DavidH (Customer) asked a question.
Are Snubber Diodes Necessary On CLICK Digital Outputs?
This might seem like a really simple question, but I currently have a C0-02DD2-D PLC. I'm driving a couple of mechanical 24VDC relays using the digital sourcing outputs. They are low power, and I have always in the past connected a snubber diode in parallel with the relay. Looking at the CLICK wiring diagram, it looks like there is a snubber diode internally(I have attached a picture).
Would I be correct to assume that an external snubber diode isn't necessary in my application? Or am I completely wrong and these Zener diodes serve a completely different purpose and can't be considered protective snubber diodes.
Hi @DavidH (Customer),
Yes, you are correct. There is some protection on the outputs.
https://cdn.automationdirect.com/static/specs/c002dd2d.pdf
If you look at the Noise Immunity in the specifications:
Tested Impulse noise 1µs, 1000V - It is a very fine print.
This is also found on the digital output cards as well.
Regards,
Garry
https://accautomation.ca/series/click-plc/
Snubbing the relay still looks like a good idea.Take a look at the arrangement of the transistor, two dioes, and relay in that schematic. Notice the blocking diode between +24V and the transistor emitter.When the transistor switch opens, the back-emf from the relay will reverse-bias that diode, and no current will flow. That allows the back-emf to develop a high voltage, which stresses the output components, especially the blocking diode. The zener diode is connected in a way that it can't shunt the back-emf across the relay. It can only shunt the transistor. It looks like the zener is there to protect the open transistor from over-voltage from the power supply, [and more. See further down the thread.]Two informative articles on relay coil snubbing:
https://www.te.com/commerce/DocumentDelivery/DDEController?Action=srchrtrv&DocNm=13C3311_AppNote&DocType=CS&DocLang=EN
https://www.te.com/commerce/DocumentDelivery/DDEController?Action=srchrtrv&DocNm=13C3264_AppNote&DocType=CS&DocLang=EN
P.S.
Actually, I'm not quite right. Since the transistor is isolated through an optoisolator,only the blocking diode would be stressed. But the blocking diode could have a very high reverse voltage rating of hundreds of volts, so it would be hard to stress it. Still, the high voltage back-emf is nevertheless going to be there without your own snubber. Maybe the CLICK can simply take it, and you don't need snubbing.P.P.S.
Truly, I'm even less right. The back-emf is negative in respect to common, which means the blocking diode would be forward-biased. So the back-emf may actually be suppressed through the zener diode, blocking diode, and the assumed bypass diode across the power supply. See my response to Tinker below.
Thanks for the responses. So from the consensus here, I take it that a snubber isn't necessary.
While we are on the topic of protecting the digital outputs, how much should I fuse each output for? The data sheet says each output is rated for 0.1A, but the maximum inrush current is 150mA.
Since I will be using a fast blow fuse, should I fuse each output for 0.1A or 150mA?
Hi @DavidH (Customer),
A fast-blow fuse, with a lower current rating than the I/O bank’s common current rating, can be wired to each common. Or, a fuse with a rating of slightly less than the maximum current per output point can be added to each output. So this would be 0.1A or less per point.
https://cdn.automationdirect.com/static/manuals/c0userm/ch3.pdf
Regards,
Garry
https://accautomation.ca/
Since we're protecting transistors, the fuse has to be incredibly fast. We know the time and current the output is designed for (150 mA, 0.01 s,) so it should be easy to find the right fuse. Look at your potential fuse type's datasheet, find the time-current graph with all the current ratings, and see if any time-current curve passes through the area less than 150mA & 0.01 s.
Here's the graph for a Littelfuse 235 Series, Fast-Acting Fuse, with our specification marked in orange. It's surprising to discover that none of the ratings fit the specification; none of the curves pass through the upper left quadrant.
That means at 150 mA, even the 0.1 amp fuse won't blow until nearly a second has gone by. I looked at the 251/253 series very-fast-acting fuses for something that would fit the spec, and only one of them does:
While a fuse might be a good idea in principle, it is no substitute for proper suppression, if the fuse were to protect the output from the coil's back emf, it would blow every time you de-energized the coil, obviously not practical.
I don't totally understand what the Zener is for, I suspect it is more to protect against power supply spikes than load back emf. I think it would be a very good idea to provide additional suppression at the coil, keep in mind that even if (big IF, I don't see how it could be) the Zener is adequate protection for the transistor, all the current is flowing thru the connecting wiring potentialy increasing noise levels. While a snubber right at the coil would keep the spike off much of the wiring.
@Tinker (Customer)
The intention is to use fuses to protect the transistors from being destroyed by, I'm assuming, short circuits, or perhaps mistakes during experimentation, but not back-emf.
I think you're right about the current flow. Some back-emf current would go through the connecting wiring, although it's not as high as it would be across a single suppression diode. I'm not sure, but most industrial DC power supplies have a bypass diode in parallel with the output terminals. Perhaps there's also one in parallel with the +24v input terminals on the I/O module. Anyways, I think this is what happens after the transistor de-energizes the relay: The bypass diode, plus the blocking diode between +24V and the transistor emitter, plus the zener diode, would all form a current path for the negative voltage generated by the relay coil.
I know a standard diode across the coil can sometimes cause a long release time on the relay. That itself can cause arcing issues over time.