Off topic = PIC -> ULN2803A -> Relay

[hillcraft]

I know this is off-topic, but I am pulling my hair out. I hope that somebody can help me solve a problem with a 18F458 driving a ULN2803A which in turn drives a number of 12V Finder 44.52 relays. I have had a couple of the ULNs destroy themselves totally.

The questions that I have are as follows:
1. Should I be using a BACK-EMF diode i.e 1N4001 across the relay with the cathode side connected to the positive rail of the relay?
2. What should be done with pin 10 (common) of the ULN? Should this be connected directly to the 12V rail, should it be left unconnected, or should it be connected to the 12V rail via a 13V zener diode?
3. Would it help to connect a Mov across the contact side of the relay against transients?
4. I have heard and seen people using transorb diodes - Is there any merit in using them, and how would you use them for transient suppression?


So esentially the question comes down to this: "How can I protect the 12V and the 220V sides of the relay?

Thanks in advance.

[PCM programmer]

The data sheet for your relay can be downloaded at the link below.
I assume you're using the top one in the chart, which is for 12 volts.
http://www.connex-electronics.com/?url=/html/products/finder/44_series.html
The data shows that the coil current will be about 50 ma which
is well within the limits of the relay driver specs.

You should add a diode across the relay coil terminals, as
as you suggested in your post. The cathode should go to
the +12v terminal and the anode to the other terminal.
I think that will fix your problem.

[hillcraft]

Do you have any idea what I should do with the common pin, pin 10 of the ULN?

[PCM programmer]

You could connect the COM pin to your +12v supply, and then the
internal diodes in the ULN2803 would suppress the "back emf".
However, it seems to me that it's better to have a diode placed
directly across the relay terminals because the path is much
shorter. If you use the internal diodes, you would have to
run a wire or a printed circuit board trace back down to the COM pin.
I don't know how far away your relays are from the driver chip.
If it's a very short distance then I suppose you could use the
internal diodes.

[asmallri]

[Mark]

We use a ULN2803A in several of our designs. We usually drive a Zettler AZ2150-1A-DEF relay. No back EMF diodes but we do use a MOV across the contacts. The COM goes to 12V with a 0.1uf cap to ground. These units have been in production for at least 8 years. There are probably 50K-75K units in the field and seem to work just fine.

[valemike]

I'm currently using a ULN2004A, which has internal diodes, and 5V relays. Nor do I have any external diodes. Haven't blown any components yet, as i'm under the assumption that the uln2004's internal diodes are enough. But I have blown a couple ICD-2s

There are twin relay drivers (aka inductive load drivers) from ON Semi called the NUD3112. Hook it right up to your PIC's pin.

[treitmey]

My most recient design is using a tpic6c595

I think of relays as a slow IO.

This is a serial to par io (like 74hc595)
and has the relay driver and diodes built in.

Drive ANY number of relays at the cost of 3 pins.

[bfemmel]

Depending on the layout of the board and the amount of back EMF generated by the coil you may find that the 1N400x series may not be fast enough to turn on before the voltage has gone over the limit of your driving part. The internal diodes of the ULN driver are a good back up to the slower external diodes. For most all of my designs I end up using a US1A diode from Diodes Inc. across the coil. Digikey carries them. They have a trr of 50ns.

- Bruce

[PCM programmer]

[valemike]

I have 5VDC relays. Though I do have an internal back emf protection diode on my darlington driver IC (ULN2004), this just protects everything connected to the 5V bus, I think. However, the 5V rail is not protected, is that true?

I use power from the ICD-2, which also supplies my 5V. The moment I plug in the ICD-2, my relays energize. Once in a while, I blow the power supply in the ICD-2. After that my 9V adapter no longer works.

Can these relays getting energized cause my ICD-2 to absorb all the inductive kick and thus fail forever?

[hillcraft]

I find what you say about the ICD2 energizing the relay quite interesting. I have a Rigol 60MHz digital storage scope. When I touch the ground side of the probe to the negative pole of the relay (even without the probe tip touching the circuit) - the relay also energizes. The relay seems to energize quite violently. The scope does not seem to affect, or be affected by) any other part of the circuit.

Maybe you and I have a similar problem.

[Ttelmah]

Key thing is, 'where does the energy go'. When you switch off the solenoid driver IC, there is a lot of energy stored in the inductor, that needs to go somewhere. The protection diodes in the driver, prevent it generating a massive overvoltage spike, and killing the driver IC, by diverting it into the 5v rail. Now if the loads on the 5v rail are small, and dependant on the capacitance values present, the result will be to raise the voltage of this rail. You can work out how much it will go up, from the inductance involved, the effective serial resistance of the diode and traces, and the capacitance present on the system. The odds are that in your case, the voltage increase is enough to spike the ICD-2. The diodes don't protect either the 5v rail, or the '5v bus'. All it protects is the 2004.
Now you can trap the energy in a lot of ways. The worst of the spike, can be stopped quite effectively by a resistor and series capacitor (the classic 'snubber' network). MOV's can also be used. However if trap diodes are used, the key thing is to ensure that there is a network present close to the driver, that can stop the 5v rail from being driven up too much.
For my own use of similar relays on 5v logic, the 'first' line of defence, is that the 5v rail for the logic, has 5.5v MOV's across it as part of the normal protection circuitry. The 5v feed to the relay drive circuit is via a small inductor, and series resistor, which limits the power that is delivered to this part of the circuit. There is then a large low ESR capacitor with good HF performance, across the supply rails close to the clamp diodes.
It is up to you to provide the energy storage, and discharge paths necessary to safely trap, and then dissipate the energy.

Best Wishes

[PCM programmer]

[valemike]

Oh well, I haven't busted the more robust ICD-U40.
Nor have I busted any other components on the board, but it's sad to know that the possibility does exist, since the board's been laid out twice already. oh well.