How is the right way to wire the relay module?
The Raspberry pi GPIO operates with 3,3V,
but the relay needs 5V to operate.
=> Open JD-VCC/VCC Jumper to isolate the two circuits
In addition no resistors are necessary.
They are already build in the relay module.
4 Channel 5V optical isolated relay module with Raspberry pi
a) Using external 5V power supply
- open JD-VCC/VCC Jumper
- connect external 5V to JD-VCC pin of Jumper
- connect GND with ground of external power supply
- connect Raspberry 3,3V to VCC pin of Jumper
- connect In1 with Raspberry GPIO-PIN of your choice
(no extra resistor necessary for GPIO → already on relay board)*
b) Using Raspberry as power supply
- open JD-VCC/VCC Jumper
- connect Raspberry 5V to JD-VCC pin of Jumper
- connect GND with ground of Raspberry
- connect Raspberry 3,3V to VCC pin of Jumper
- connect In1 with Raspberry GPIO-PIN of your choice
(no extra resistor necessary for GPIO → already on relay board)*
*Current flows from 3.3V, delivered by the Raspberry through an on board mounted resistor to the diode of the optical couppler and afterwards into the GPIO-PIN of the Raspberry when the GPIO-PIN is set to LOW.
At this side of the circuit no extra ground needed!
I connected the 4 relay module to external power, let the jumper and connected the In to a GPIO pin in raspi 4 and everything OK.
In this case the raspi GPIO pin will drive in HIGH will have a 2.7v difference between diode and then when LOW will have the 5v difference. Any comments?
Sorry, I currently have no relay module where I could try to reproduce your finding.