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This is a follow up to this question, but isn't specific to the other one. This is both an ohm's law question and a 'how does one interpret data sheets' question.

I want to minimize the current I use to hold open a relay via a trigger (which only requires 1.2mA) that I need to isolate. If I use, say, an NTE3083 optoisolator to trigger this relay, how do I pick the proper resistor for the optoisolator's input? The datasheet says the NTE3083 will trigger as low as 1mA.

But the datasheet doesn't give a definitive forward voltage. It gives a 60mA forward current, and then has a table that also mentions 1.15V-1.5V and 20mA as part of its characteristics.

If my source voltage is 12V, and I want, say, 2mA to be conservative, how do I get there from here?

Do I just drop the voltage and current (R of the optoisolator would be constant?) by scaling down current and voltage? So if I want 2mA, divide by 10: 20mA/10 and 1.15V (or 1.5??)/10? And then calculate the resistor?

((12 volts) - (.1 volts)) / (2 milliamperes) = 5950 ohms

Is that the right approach?

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the datasheet doesn't give a definitive forward voltage

It does: The 1.15 Volts typical, 1.5 Volts maximum specified at 20 mA is the LED's forward voltage specification: It never is a precise voltage, LED voltages can vary greatly between units even within a single batch.

For calculating resistance for minimum current, in the absence of input Vf versus current graphs in the datasheet, use the minimum forward voltage rating for calculations: Vf falls with decreasing current through the LED, but not linearly as the question indicates.

So, for 1 mA: (12 - 1.15) / 0.001 = 10.85 kOhms, use a standard value resistor of 10k.

Note that even if one uses the maximum Vf from the datahsheet, the results are 10.5k, not much difference.

Note:

With the CTR of the device specified as 200% minimum, 400% typical, a 1 mA input current would allow a nominal 2-4 mA of Collector Current to flow at the output if there were no other current limiting on the output circuit. This may not be sufficient to hold open the driven relay.

The input current will need to be increased to meet the required output current for the relay. Hence the above 1 mA "desired" current as specified in the question may well be invalid and irrelevant.

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  • \$\begingroup\$ The relay board in question has a "sensitive trigger" that only requires 1.2mA to trigger. I'll update that in the question. \$\endgroup\$ – rrauenza Jun 19 '13 at 14:10
  • \$\begingroup\$ @rrauenza Then you're all set, my answer covers what you need, I believe. \$\endgroup\$ – Anindo Ghosh Jun 19 '13 at 14:52

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