Optocoupler not giving me full voltage!

[Telemachus]

I am using 4 channel optocouplers to isolate a low voltage microcontroller PIC16F1936 from a much higher voltage LED array, and would like to keep the two power sources isolated.

I was using the output of the IC
(datasheet here: https://www.electro-tech-online.com/custompdfs/2011/10/41364E.pdf)

to drive the optocoupler (datasheet here: https://www.electro-tech-online.com/custompdfs/2011/10/LTV-8x6.pdf)

with a 1k ohm resistor in series, and although it works fine, the amount of current moving through the transistor side is quite limited, making the lights far too dim. I dropped the resistor down to 500 ohm and 220 ohm, and although the lights are brighter, they are nowhere near what they would be if I connected right to the power source on the higher voltage side.

I have scoured the datasheet on the PIC, and although is states it has "high-current source/sink for direct LED drive" it won't tell me how many mah it can source.

If I assume 40-50mAh (is this too high?), with 1.2 voltage drop on the IR LED in the optocoupler, and 4.5V out from the IC, then I could potentially go all the way down to a 68 or 100 ohm resistor, and hopefully get my full current flowing on the other side.

I can continue the trial and error method, and other than the wasted money in blown optocouplers and resistors, there's not much to lose, but I thought I would throw it out to the experts and get some opinions.

Thanks

 

[ronsimpson]

There are different types of the LVT-8x6. The L,A,B,C,D each have a different current transfer rate. Input current to output current. If you put 1mA into the IR LED you might get 0.5, 1, or 2mA on the transistor side, depending on which version you have.



40-50mA seems high. If you measure the voltage across the resistor on the PIC side you can find the current.

 

[Telemachus]

Ok, great tip on using Ohm's law to my advantage, thanks.


Interesting about the Rank type, I didn't quite gather what that was talking about, and didn't see a rank code on the chips.

The part number at Digikey is 160-1364-5-ND
(http://search.digikey.com/scripts/DkSearch/dksus.dll?x=0&y=0&lang=en&site=us&KeyWords=160-1364-5-ND)

and it mentions
Current Transfer Ratio (Min) 50% @ 5mA
Current Transfer Ratio (Max) 600% @ 5mA

So I am assuming I am looking at the "no mark" type. I'll recheck a chip when I get home.

50-600% seems like a huge range, I assume the graphs are helping me determine how that is calculated...

 

[Telemachus]

Is it safe to assume that I can keep dropping the resistor as long as no more than 50 mah of current are let through, so to maximize the current passed on the opposite side?

 

[audioguru]

Read the datasheet of your PIC to see the max allowed output current. Some PICs have a max allowed output current of 25mA and then the output voltage drops about 1.2V.

 

[alec_t]

Is it safe to assume that I can keep dropping the resistor as long as no more than 50 mah of current are let through

No: it's not safe. Excess current could harm the PIC or the opto-diode. I suggest you supply only a few mA to the opto-diode (so the PIC won't be significantly loaded) and use the opto output to drive the base of an external transistor which controls current to the LEDs.

 

[BBriano]

Analog Devices has a wide range of digital isoaltors that will solve your problem without the need to worry about the current into the optocoupler. The ADI iCoupler isolators have CMOS inputs. For a low voltage functional isolation that you seem to be looking for I would suggest the ADuM744x quad low-cost isolator. For full disclosure, I am the iCoupler application manager at ADI.
**broken link removed**