i am using a 4n25 for a high voltage detection circuit. it has a maximum reverse voltage of 5V. does this mean that the LED inside it can only take in a maximum of 5V in the reverse direction before burning out?
in the below picture, i am using the high output (10.5V) from the TL072 to trigger a low output from the 4n25. a low output from the tl072 will ground the led inside the 4n25 and trigger a high output. whereas, the high output from the 4n25 will not ground it and thus the led will be off and the 4n25 output low.
but this creates a negative voltage of 0.51 across the led. does this mean this negative voltage is safe as long as it is not above the max reverse voltage of the 4n25 which is 5v?
should i put a diode somewhere for more safety?
Best Answer
If you connect the LED anode to the same voltage as the op-amp rail, it will be impossible for the LED to be reverse biased at all.
If you use a lower supply to the LED anode (5V, say), then you need to limit the anode-cathode voltage to no more than 5V, or 10V max for the anode. You could do this by using a reverse-connected diode across the LED which will limit the reverse voltage to one diode drop, about 0.7V for a ordinary switching diode like a 1N4148.
The issue you’re actually encountering is that the op-amp won’t swing any higher than VDD-2V. This is leading to leakage current on the LED leaking back to the op-amp, which can only swing to about 10V, so the LED is still being biased at about 2V. LED Vf is 1.3 ~ 1.5V.
Add 3 series diodes to the LED (1N4148 type is ok). This will shift the forward bias voltage down by about 2.1V, resulting in an overall forward voltage of (1.3~1.5+2.1) = 3.4 ~ 3.7V.