Optocoupler calculation

[Burner1]

I am working with an Optocoupler #PS2501 for an automotive application. I originally set up the circuit with the feed side of the optocoupler to first go through a voltage regulator (7812A) to provide a stable voltage. Then divided with resistors to get the proper voltage for the optocoupler.




It is not working and I know why. I am not getting enough current through because of my voltage divider using large value resistors.

I can play around with making it work with other combinations (And I have) but what I would really like to do is to know and understand my target for the PS2501.

I look at the datasheet and I simply don't know what value I am targeting. I would say I am google educated and I do not completely understand what I am looking for in the datasheet. Am I targeting the forward current for the diode of 80ma?

 

[chuckey]

80 mA seems high, I would calculate for a current of 20 mA through the LED. if it is used digitally, i.e. the 12V is on or off, then a simple series resistor would do ( R = 12/.02 = 600 ohms), make it 680 ohms to compensate for when the battery is fully charged. The forward current is 6 X that of the LED current so at 20 mA LED current the collector current could be about 120 mA, but you will be getting close to the max power dissipation of the device (allow for a HOT car!!).
Frank

 

[Burner1]

Thanks. On the Datasheet the forward current; is that a maximum value?

 

[chuckey]

yes, but look at it back-to-front. The transistor can only handle 50mA (the current must be limited externally). So with a X6 current gain from the LED current, any current in the LED of over 50/6 ~8.1 mA is not doing anything more to the output transistor. All its doing is to allow for the external voltage to go much higher(and deliver more current) into the LED without burning it out. Also its best to keep the power dissipation as low as possible else the chip will over heat.
Frank

 

[Burner1]

Thanks Chuck. I am trying to absorb what you wrote.

Forgive me for being dense,

I think I get that If = 10ma meaning this is the needed current to triger it.
It then shows typical and max voltage of 1.17/1.4

Power or whats would be 10 x 1.17 or 10.17.

So if voltage is say 5v, would I then be targeting 10.17 /5 = 2.034ma

What I am saying, if I understand the spec sheet, 10ma at 1.17v would that be the same as 2.034ma at 5v?

 

[chuckey]

" Power or whats would be 10 x 1.17 or 10.17. ". current is 10 mA (.01 A), voltage = 1.7 so power is 17 mW or .017 Watts.
" So if voltage is say 5v, would I then be targeting 10.17 /5 = 2.034ma ". If you have 5 V you only need 1.7 V, so you must "loose" 5-1.7 = 3.3 V. The current will be 10 mA, so R = V/I = 3.3/.01 = 330 ohms. So you need to put a 330 ohm resistor in series with the LED when driving it from 5V.
Frank

 

[Burner1]

Thank You

Gary