I want to use an optocoupler for a signal on the output. Lets say there's 5mA on the input. Does this mean that with a current transfer ratio of 50% that the output is limited to 2.5mA or that I Should be driving a 2.5mA load thru the output?

So does it matter how much current (within specification) the input gets to drive a tiny current on the output?

Thanks

 

If you think of it like a transistor, so that the LED connection is the base, then the CTR is equivalent to the Hfe.
So a 50% CTR is like having a transistor with a gain of 0.5 ( which is a bit of a rubbish transistor).
Similar to HFE on transistors CTR varies all over the place, with current, with temperature annd there is a wide variation between devices; and it takes rather more current than you might expect to get the output transistor switched fully on.

 

If you think of it like a transistor, so that the LED connection is the base, then the CTR is equivalent to the Hfe.
So a 50% CTR is like having a transistor with a gain of 0.5 ( which is a bit of a rubbish transistor).
Similar to HFE on transistors CTR varies all over the place, with current, with temperature annd there is a wide variation between devices; and it takes rather more current than you might expect to get the output transistor switched fully on.

Ok thank you Ian for your reply, I think you know where I'm going with this from my other post but I will try it out anyway. Cheers

 

So does it matter how much current (within specification) the input gets to drive a tiny current on the output?

It depends on the circuit.

CTR degrades over time. Good designs will accommodate degradation, bad designs won't.

 

It depends on the circuit.

CTR degrades over time. Good designs will accommodate degradation, bad designs won't.

Do you see any problems with this?

The signal has a mixed P-N-P pulse, so I'm blocking the positive with the diode and hoping to catch the negative.

 

Hi dandy,
What are the details of the 50Hz source?
E

 

Do you see any problems with this?

Can't say without knowing more about the input signal and how the output is being used.

 

Can't say without knowing more about the input signal and how the output is being used.

 

Hi dandy,
What are the details of the 50Hz source?
E

Please see above

 

I should point out that this is a DC pulse from a generator ignition coil.

 

the output tied to MCU input,

 

I would forget the opto and use a comparator with a high input impedance, and an open collector output.

 

I would forget the opto and use a comparator with a high input impedance, and an open collector output.

Why when the opto provides total isolation and fewer parts.

 

I would forget the opto and use a comparator with a high input impedance, and an open collector output.

Im not sure thats necessary

This is the opto output as described above - take the values with a pinch of salt...

 

That doesn't look correct. Did you separate the ground connection on the opto output from the ignition circuit ground?

 

That doesn't look correct. Did you separate the ground connection on the opto output from the ignition circuit ground?
View attachment 300491

sorry

 

yes they are separate grounds for testing, however they will be common in application

 

yes they are separate grounds for testing, however they will be common in application

If you have a common ground, then why do you need an isolator, there's nothing to isolate?

 

If you have a common ground, then why do you need an isolator, there's nothing to isolate?

I don't need an isolator it's just the way its configured its an easy option, It was a suggestion from another thread and i've just experimented. I dont know much about electronics so here I am

 

I don't need an isolator it's just the way its configured its an easy option, It was a suggestion from another thread and i've just experimented. I dont know much about electronics so here I am

If you have an opto-isolator with the diode cathode joined to the emitter, then you might just as well use an NPN transistor.