# Relayshield Question

Hi,

im a bit too scared to just try. Since some things r still not 100% obvious to me. I hope u can put some light in the dark.

It says:
5V 8-Channel Relay interface board, and each one needs 15-20mA Driver Current

But that doesnt account for the max 40mA per pin and 200mA all pins rule, as the Inputs only deal with the small current amounts of that opto led, right?

So whats the limit for the amount of relays and current? 300mA USB can give to the arduino? Or if i power the ardui via powerjack with 9V max 500mA, would be 500mA the limit or even more, like 1A of a 7805?
edit: or should my question be, how much current can the powerregulator of the ardui handle? cant find something in the "hardware section"

and is this true: if i use powerjack with 12V i get 12V at the VINpin of the ardu, which i could use to power up something else like a 12V relay?

In answer to the basic question, relay data sheets for that type of relay suggests somewhere around 0.4 watts to drive the relay coil. Considering that you have bought the 5 volt version, that implies each relay will require around 80mA to drive it. (0.4/5), times 8 for all relays being energised, equals 640mA at least. That being the case, and presuming you want to use this board, I'd suggest you couple a 5 volt regulator (with heatsink - it will get hot since it will be required to dissipate around 5 watts) to say a 12 volt supply and feed the relay board power from this source.

as the Inputs only deal with the small current amounts of that opto led, right?

Right.

So whats the limit for the amount of relays and current?

The resistance of the coil wires. You don't need any extra current limiting if you are driving the coil at the rated voltage.

Or if i power the ardui via powerjack

Then it is up to 1A, the rating of the regulator and the series diode.

if i use powerjack with 12V i get 12V at the VINpin of the ardu, which i could use to power up something else like a 12V relay?

Yes but the bigger the voltage the more heat you dissipate in the regulator and this limits the current you can draw from the 5V rail.

thank you, so i was right to ask first instead of just trieng it

Whats this:
"5V 8-Channel Relay interface board, and each one needs 15-20mA Driver Current" about?

As i know opto LEDs need something like 1-5mA to work properly, and also that mini green LED, but not 15-20mA? Or did they mean a relay would need 20mA instead of 80mA?

I have another question:

i have a powersource that can switch between 12;9;7,5; and 6V.
and these specs:
Power: DC 9W
Current: DC 500mA

so i guess those 500mA wont be enough if i need 600mA+
But is 500mA really the limit? As using 7,5V would be 9W/7,5V = 1,2A and not 500mA, or is it meant in a different way?

"5V 8-Channel Relay interface board, and each one needs 15-20mA Driver Current" about

A lot of opto couplers need bright LEDs to transfer the current over, it is the normal sort of current.

But is 500mA really the limit? As using 7,5V would be 9W/7,5V = 1,2A and not 500mA, or is it meant in a different way?

Yes it is because with a linear regulator the current is common. So 500mA from the supply is needed to give 500mA in the load, the excess power is wasted as heat in the regulator.

Those sort of calculations only apply if you are using switching mode DC to DC converters and then you throw in an 80% efficiency term as well.