is that better?

Much better, but you have a resistor in the battery negative lead which should be a direct connection.

Also note that the power supply needs to be able to supply the input current and the relay current for each channel. That is (11 channels) * ((20mA input current) + (70mA relay current)) = 990mA. You should allow a margin above that so 1.5A as a minimum.

 

Much better, but you have a resistor in the battery negative lead which should be a direct connection.

how should i do?

Also note that the power supply needs to be able to supply the input current and the relay current for each channel. That is (11 channels) * ((20mA input current) + (70mA relay current)) = 990mA. You should allow a margin above that so 1.5A as a minimum.

i got you.. by the way 5v / 1.5A would work? or it should be 12v / 1.5A something like this?

 

how should i do?
i got you.. by the way 5v / 1.5A would work? or it should be 12v / 1.5A something like this?

Where the resistor is just use a piece of wire - zero resistance.
If you use the 5V relay boards then you should have a 5V power supply. Go for a minimum of 1.5A, that will work fine.

 

Where the resistor is just use a piece of wire - zero resistance.
If you use the 5V relay boards then you should have a 5V power supply. Go for a minimum of 1.5A, that will work fine.

i see now what you mean about direct connection.

thank you very much.