Oh, I think I understand what you are seeing. Thanks for the photo.
You are correct in saying transistors or FETs can provide buffering and voltage translation. The transistors in this case appear to be on the other side of the optoisolator on the JD-VCC rail. It is not for buffering of the GPIO but instead they act as a switch to interrupt the current through the coil in the relay (because of the high current inductive kick when switching it on/off could blow up the isolator otherwise). The buffer is the optoisolator but it is for buffering the IO signal for on/off.
The PI side is on the opposite side of the optoisolator on the VCC rail.
The way the power on the two sides are decoupled from each other is by the two pin jumper. When the jumper is in place the VCC rail is shorted to the JD-VCC rail. When the jumper is removed, you have the VCC driving the GPIO side and pin 2 of the two pin jumper is used to provide power to the JD-VCC rail (opposite side of the isolator, relay coil, transistors, etc.) If you remove the jumper, the JD-VCC is effectively powered-down until power is supplied to pin 2 on the two pin header.
There is a schematic I had posted earlier in the thread which is easier to decipher than looking at the PCB.
There is no active buffering on the IO entering the PCB. What you have in series is an LED, then the transmitter side of the isolater (also an LED), a current limit resistor, and then VCC. What you do get is a bunch of voltage drop from the LEDs and the limiting resistor. The more I look at it the more Iâm thinking it may be sufficient to ensure any voltage appearing at the uP is generally below the 3V3 rail (when both rails are driven to 5V).
edit: With the close-up photo, I found the datasheet for the optoisolator.
http://henrysbench.capnfatz.com/wp-content/uploads/2015/05/817C-Optocoupler-Datasheet.pdf
The transmit side LED notes a forward voltage of 1.2V. Just going by that, youâd have 5 - 1.2V = 3.8 volts. Then you have another drop from the second LED along with the current limit resistor (which has no value in the schematic).
With the additional drop, Iâm guessing the voltage will below the rail. But, Iâd measure it to make certain thatâs actually the case (for longevity of the device). When using 5V on both VCC and JD-VCC, that is.