Solid State Relay from CMOS pin
Figure 1. Extract from the PGA2505 datasheet showing the likely limitation for this application.

The electrical characteristics don't seem to give a maximum source or sink current but does quote (1) output voltage at a measy 200 µA when high and (2) sinking 3.2 mA when low. I wouldn't push the chip beyond that.

You're going to have to buffer the output. You can choose between a digital buffer, a transistor or an FET.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 2. Darlington drive for LED.

Given that the specification lists the output voltage at a 200 µA drive current I suggest using a Darlington pair to step up the drive. R1 limits the base current to less than 100 µA. With a gain of 80 minimum this should be enough to drive the LED at the min 8 mA turn-on but Q2 ensures that we have enough drive as its gain effectively squares the current gain of the combination.

The Darlington pair will have a 1.3 V drop (2 x \$V_{BE} \$) and the LED 1.4 V leaving 5 - 1.3 - 1.4 = 2.3 V across R2. If we drive at 10 mA (25% over the minimum) we get \$ R2 = \frac {V}{I} = \frac {2.3}{0.01} = 230 \Omega \$. 220 is the next lower E12 value.