Safe forward voltage on an optical solid state relay?

Thread Starter


Joined Mar 26, 2021
I've got a dozen small opt solid state relays that I am using with an MCU that will be ground switched with NPNs or a darlington array (fewer pins needed). The optical solid state relays are Omron G3VM-31AR. The link to the data sheet is below. In my design I plan on using 5 vdc coming from a separate power supply to switch them with the transistors triggered from the MCU gpio pins. When I look at the data sheet for these Omron SSRs I cannot figure out what the max forward voltage is to trigger the relays. Is 5V too much for the LED trigger?

The datasheet only indicates a maximum forward current of 30ma and a maximum reverse voltage of 6 volts. In the next section it references testing at 10ma with a min forward voltage of 1.5v and max of 1.8v, but I think that is just for trigger voltage.

Here is a link to the datasheet. Thank you in advance, for what seems to me like a simple spec question. I am just trying to figure out if I need a current limiting resistor for the solid state relays.
The optoisolators datasheet tells us that the forward voltage, (VF) drop of the input light emitting diode is 1.2 volts and the maximum forward current, (IF) is 50mA. Thus a series current limiting resistor with a value between 126 and 380Ω's can be used.
For reliability, you have to design for worst case.
You have a 5V supply.
There is a maximum Vf - call it 1.5 for now.
And you need to supply 3 mA (max current required)

Now, you have another problem. A transistor, you have do that one as well. You have to look at beta(min/max), Vce(sat). You might need to look at the curves. You need about 10x more base current than calculated. You;ll need a base resistor. You also have to look at the specs of the output po

generally, you'll need to have a ressitor to ground to absorb the leakage current and ensure the transistor turns off.

You can change to a FET. Same deal. a gate current and a resistor to absorb leakage. graphs are more important.
You;ll need what's called a LOGIC FET. You;ll need to look at Vgs max and Vth. Vth is where the FET just starts to conduct. ou have to look at the curves to figure out what voltage you need to ensure the FET is fully on at the operating current.

FET gate series resistors are usually around 200 ohms. You need a high value resistor to absorb the leakage current. e.f, conduct a few uA probably.

If your concerned about power from the supply, you might have to minimize that.
Temperature can change the numbers.

The gate series resistance is there because the gate has capacitance. When operating at high frequencies, you would need a gate driver. You also have to absorb the leakage current or the device would turn on by itself with no connection.

the lower the supply voltage, the more you have to do.


Joined Jun 5, 2013
The datasheet lists a recommended current of 10ma and an operating range of 5 to 20 ma. Design for 10 ma using the typical Vf and you will be fine.

For example: Taking the first one in the table. Typical Vf at 10ma is 1.33V

So your resistor calculation is (5 - 1.33) / 0.010 = 367Ω.

Just the check: The max and min Vf for that model are 1.18 and 1.48

If = (5 - 1.18) / 367 = 10.4 ma
If = (5 - 1.48) / 367 = 9.6 ma


Thread Starter


Joined Mar 26, 2021
I really like these little opto SSRs, but the Vf is really narrow. I prototyped this up on the bench this morning with different resistors in the 150-400 ohm range and 300 seemed to be closest to the sweet spot for 10ma, so the math above is pretty close. As a science project, I put the straight 5V on the poor little opto and it pulled 400ma, so obviously I can't do that (it survived, but I won't be using that opto again). Thanks!