Considering LED as a diode, -- what is, in amperes, its reversed current (the one that is usually denoted as is, or io), i.e. the coefficient in the equality:
i(v) = is (exp{qv/kT} …} -1)
For some of my calculations, I "would like" it to be of order 0.1 micro-ampere. Is this realistic?
I assume that when the LED lamp works, v (in the degree of the exponent) is 0.5V. Is this correct?
I assume for the series-resistor, that limits the diode's current, 10 Ohm. Is this estimation good, or should I take more?
It depends on the LED, the best thing to do is to download the LED datasheet from the manufacturer.
I have been using some LEDs from Marubeni that have a maximum reverse current of 10 uA (microamperes). When they are directly polarized their voltage drop is about 1.5V .
It's always a good idea to limit the current but it depends on the application and the amount of power you need from the LED.
If it is a simple application a resistor may work, if you need a very good to control of the current , a constant current source mey be nedded.
Please consider downloading the catalog from kingbright (http://www.kingbrightusa.com/[email protected]), where you can find some I/V characteristics for a large variety of LEDs. In my oppinion 1microamp is a good value for the reverse current and around 1.5 for the treshold voltage
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