hello;

my question is on optocoupler (fairchild semi mocd207m) is, on the fairchild datasheet for this device, states that the forward voltage drop is 1.25 with a max of 1.55. however, the min is just a dash line.

i'm using a dmm, which is set to diode and I'm getting 1.03. and when i check the reverse i'm getting .701. i believe that i have a bad opto. the reason why is, when checking the forward voltage drop i should not be below 1.25 and when checking reverse, i should see an open.

is my logic correct?

also, should i be calucating for the min? if so, do i just subtract the max from the nomial to get the min?

thanks;

 

That is the conduction drop across the IRLED junction. The IRLED's are good for up to 60 ma. Just subtract the average drop from the applied voltage and use that voltage to figure the limiting resistor. Never exceed 50 ma to leave some headroom - usually you get good results with about 30 ma current.

Arrange both emitters to go to ground and pull the collectors up separately with resistors that holds current through each transistor to under 5 ma. You may have other needs - I just use the critters to go low when there's a signal, and high when not.

 

thanks for your quick reply.

however, i'm only checking the device (opto) to verify that i can get the forward voltage drop as it states on the datasheet, which is 1.25.

and when i check the reverse voltage drop, it should be an open.

i'm not even close to what the numbers are on the datasheet. i be checking the reverse voltage, i'm seeing a diode drop of .701. it is my understanding that when checking a diode in reverse, it should be an open. is this correct?

thanks

 

With LED's, 1.50 is the forward junction figure; + probe on anode, - to cathode. Reversing the probes, you should show open circuit.

Cheers, DPW [ Spent years making heaters out of op-amps.]

 

thanks... so there is no min for a forward voltage drop. right? if there was, would one caluclate it as subtracting the max from the nonimal?

thanks...