# The 4N35 optocoupler and its input voltage.

#### MATT838383

Joined Jul 31, 2019
132
Hello everyone! i have a question about the 4N35 optocoupler and trying to find the input voltage (diode) on it! the datasheet notes a 7v revers voltage! thank you to light me!

#### MATT838383

Joined Jul 31, 2019
132
Hi bertus so the input voltage pin 1 and 2 doesnt exeeds 1,3V? is that?
matt

#### bertus

Joined Apr 5, 2008
21,653
Hello,

Have a look at page 154 of the datasheet.
The absolute max is about 1.7 Volts at -55 °C, according to the ELECTRICAL CHARACTERISTICS table.

Bertus

#### ronsimpson

Joined Oct 7, 2019
1,619
A LED in the opto is a current device not a voltage device. Example. A light bulb (filament type) you apply 12V and it works. A LED you apply current not voltage. So use a current limiting resistor. If you have 12V use a resistor to limit the current to 1 to 10mA and the 1V drop just happens. Do not apply 1V from a power supply.
the datasheet notes a 7v revers voltage
That is reverse voltage and the LED will not light with reverse voltage.

• DickCappels, SamR and bertus

#### crutschow

Joined Mar 14, 2008
28,155
The input is an LED, and like all diodes, it will drop the voltage it needs to operate from a current-limited source.
So typically a resistor is added in series with the input to limit the current.
Thus, for example, if your signal source is 5V and you want 10mA through the diode to turn on the opto, you would add a series resistor of nominally (5V-1.1V) / 10mA = 390Ω.

#### MATT838383

Joined Jul 31, 2019
132
ok thanks to everybody! so on a diode input it doesnt matter the voltage only the mAmps are calculated!

#### crutschow

Joined Mar 14, 2008
28,155
ok thanks to everybody! so on a diode input it doesnt matter the voltage only the mAmps are calculated!
Only to the extent that its voltage affects the determination of the value of the current-limiting resistor (or circuit).

#### eetech00

Joined Jun 8, 2013
2,644
ok thanks to everybody! so on a diode input it doesnt matter the voltage only the mAmps are calculated!
Almost.....The chart in post #2 is showing the typical operating characteristic for VFwd (forward voltage (DROP)) vs the IFwd (Forward Current) across the LED. So you have to take the voltage drop across any current limiting resistor into account when calculating the resistor value to obtain a specific LED operating voltage/current.

#### MrChips

Joined Oct 2, 2009
24,593
Almost.

In theory, an ideal current source has infinite source resistance, but you would have to make the source voltage infinitely large as well.

Therefore a compromise is a sufficiently large source resistance with sufficiently large source voltage. When we apply this to the load, the voltage at the load is negligible compared to the source voltage.

For example, suppose we need to power an LED at 2V @ 10mA from a 20VDC source.
If we apply Ohm's Law, the required series resistance is 18V / 10mA = 1800Ω

Now, suppose that we ignore the voltage across the LED.
Required series resistance is 20V / 10mA = 2000Ω

If we had used a 2000Ω resistor instead of the 1800Ω the LED current would be 9mA instead of 10mA, nothing to worry over.

The bottom line is, if you make the series resistance and hence the supply voltage large enough, the circuit will be very forgiving with any uncertainties in the LED current and voltage.

#### crutschow

Joined Mar 14, 2008
28,155
In theory, an ideal current source has infinite source resistance, but you would have to make the source voltage infinitely large as well.
That's if you use a resistor to give a constant current.
You can, of course, make a near ideal current-source (within limits) using an electronic circuit, but that would be more complicated than a simple resistor.

Below is the LTspice simulation of a simple, somewhat stiff, constant-current source using two PNP transistors:
Note that the LED current (green trace) is fairly constant with a change in the input voltage (horizontal axis) once the output voltage is above the LED forward voltage of about 1.8V (yellow trace) at about 3V input voltage.

For an input voltage of 3V to 10V the circuit has an equivalent output impedance of about 7.8kΩ Last edited:
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#### MATT838383

Joined Jul 31, 2019
132
thanks to every one for showing me your theory( crustshow mrchips) now it looks clear!
best regards!

#### MisterBill2

Joined Jan 23, 2018
9,775
Note also that exceeding that 7 volt reverse voltage results in immediate destruction of the LED, leaving the device useless.

Joined Jul 18, 2013
23,897
Hello everyone! i have a question about the 4N35 optocoupler and trying to find the input voltage
Just as an aside note, re the output, you can use either the emitter or collector output, i.e. inversion, sink or source.

#### Papabravo

Joined Feb 24, 2006
17,228
Note also that exceeding that 7 volt reverse voltage results in immediate destruction of the LED, leaving the device useless.
I'm curious about what 7 volt reverse voltage specification are you talking about? Which component can potentially be reverse biased?

#### Ian0

Joined Aug 7, 2020
3,738
I'm curious about what 7 volt reverse voltage specification are you talking about? Which component can potentially be reverse biased?
The LED.

#### Papabravo

Joined Feb 24, 2006
17,228

#### MisterBill2

Joined Jan 23, 2018
9,775
In post #1 the TS comments about seeing a reverse voltage of 7 volts, and wonders about that. At some point I state that exceeding 7 volts reverse will destroy the LED. the diodes on LEDs do not have high reverse voltage limits and so are quite easily destroyed.

#### Papabravo

Joined Feb 24, 2006
17,228
In post #1 the TS comments about seeing a reverse voltage of 7 volts, and wonders about that. At some point I state that exceeding 7 volts reverse will destroy the LED. the diodes on LEDs do not have high reverse voltage limits and so are quite easily destroyed.
Well since the TS never provided a schematic of what he was doing so it would be hard to know what his observation means. I thought you were talking about the circuit in post #11. My mistake.