March 23, 2019, 09:01:09 am »
No idea, it's not spec'd.

It may just be the raw photodiode current, since the photosensitive area is, I forget what it is -- is it a thick collector region beneath a thin emitter and base, so the photodiode is effectively the C-B junction, which in turn causes B-E current and amplification?  Or is it the other way around, thick emitter or base near the top?

In any case, you'll have to characterize it yourself.

Tim
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Offline David Hess

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Re: Optocoupler Reverse Active mode
« Reply #2 on: March 23, 2019, 03:26:22 pm »
I know I measured it once but I do not remember how much difference there was between the photodiode collector-base current and base-emitter current.  I was interested in using standard cheap 4N35s as low voltage current sources for single supply applications and they worked great.  Want a true zero input or output with a single supply?  This works great although it costs power.

The phototransistor will happily work with the collector and emitter reversed but the reverse beta (current gain) is low limiting current transfer ratio and of course the reverse collector-emitter breakdown voltage is only like 5 volts.
 


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