OPTO Specification Confusion

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ElecDesigner

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I have designed in lots of OPTOs in the past and an looking to use the ACPL-K54L. I'm now confused about one of the specifications on the datasheet of this particular part.


Top of page 10: CTR. So if I take the worst case (min) CTR of 93% and for absolute worst case assume a 0.88 multiplier (max temp 75deg) I get a worst case min of about 82% and can design on that basis.
However, there is a further, lower, CTR of 53% listed on the datasheet for condition Vo = 0.5V. I don't currently understand the logic of this or am i misreading the information.
 

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ElecDesigner said:
However, there is a further, lower, CTR of 53% listed on the datasheet for condition Vo = 0.5V.
Click to expand...
Nothing special. The same happens to every simple bjt.
--> the more you go to output_saturation (lower V_CE) the less the CTR will be.

Simple question: What is your situation:
* I_F
* I_O
* V_O (limits)

Btw: I had optocouplers that reduced their CTR to 30% of the initial CTR during a couple of years. Operated well within specification.
Surely not every optocoupler is that bad.

Klaus
 

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I'm still confused as it feels the wrong way round to me.
CTRmin 53% for Vo = 0.5V
CTRmin 93% for Vo = 0.4V

Surely this is the opposite of what you mention?

My Vout max is about 2V (2mA) and I was looking to select a resistor to limit I/F.
I dare say I can put loads of margin on so not a problem, I was just drawn to these specs.
 

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Hi,

you need to read the whole lines:

CTRmin 53% for Vo = 0.5V (for total TA = -40°C ... +105°C, as written above)
CTRmin 93% for Vo = 0.4V for TA = 25°C

VO = 2V is quite huge. Not a standard logic level. Sure about it?

Klaus
 
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Ok thanks. Its obvious now, regarding the specs. For some reason I didn't factor in the temp stuff.

The output is connected to a 4000 series schmitt at 12V power even 2V gives margin.
 

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Rule of Thumb for any BJT
hFE starts drop when Vcb drops < 2V at higher current and < 0.6V at low current when Vbe=Vce @ approx 1mA for IC

Same for CTR which is hFE * LED coupling loss = CTR

So if hypothetically if infrared coupling is factor 1% and hFE is 100 your CTR is 1 or 100%. It is this product that yields CTR.

Consider the old 2N2222 curves, ( which they don't publish much anymore)

See how computed hFE drops as Vce reduces in the Y axis.



Bottom line

Use a low Ic value to improve CTR and adjust Rc accordingly to meet Rc Cp=T rise time
e.g. 1 to 5 mA The datasheet uses 3 mA for spec, so this should be optimal for most users

For a 2N2222 the optimum hFE is 30 mA,
 
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