How to use a solid state relay in a dip package to drive a 24 vac relay

Question

We have a fan coil unit in our server room where I work that utilizes a microcontroller to control temperature and humidity. I have an HVAC and electrical background so I am familiar with instrumentation and controls but I am still on a learning curve with electronics and semiconductors. The project I am working on is building and programing a microcontroller (which operates on 5vdc) to replace the unreliable existing one that is already in place. I have completed the programming and touch screen interface for the controller and am now in the stage of designing a shield (interface board) for the controller so it can operate several 24 vac contactors, relays and a few pump down liquid line solenoids that are on the fan coil unit.

Originally I was going to use triacs with octocouplers but found a solid state relay (AQW212EH) in a DIP package that would allow me to install more outputs on the shield. Here is where my lack of experience comes in. Looking at the data sheet for this relay, pins 1,2,3 & 4 are used for the inputs and 5,6,7 & 8 are used as the outputs. While physically looking at the relay there is a small circle in the corner of the dip package. I assume that this is the anode of pin 1.

So with all of this information and assuming I wired this circuit correctly going off of what I assumed was pin 1 and looking at the Data sheet for the solid state relay I came to the conclusion that connecting 5vdc through a 560 ohm resistor to the input side of relay (anode), the LED reverse voltage of 5V, the Peak forward current of 1A and the LED forward current of 50 mA would be more than sufficient to safely drive the relay. On the output side I connected 24 vac to what I assumed was pin 8 and then pin 7 was connected to the coil of a 24 vac 30 amp contactor. This in theory I assumed would drive the contactor, but alas I let the smoke out of my relay.

Can someone tell me if pin one on the relay is what I assumed it was? If it is, am I using the correct relay for this application (do I need some sort of snubber for this circuit)? If this is not the correct relay to use does anyone have a suggestion on which one I might try?

Answer 1

A couple of things.

1) Your 24 Vac contactor requires how much current? You don't say. But my past experience tells me that the coil on the contactor uses probably about 20 VA. 20VA / 24V = 0.83A. That means that your SSR is too small for that contactor.

Also be aware that contactors with AC coils have significant inrush current. If the datasheet for the contactor doesn't give you the inrush current for the coil, you will have to measure it yourself. Pick your SSR to have a peak current rating higher than the inrush current for your contactor.

2) From the datasheet, the LED in the SSR needs at least 5mA. 5V - 1.5V = 3.5V / 5 mA = 700 Ohms. Your 560 Ohm resistor is a good value to use.

In general, you are on the right path. You do need to find a different SSR to use for your contactors but there are suitable parts available.

[Edit]

I mention that contactors with AC coils have significant inrush current. The datasheet for the contactor will probably use terms like "sealed current" and "un-sealed current". The "sealed current" is the current that the coil draws when it is fully engaged (the armature is "sealed" to the stator coil).

The "un-sealed" current is what the contactor coil draws when the armature is fully extended from the stator coil. This current is significantly higher than when the armature is sealed.

If your contactor datasheet doesn't give you the un-sealed current rating or if you don't have access to the datasheet because this is already-existing equipment, you can easily measure the un-sealed current yourself. Simply block the armature so that it can't pull into the coil and apply power. Be sure to block it right at the top end of its' travel - the current begins to drop as soon as the armature starts to enter into the coil.

But be really quick when doing this - the coil will burn up if you leave power applied too long while the armature isn't sealed. It shouldn't take more than a few seconds to measure the current and the coil won't be harmed at all - so long as this does take only a few seconds.

Answer 2

What you want to look at is the forward voltage of the LED, not the reverse breakdown voltage; The latter is useful if you have the diode in an application where it may encounter sufficient negative voltage with respect to its cathode terminal. If this happens the diode will begin conducting in the reverse direction.

Also note that the parameters you have posted in your question are the maximum suggested operating conditions, not the nominal that you should be using under normal conditions.

Since your supply is 5V as you mentioned, the LED will drop 1.5V, so 5-1.5=3.5V will drop across your resistor. The screen capture above shows 5mA typical forward current, so a few simple calculations and:

$$R_{series} = \frac{5-1.5}{5mA} = 700 \Omega$$

Your resistance of 560 will give you 6.25mA of forward current which is within the operating specifications of this device.

As for the orientation question you asked, yes. Pin 1 here

Is the same as the pin nearest the indention shown here