[aegfurn]

I read somewhere on this site that you should wire in a diode when using a relay. I want to use a Grayhill 70s2 solid state relay to avtivate a 24v pneumatic valve from one of my parallel port pins. The info for the relay says that the input is optically isolated from the output, to protect from "transients". Do I still need the diode?

[Bill Berkley]

I'm new to milling but I have designed and installed a lot of computer control systems.
So here are some basics. Whenever a control voltage to a coil, such as the solenoid on your valve, is turned off the collapsing magnetic field will generate a high voltage spike of oposite polarity. This spike will usually exceed the safe operating voltage of your control relay. So it is always a good idea to protect from it. If your solid state relay switches a DC voltage a properly sized diode will do the trick. The diode should be placed as close to the solenoid as is practical. Just to give it an extra instant to do its job. It also needs to be connected so that it doesn't conduct when the control voltage is applied to the coil but only the spike which will be oposite polarity of the control voltage.
Hope this helps.

[aegfurn]

Thanks Bill,
Just to make sure I understand you correctly, I should wire it like the picture?

[Al_The_Man]

Just a couple of points till Bill gets back to you, are you sure the printer port will be able to drive the SSR? check the specs, as they usually require about 8~20ma, also is the 70S2 a DC output type? The ones I have come across are AC output.
Also the diode needs to be corrected, leave the end connected to the Solenoid and connect the other end to common, then connect directly from SSR out to Solenoid. ( diode + and SSR out connected to Solenoid ).
The diode will then be in parallel to the Solenoid.
Al.

[Bill Berkley]

The diode actually has to go between the coil terminals on the solenoid. What it does is switch the spike back through the coil before it can get to your relay. I'll figure out how to make a drawing and attach it to my next post, in a little bit.

[Bill Berkley]

aegfurn
I have attached a drawing that I think will help.

Bill

[kreutz]

Magnecraft's 70S2 series include DC and AC load SSRs. In a DC circuit, Diode protection is necessary when driving inductive loads . Make sure you check the data-sheet for the exact model number you have.

If you are going to drive the SSR from the parallel port the safest way is to connect the anode of the internal LED to +5Volts, and the Cathode to the parallel port output pin, then you will drive it "Active low", the output pin can sink more current (output Low) than the current it is able to provide when in "high" status. Make sure your SSR model does not need more than 20 mA, otherwise you will need a buffer.

[gar]

070223-2112 EST USA

kreutz:

When you present a question like this you need to provide more information.

First, is your solenoid supplied from AC or DC. I believe all the responses assumed it was excited from DC. Which may be correct, but I have no a priori way of knowing.

Second, with many (close to most) solid state relays you need to know whether your are switching a DC or AC load to select the correct kind of SSR.

In the past SSRs for AC loads were usually SCR or Triac types. These require the load current to drop below some holding value before the SCR or Triac will turn off even though the input excitation to the SSR was removed. This is accomplished automatically in an AC circuit, but not a DC circuit.

In the past for DC loads the output switch was typically a bipolar transistor. This are not suited for switching AC loads.

Third, we will assume that you are switching a DC circuit.

The above comments are correct that you need to limit the peak reverse voltage generated on turning off an inductive load.

I would describe the diode method as placing a shunt diode across the relay coil at the relay coil. Consider this combination as your relay coil. The important point here is that when you apply excitation to the coil that the + side of excitation goes to the cathode of the diode, and the - source to the anode of the diode. This applies a reverse bias to the diode and the only load on the source is the coil resistance. Virtually no current flows in the reverse direction thru the diode. If wired opposite to this the diode would be a short on the power source.

After a current is flowing in the coil there is energy stored in the magnetic field. When you break the excitation circuit to the coil, then the coil wants to keep the current flowing in the same direction, and thus instantaneously the coil voltage reverses and the coil is the source of energy rather than the load. This causes current to flow in the forward direction thru the diode, but the diode limits the voltage to the diode forward voltage drop. This prevents excessive voltage being applied to your switch.

In a DC bipolar transistor type SSR you have to observe the correct direction of current flow because it is a one-way device.

.

[kreutz]

kreutz:

When you present a question like this you need to provide more information.

First, is your solenoid supplied from AC or DC. I believe all the responses assumed it was excited from DC.

I did not ask the question, but I also assumed it was DC because of the diagram he posted.

If I am mistaken, and the original question refers to driving an AC solenoid, then using a snubber diode is not the solution, you will need to use an RC snubber across the solenoid terminals.

[holbieone]

The info for the relay says that the input is optically isolated from the output, to protect from "transients".

i think you answered your own question