# Ohm's Law and SSR

Ok, so I'm trying to hook up a solid state relay, Carlo Gavazzi RS1A23D25, to my shiny new arduino so I can switch a 110/120v line.

http://www.ttiglobal.com/assets/108/RS1AxxA.pdf

My research shows that an arduino data pin will output up to 40mA of current at 5v. Now, my SSR lists:

Input current @ max input voltage [ch8804] 12 mA
Control voltage 3-32 VDC

So, I whip out my shiny new-to-me Ohm's Law and figure 32v*12ma = .384mw
Ok, so now I need to know, how much power is needed at 5v to operate the SSR?

.384mv = 5v * 76.8mA

Is this correct? Do I need 77mA at 5v? Is this a max value or what? Does anyone think I can (or should) power this direct off an arduino pin, or do I need to drop a transistor in there and power the SSR off a 12v line?

Thank you!!

well, you can always try to run 5v through it and check the current consumed. if it's around 80mA, I guess you could simply use 3 output pins to be on the safe side, and you'll get your 120mA.

Your calculations are correct. The input should draw 77mA from the source. If you add a simple 3904 transistor and a resistor you should be fine.

Handi: Wooo, I hadn't thought of that! Could there be issues with how fast the three lines get turned on in sequence?

jes1510: Thanks for confirmation of my calculations! Would that resistor go between the arduino digital pin and the transistor? (Yeah, I'm really a noob. I've never taken a physics class even. I need to go and audit one at the local college.)

Thanks for confirmation of my calculations!

Sorry you are wrong.

First of all the calculations:-

32v*12ma = .384mw

Ohms law is E = I * R or if you want the resistance R = E / I
(I guess you didn't take math either )

So it should be 32V / 12mA = 2666 ohms or 2.66K

So at 5 volts this input will draw I = E/R = 5/2666 = 0.0018 Amps or 1.8mA

Well within the current capacity of an Arduino

However, that assumes that the input of this device looks like a resistor which it will probably not be.
So your method is wrong as well. These sort of devices have a none liner characteristic.
It is more likely that it will draw closer to 12 mA at 5V but you will still be OK to connect it directly.

Sorry you are wrong.

First of all the calculations:-

32v*12ma = .384mw

Ohms law is E = I * R or if you want the resistance R = E / I
(I guess you didn't take math either )

Actually, I minored in Math. (Where's a exploding brain smiley when you need one...)

But, but, but... Doesn't P = E * I ?

And why do we get two different answers solving for Power versus Resistance?

However, that assumes that the input of this device looks like a resistor which it will probably not be.
So your method is wrong as well. These sort of devices have a none liner characteristic.
It is more likely that it will draw closer to 12 mA at 5V but you will still be OK to connect it directly.

Ok, so Ohm's Law is more of a limit? Or does it completely not apply here, and I should just try and hook it up and measure it?

Thanks!

C

I think your SSR uses a current regulated optocoupler on the input so the input load should be no more than 12ma at 5v.

The data sheet linked in the first post is for SSRs with AC inputs of 80 volts or more, I assume you have [u]this[/u] 3-32vdc input relay

But, but, but... Doesn't P = E * I ?

Indeed it does. Now I see you were even more wrong in what you were trying to do.
You were working out the power you needed at 32 volts and using that power to see what current you needed to produce the same power at 5V.

Life, and electronics doesn't work like that. The current is pushed round by the voltage into the load (resistance). Normally you control only the voltage and the current is what it is, given the load. You can't make a 5V output give any more current into a given load. Voltage, current and load setting any two forces the third.

mem is quite right in what he says. Ohms law only applies to liner loads.

That is a load who's voltage / current characteristic when plotted is a straight line. Things like semiconductors (LEDs diodes & transistors) are non liner and ohms law can only be applied to a specific point on the voltage / current curve. Like an LED dropping a certain voltage across it when it is on that does not change very much with the amount of current flowing through it.