I am trying to trigger this SainSmart 4-Channel 5V Relay module made for Arduino :
It has opto-isolated inputs that are triggered by micro-controllers. The inputs are TTL logic active low. The specs have a misprint saying that they need 15-20mA driver current. This is not true. I measured and it only needs about 1-1.5mA driver current. Other people on the web have confirmed this as well. The connections on this relay module are IN1, IN2, IN3, IN4, GND, Vcc, and JD Vcc. The coil switch side takes about 66 mA each when active.
The relay comes with a jumper on the Vcc and JD Vcc. The Vcc pin power is for triggering the IN pin side and the JD Vcc pin power is for the switch coil side. There is some debate on the web whether the jumper should be removed and separate isolated power used. My question is can I leave this jumper on or should I try to separate? Is it safe for my control unit with this jumper still connected or does it need to be separate in order for the opto design to work?
I am using an Expert Sleepers ES-5 to control the triggers. It sends 5V gates out. I really wouldn't know where to get a 5V Vcc off this control unit if I had to remove the jumper.
I can think of two reasons to separate relay supply from logic supply in this case:
The relays are drawing 66mA when active. Five relays active at once would draw 330mA which, depending on what else you are driving could push a 500mA regulator into the red.
The back-EMF (even with a flyback diode) from a relay's coil can be very noisy. Not a great thing for a digital supply line.
I would try to get a separate supply for the relays. Even going so far as building something with a 7805 regulator and a couple of capacitors.
You must give the unit its spec'd current or you'll probably melt the transistor -
that current is necessary to saturate the switching transistor so its Vsat is nice and low
and all the voltage is across the relay coil.
If you can't separate the supplies add some 100uF or so of decoupling at the module,
should help nail any switching noise.
Is there a way to measure the necessary current? The specs are said to be wrong. I emailed the company, but their customer service is pretty much nonexistent.
Give it 15 or 20mA, that's a standard value for opto-isolators - remember the output
current from the opto isolator depends on the input current, and that it's driving
a switching transistor that needs saturating...
hmm.. i thought that the whole design of opto isolators is not input current dependent on output current. the input just turns on a led which uses the light as a switch to activate the transistor side that draws enough power. the 1.5mA trigger side seemed to work just fine. i'll try upping the amps on the trigger side, but i'm pretty sure the coil draw is always 66mA as long as that led switch turns on. i could be wrong though. let me do some testing.
The only drawback of the USB device you quote for the relay supply is that you need a USB connector to draw power from it. Other than that, it should work fine.
this website says : remove jumper and don't connect any GNDs except the JD-Vcc.
interesting. never seen that before.
so for my setup:
remove jumper
the uZeus power would only send +5V to Vcc.
the ES-5 only send the positive tip out of the jack to IN.
the USB power supply would send +5V to JD-Vcc and its GND to GND
Hmm.. would that work? With no GNDs from the first two?
