I know the basics of how an opto-isolator (opocoupler) should work, but I'm trying to get a better understanding about them, specially on the output side (transistor).
I have a PS2505-1 AC opto-isolator wired on my bread board as in the picture below. The current going in is 10.17mA measured with DMM. I'm using a cell phone charger as the voltage in on this side and it is stable at 5.11V. When I calculate this (5.11V - 1.17Vf)/(386Ohms) = 10.2mA mathematically which is pretty good and close to what I measured with the DMM, not a big deal here.
So now on to the photo-detector side. The LED has Vf = ~1.95V (RED standard LED). On this side I'm using a 9V battery and its output is ~ 8.45V. So I calculated (8.45V - 1.95Vf)/(266Ohms) = 24.4mA mathematically speaking unless I'm missing something, this would be Ic = 24.4mA right?
But my question is when I measure the current with the DMM on the transistor side. I only get 16.15mA, I'm out ~ 10mA, not too bad, but I want figure out what's going on. So what I did next was measure the voltage Between the Base and Collector (pin 3 and pin 4) and I get 2.18V. So I did (8.45)-(2.18Vce + 1.95Vf)/(266Ohms) = ~16.2mA. This works out. So the question is about the voltage that is present at the collector and the emitter. Looks like this is another voltage drop? does it vary depending on the Ic? Maybe I'm missing something from the datasheet and need to understand it.
datasheets/ps2505.pdf
Any help with this will be appreciated.
thanks,
-Jose
I have a PS2505-1 AC opto-isolator wired on my bread board as in the picture below. The current going in is 10.17mA measured with DMM. I'm using a cell phone charger as the voltage in on this side and it is stable at 5.11V. When I calculate this (5.11V - 1.17Vf)/(386Ohms) = 10.2mA mathematically which is pretty good and close to what I measured with the DMM, not a big deal here.
So now on to the photo-detector side. The LED has Vf = ~1.95V (RED standard LED). On this side I'm using a 9V battery and its output is ~ 8.45V. So I calculated (8.45V - 1.95Vf)/(266Ohms) = 24.4mA mathematically speaking unless I'm missing something, this would be Ic = 24.4mA right?
But my question is when I measure the current with the DMM on the transistor side. I only get 16.15mA, I'm out ~ 10mA, not too bad, but I want figure out what's going on. So what I did next was measure the voltage Between the Base and Collector (pin 3 and pin 4) and I get 2.18V. So I did (8.45)-(2.18Vce + 1.95Vf)/(266Ohms) = ~16.2mA. This works out. So the question is about the voltage that is present at the collector and the emitter. Looks like this is another voltage drop? does it vary depending on the Ic? Maybe I'm missing something from the datasheet and need to understand it.
datasheets/ps2505.pdf
Any help with this will be appreciated.
thanks,
-Jose
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