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Reading different design rules regarding optocouplers used as switch I have a question about the current limiting resistors R4 and R6

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One aspect is the protection of the devices (e.g. ~200Ohm for a standard LED over a 3V voltage drop @ 15mA). How can I calculate the correct values?

For the 4N25 the forward voltage is 1.5V and my digital pin can produce up to 20mA => $$R_4 = (5V-1.5V)/20mA \approx 175\Omega $$ where for the Mosfet can switch logic input (5V) but I do not know which value describes the limiting factor for calculation of the resistor R6?

My attempt is with \$V_{GS} = 2V\$ @ \$I_D = 250\mu A\$ with following consideration $$ R_6 = (5V-2V)/250\mu A) \approx 12k\Omega$$

Edit: the datasheets 4N25 IRLZ34NPBF

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  • \$\begingroup\$ If your digital pin can produce 20 mA, this might be the absolute max rating and no way would there be 5 volts on that pin. You need to be more conservative. Also, when driving a MOSFET via an opto connected to 5 volts, R6 can be a short circuit. \$\endgroup\$ – Andy aka Jan 28 at 12:57
  • \$\begingroup\$ Do you mean the digital pin PR_SIG? On my uC I measure 5V when I set the output to HIGH \$\endgroup\$ – v3xX Jan 28 at 13:09
  • \$\begingroup\$ What do you measure on PR_SIG when driving a load that takes around 20 mA? \$\endgroup\$ – Andy aka Jan 28 at 13:11
  • \$\begingroup\$ I measure the voltage from PR_SIG to Pin1 of the opto against GND without a resistor. \$\endgroup\$ – v3xX Jan 28 at 13:19
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    \$\begingroup\$ As you have described, there is absolutely no way that you can measure 5 volts driving the opto directly. It's a forward biased photodiode and might take over a 100 mA to get it above 2 volts. You run the risk of damaging the opto without a resistor. \$\endgroup\$ – Andy aka Jan 28 at 13:36
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You should start at the mosfet Q2.

I think there is no need to drive the mosfet on a low voltage (2V). Always drive the gate at a voltage that is higher than the \$V_{GS}\$ given in the datasheet. A good guidance are the conditions for \$V_{GS}\$ used to specify the \$R_{DS(on)}\$.
When the mosfet is turned on/off at (very) low frequency, just drive it with the highest voltage available. (When driving at higher frequency, you may consider a bit lower voltage to reduce switching losses).
In your case this implies setting R6 to 0 Ω.

UPDATE With updated question, the following does not apply anymore
Note that this also implies R5 becomes superfluous as it is now parallel to R7.

Regarding the optocoupler: don't short the base connection to the emitter, but leave it open, or use a high ohmic resistor to control the sensitivity of the optocoupler.
UPDATE With updated question, this issue is addressed.

In order to get a \$V_{GS}\$ of 5V, (assuming R6 is shorted and R5 is removed), you need a current through the optocouper of only 5V / 10 kΩ = 0.5 mA.

Next, check the datasheet to find the minimum CTR: for the 4N25 it is 20%.

This means (in worst case) the optocoupler needs a forward current of 0.5 mA / 20% = 2.5 mA.

According to the datasheet, Fig 3, the forward voltage is about 1.1V.

So, worst case R4 should be equal to:

(5V-1.1V) / 2.5 mA = 1.56 kΩ

It may still work with a (bit) higher value as the CTR is typically higher.

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  • \$\begingroup\$ I corrected the circuit Pin6 and Pin4 were connected \$\endgroup\$ – v3xX Jan 28 at 14:13
  • \$\begingroup\$ @v3xX as you see, you don't need to use the whole 20 mA the uC can deliver. Just 2.5 mA is enough. \$\endgroup\$ – Huisman Jan 28 at 14:14
  • \$\begingroup\$ For the output of the optocoupler, you need to add some current to allow for charge injection into the gate of the MOSFET during switching. I would suggest up to 1mA extra (I haven't run this in simulation - just a rough guess). \$\endgroup\$ – Peter Smith Jan 28 at 14:15
  • \$\begingroup\$ Ok. The max current was a mistake on my part. I am still not understand why I should remove R5 from the base. In my guidelines it is recommended not to leave it floating \$\endgroup\$ – v3xX Jan 28 at 14:17
  • \$\begingroup\$ Since the switching voltage of my Mosfet is 5V and this is provided by the +5V-Powersupply I do not need a limiting resistor. Correct? \$\endgroup\$ – v3xX Jan 28 at 14:22
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First of all, I think shorting the photo transistor's B and E will make the part useless. Please check the Pin 4 and 6 connection.

Second, if you target the LED IF = 1mA to 10mA, you will get 100uA to 5mA IC per CTR number from the datasheet. Once you correct your schematic, you can figure out the resistor values accordingly.

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