Hi, I have a couple of these voltage stabilizers, the description says input is AC or DC and output is 5v. Does anyone know which end is input and which is output? The board doesn't say and I don't want to guess. 
There's also some header pins, 10 of them, anyone know what they are for? I want to run 120v AC into this and have 5v DC out.
http://www.dx.com/p/l7805-ac-dc-voltage-stabilizer-regulator-module-black-green-219444#.VJEtOSuaRrg
Three clues:
I'd say the + on one end signifies the DC input end.
The four diodes on the end would be the rectifier so that's the AC end.
You can trace the right hand pin of the 7805. it's the 5V out so see where it goes.
I'll guess the headers are just pairs of +- outputs?
robsworld78:
I want to run 120v AC into this
Except the link says input is up to 20V....
JimboZA:
Except the link says input is up to 20V....
It can indeed work with 20 VAC, but since the 7805 is a linear regulator it has to dissipate a lot of heat.
So my suggestion of 9VAC.
HansjcOtten:
So my suggestion of 9VAC.
A comment which in the context of this thread is 100% irrelevant: the OP's 120V requirement is already outside the maker's 20V spec, let alone more than 9V. (And any way it has a heatsink the size of a house on there.)
So this won't work then, darn... What I'm trying to do is power an 8ch relay board using 120v. I built a power bar with 8 outlets and want to use a rj45 cable but of course it only has 8 wires. So I want to power the board independently so I don't need the extra wires.
Does anyone have any ideas on what would be best to use? Trying to stay on the cheap side and small. I don't really have room for a small power supply and voltage regulator, hoping one board can do it.
Hi, DONOT use rj45 cable for 120Vac.
Use wire that is suitable for mains wiring, the rj45 wire or the rj45 connectors will not have the correct insulation or current carrying capacity.
I'm sorry but how much do you know about electricity,volts, amps, watts and ohms?
Can you please post a copy of your circuit, in CAD or a picture of a hand drawn circuit in jpg, png or pdf?
Tom.......
No, no, I wouldn't put 120v through RJ45, the RJ45 is only for the signal from the arduino. I don't want any power coming from the arduino to the 8ch relay board. Sorry I'm bad at explaining.
I attached some pics of it, (haven't run the hot wire yet) the blue wire is the cat5 and that's what I'm want to use to turn the relays on and off. Problem is there's only 8 wires in there and I need one more for 5v. So instead of running another wire or going to 10 conductor I think its better to put a 5v power source in the box to power up the relay board.
Ideally it would be nice to connect to the 120v AC which is already in the power bar and convert that to 5v DC. I don't have much room to work with.
You have another problem that you have overlooked... you will need at least 9 wires between the arduino and the relay board, 8 for signals and 1 for common (either +5 or ground depending on the relay board).
I was reading something that made it sound like I didn't need a common ground? The link below is the board I have.
Hi, this location, I think has the same relay board, and it has more info than the banggood site.
It also has a CIRCUIT DIAGRAM of the relay circuit and how it is to be powered on the 5Vdc side..
It is also LOW active, that means that the arduino output will have to go LOW to activate the relay.
Please note that it will need a separate supply, not the 5V from the arduino as it cannot supply enough current to power the coils.
Tom.....
robsworld78:
I was reading something that made it sound like I didn't need a common ground?
Indeed it does not. To make use of the opto-isolators, you connect your relay power supply (a switchmode 5V module such as this) y\to "GND" and "JD-Vcc", removing the link between "Vcc" and "JD-Vcc", and your control wires go to the Arduino as well as a common +5V connection to the Arduino from "Vcc".
The per-relay circuit of the board should be this:
But is there a way to do this with just 8 wires? I don't have 9.
I did get the L7805 hooked up, input is on the side with caps, LED is output. I guessed the other way too. It works and powers up the relay board but yeah no more click since no power from arduino.
Would this be good to power the relay board? Its a 12v 1amp adapter and a DC-DC step down adjustable module?
Or is it better to use a 5v regulator like the following? If I used this is it as easy as connecting 12v to one pin, ground to the other and 3rd pins gives 5v? Or do I need other components to make this work?
Did you read what I said?
Paul__B:
a switchmode 5V module
Sorry, I was in a hurry, omitted the intended link. Should have said this.
You want 5V, right? You use e a 5V supply.
And just to clear something up,
JimboZA:
I'll guess the headers are just pairs of +- outputs?
Looking at the illustration of the reverse of that PCB, they clearly are.
Paul__B:
Did you read what I said?
Yeah but it sounds like you said I still need 9 wires? The 9th would be 5v from the arduino.
On the power supply I was worried the voltage wouldn't be 5v, usually they are higher than rated in my experience. I just cut a connector of one I have and its 5.13v, is that ok? How high can the voltage go till I put the relay board in danger?
robsworld78:
On the power supply I was worried the voltage wouldn't be 5v, usually they are higher than rated in my experience. I just cut a connector of one I have and its 5.13v, is that ok? How high can the voltage go till I put the relay board in danger?
For the relay board, you probably do not want to operate it much over 10V.
Yes, 8 relay inputs, one common, in this case, the supply positive. Total 9. How else could it work?
Nice, that's lots of wiggle room for voltage, 5v adapter will work nicely.
I was hoping there would be some fancy wiring scheme or components that would make it work with 8 wires. 
Someone needs to work on that.
Thanks for all the help everyone.
robsworld78:
I was hoping there would be some fancy wiring scheme or components that would make it work with 8 wires.
Well, there is of course - you have to use a shift register on a piece of stripboard next to the relay board, reduces the cable to five wires - Gnd, Vcc, Clk, Data, Latch.
I'd say the + on one end signifies the DC input end.
Correct, except I think you mean DC OUTPUT end. (Since the INPUT is the AC end)
And any way it has a heatsink the size of a house on there.)
That's a good thing for the reason you already stated. (without it you can cook an egg on that thing).
(a very small one)
