lloydi12345
Member
I am trying to tap at the cathode and anode of an LED and it's voltage difference is 1.7v when it is on while 0.4v when it is off. Is there any way I could turn on an optocoupler like 4n30 with only 1.7v?
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Sure, substitue the LED inside the optocoupler for the existing LED. Or wire the optocoupler LED in-series with the existing LED.
Or put a resistor in series with the existing LED, then put that LED+R in parallel with the optocoupler LED.
It depends on the comparative forward voltages of both LEDs, but with some messing around you sould be able to get both LEDs lighting up, although at a lower brightness than one LED.
I am trying to tap at the cathode and anode of an LED and it's voltage difference is 1.7v when it is on while 0.4v when it is off. Is there any way I could turn on an optocoupler like 4n30 with only 1.7v?
You are measuring the voltage across the diode section while it is ON. Perhaps you might have used 5V DC to drive the diode section, using a series resistor. In such case the additional LED can well be wire in series to it. you might need to reduce the series resistor value to manage the needed current , say 10mA or so.Sorry MikeMl, I don't get it. What do you mean substitute the LED inside the optocoupler? If I wire it in series will 4n30 function with 1.7v??
Hi,
The 4N30 d/s states 1.2Vfwd to 1.5Vfwd at 10mA.
Providing the 1.7V source can supply ~10mA , with a say a 47R series resistor, assuming the 4N30 has the Vfwd of 1.2V it might work, BUT I doubt due the drop across the resistor that drives the existing LED, it will not be able to source sufficient current to fully drive the LED and 4N30.
You are measuring the voltage across the diode section while it is ON. Perhaps you might have used 5V DC to drive the diode section, using a series resistor. In such case the additional LED can well be wire in series to it. you might need to reduce the series resistor value to manage the needed current , say 10mA or so.