# Current limit resistors for Optocoupler controlled Mosfet

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

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

• 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. Commented Jan 28, 2020 at 12:57
• Do you mean the digital pin PR_SIG? On my uC I measure 5V when I set the output to HIGH
– v3xX
Commented Jan 28, 2020 at 13:09
• What do you measure on PR_SIG when driving a load that takes around 20 mA? Commented Jan 28, 2020 at 13:11
• I measure the voltage from PR_SIG to Pin1 of the opto against GND without a resistor.
– v3xX
Commented Jan 28, 2020 at 13:19
• 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. Commented Jan 28, 2020 at 13:36

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.

• I corrected the circuit Pin6 and Pin4 were connected
– v3xX
Commented Jan 28, 2020 at 14:13
• @v3xX as you see, you don't need to use the whole 20 mA the uC can deliver. Just 2.5 mA is enough. Commented Jan 28, 2020 at 14:14
• 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). Commented Jan 28, 2020 at 14:15
• 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
– v3xX
Commented Jan 28, 2020 at 14:17
• 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?
– v3xX
Commented Jan 28, 2020 at 14:22

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.