The "low level inputs" hook to Arduino output pins. The GND pin connects to Arduino ground. The two screw terminals near the low level inputs provide 12V for the relay coils. You will need a 12V supply and they don't bother to say how much current. When the board arrives, search for the relay model number to find out how much current each coil draws and multiply by 16.
So here are some specifications that we can all use:Overview:1. The 12VDC input requires > 500mA.2. The drive to each control input pin must "sink" 3mA when low (low = relay ON).***** Input Power (12 VDC input)*****- About 8 mA is required with all relays off.- Each relay requires about 30 mA when on.- So max supply current is 8 mA + (16 x 30 mA) = 488 mA (actual measured was 500 mA)- Because one may use the board's +5 VDC output (2 pins) to power an Arduino/PIC circuit, use a 12V power supply that can provide MORE than 500mA (depending on your circuit's requirements).- Note that the switching regulator on the Relay Board should somewhat efficiently (say 70%?) convert the board's 5V power usage to 12 V power input requirements. For example: 200mA at +5VDC (1 Watt) does NOT mean the +12V supply needs to supply an additional 200 mA also. This is because 1 W of power from the +12V supply only requires about 83 mA ( 12 V x 83 mA = 1 W ); however at say 70% efficiency of the 5 V regulator, this goes up to about 120 mA (83 mA / 0.7) but NOT the full 200 mA.NOTE: The best way to discover what 12 V supply is needed (its max current rating) is to ACTUALLY MEASURE the 12 V input current while using a "test supply" that can more than handle worst case (with all relays ON) then buy the supply that meets your needs. Always use a modern "switching" supply (wall wart) because they are smaller, way more efficient, generate little heat, and normally use much less "vampire power".- The baord's LM2576 (+5V) voltage regulator is rated at 3 Amps; however, one should not push it this hard. The circuits powered by the 5 V supply on the Relay Board appear to only be the LED side of the opto-isolators. Driving an input control line low turns on an opto-isolator LED ... turning on its relay. Each opto-isolator LED seems to require about 3 mA (for a total of 3 mA x 16 = 48 mA). This should leave you with at least many hundreds of mA available to power your circuits off of the relay board's 5V output pins (two of them on the connector).***** Input control pins *****- Grounding an input control pin (logic low) turns on the associated relay.- The circuit driving the input control pin must be able to "sink" (drive logic low) about 3 mA of current (easy for most PIC/Arduino output pins).*** CAUTION *** When a pin is NOT driven low, it "floats" to nearly the +5 V that drives the opto-isolators. This means that the driving circuit (Arduino/PIC) must either be also powered by +5V, or if powered by the now common 3.3V (or less!), its output pins must be "5 Volt Tolerant" (see your micro-controller pin specs). Another option is use of a "5V tolerant serial port expander" chip like an MCP23018 (I2C interface) or MCP23S18 (SPI interface) ... where just a few micro-controller pins give you 16 I/O pins. These can be powered by 3.3 V or 5 V. They are a bit complex, but a simple "software bit banged" I2C or SPI interface can be used to control them. Finally, one could use little signal transistors (2N3904) for this isolation from the 5 V (MCU pin -to- a say 2.7K resistor -to- transistor base, emitter to ground, collector to relay board input control pin).
Beware. 16-channel relay board with optos, but without opto isolation.Leo..
The Arduino will be powered by a computer's USB port...
Nope.This relay board is hardwired to Arduino's 5volt rail.Depending on which Arduno, it could backfeed your computer's USB port.Leo..
So that all being said how should I best approach this given that I need to send dat to the Arduino from my computer / eventually a small computer for the car?
the answer is easy.un-solder those opto's and put then on a separate board. or find the common ground or common power and actually isolate them from the relay power supply. this should not take too long and would be pretty straight forward.it is very odd that the relay board designers did not bother to offer two power options, one from on-board and one from off-board.