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cdarga (Automotive)
21 Nov 08 14:28
Hello,

Let me start out by mentioning that I am in NO WAY an expert in the electrical dept...so please forgive me.

I have a DAQ system with an open collector output with a 100k pull up resistance (not really sure what that means). I have 2 wires: COM ground (emitter?, pink) and alarm output (collector?, yellow). When I measure the voltage across these, I get 5V with the alarm in an "off" condition. When the alarm is activated, the voltage between these goes to 0.

What I need/want to do is use this output to send a 20V signal to a test machine for a shutdown if the alarm is activated. I tried to use a normally closed relay, but I am unable to activate it using this output signal. I'm at my wit's end guys (and/or gals). Any help would be greatly appreciated.

Thank you.

Christian
cranky108 (Electrical)
21 Nov 08 14:57
I'm a power guy myself, but if I recall an open collector can sink current to common, but can't supply DC for a positive output, hence the pull up resistor.
If the relay is connected collector to the positive DC, and the impedance won't allow an over current into the open collector on a zero output (not zero current), then the relay should try to pickup.
The pull up resistor won't matter to the relay. However, a freewheeling diode maybe required across the relay to prevent back emf.
Helpful Member!  BobM3 (Mechanical)
21 Nov 08 15:28
zappedagain (Electrical)
21 Nov 08 16:33
That diagram looks pretty good.  Make sure your open collector can sink more current (Iol - low output current) than the relay requires, otherwise it might not work reliably.  

A regular diode will work as well as the zener diode in this case, and if you have a different voltage relay you don't have to tie it to the 20V supply.  If you use a 5V relay make sure to add a bypass capacitor (10-100uF or so) by the relay so it doesn't clobber your 5V supply.  
 
cdarga (Automotive)
24 Nov 08 9:35
Thanks so much BobM3! That diagram is exactly what I needed. Thanks to the others who put in their input as well.

Now, 1 more question, there is a diode already implemented in the DAQ device (no specs given for it). Since I'm not sure of the specs on the diode in the DAQ system, it probably wouldn't hurt to include the diode you mentioned anyway, right? It's wired: +5V -->diode-->100k resistor, all internal to the DAQ system (shown in the attached schematic).

I attached a schematic. It's Bob's circuit (thanks) with a couple of things added. I want to fully understand what I need to do before I wire anything up. Please excuse the crudity of it. Hopefully one of you guys could tell me if it should work or if I'll blow myself up...ha.

Just to reiterate, when the DAQ sends out an alarm signal, I want it to power the relay and send the 20V signal to shut down the machine, so I think I need a normally open relay. At no alarm condition, there is 5V between the collector and the common (from the DAQ).

Thanks so much everyone for the help!!

Christian

 
BobM3 (Mechanical)
24 Nov 08 12:05
Looks o.k. to me.  You'll still need to protect the DAQ's transistor with the zener or some other transient protection.
cdarga (Automotive)
24 Nov 08 13:10
Thanks again Bob.

The rating on my transistor for max current (connector current) is 0.5A.

For the zener diode, would any size below 0.5A work? For example, I see a diode that's rated at 24V, 5.2mA. This means the most current that could flow "backwards" would be 5.2mA? This is obviously under the max rating...so this should work?

Christian
Helpful Member!(2)  Skogsgurra (Electrical)
24 Nov 08 17:03
Sorry, but I feel like we need to tidy up a few things here.

First. The 100 kohms plus the diode could as well be non-existing if you add a relay coil and connect it to an external DC source higher than the internal voltage (+5 V).

Second. The rating of the zener (current-wise) does not mean that the rated current (5.2 mA) flows in it. There should not be any current in the zener until you turn off the transistor. There will then be a short decaying current circulating in the coil/diode combination until losses have consumed the energy in the coil.

Third. It is rather unwise to use a zener diode. Mainly because it is an odd component, but also because the zener property will never come to good use in this application. A standard 1 A diode will work very well. Use the 1N4007 for example. Or 1N4002, which is also adequate, if you can find one.

Fourth. About the zener again. There is a very narrow tolerance between the zener voltage and the 20 V coil voltage. The latter isn't often very well regulated (if you are not using a stabilized power supply, which seems umlikely here). That means that you can easily overheat the zener if its zener voltage is on the low side (quite wide tolerances there) and your 20 V rises a bit. Zener voltages are usually given a bit down on the "zener knee", so a diode cannot be regarded as non-conducting when working close to rated voltage. The risk is all the bigger since there is no current limiting resistor in series with the zener. Operating a zener without any current limiting device, like a resistor, is begging for trouble.

To sum up: Put an ordinary diode across the coil. Catode up - as Bob has shown.

 

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

biff44 (Electrical)
15 Dec 08 21:53
I don't like that circuit so much.  You have a 24 volt relay/power supply hooked to a 5 V supplied chip.  If for any reason the voltage on the "open collector" output goes much above +6 volts, the transistor can breakdown and blow up.  I would take your open collector output and have it drive another transistor inverter stage, but use an external transistor that has at least a 60 volt VCE0.  Don't forget, switching coils can have big voltage spikes on them.
Skogsgurra (Electrical)
15 Dec 08 22:47
biff44

You are overly pessimistic. I have never (40+ years in drives and automation) seen what you describe. And I cannot see the risk either. Also, there is a transient protection in the diode. No kickbacks above diode fwd drop here.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

LionelHutz (Electrical)
16 Dec 08 8:28
Agreed with Gunner. Use a normal diode.

The diode internal to the module is to prevent current from flowing to the 5V supply when you connect the module up like you are planning to do. If that diode was not there then current would flow from the 20V through the relay coil and 100k resistor into the 5V supply. If you do this on a number of outputs and you could end up pulling the 5V supply greater than 5V and destroying the module.

If you look in the module spec you should also find a maximum voltage for the output transistor. Make sure it is greater than 20V even though I'd be very surprised if it is not.

 
zekeman (Mechanical)
2 Jan 09 14:48
The  100K "pullup" does nothing for you. Why unecessarily tie your 5 volt supply to the external circuit. The collector is already pulled up by the relay coil.
The circuit will work just fine without it. But as others said , all you need is a 1 amp diode. e. g. 1N 2002 or 1N 2003 to protect the transistor on opening the relay; at that time the diode will allow the current to safely decay, with a voltage rise of forward voltage diode drop + 20 volts. There is no further need or reason to protect the transistor.
roydm (Industrial)
19 Jan 09 9:03
I agree with Zekeman "the 100K pullup resistor does nothing for you" however if you were feeding the output into another high impedance input referenced to zero then it might. The manufacturer has included it as cheap insurance.
biff44 (Electrical)
23 Jan 09 19:59
Well, 2 comments:
1) I just worked on an engineering failure analysis where a client had a similar system, only using power fets in what I guess you could call an "open drain" configuration. 10% field failure rate due to the fets blowing up.  A major problem that was going to lose this guy his client.  So maybe you guys do not get out enough?

2) what is the breakdown voltage of the open collector transistor in the DAC chip?  If it is an integrated circuit, it might only be an 8 volt Vceo.
ScottyUK (Electrical)
24 Jan 09 2:52
Biff,

Anybody can design a bad circuit from good components. Open collector interfaces have been around since before I was born; if the topology were inherently prone to failure, as opposed to sometimes failing because of bad design, it would have fallen out of favour years ago yet plainly it has not.
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

Johnee2 (Electrical)
25 Jan 09 11:47
Hi,  

I've been lurking but felt compelled to chime in.  Since you said this is an output for monitoring an alarm & shut-down conditions (safety related) I strongly agree with the //conservative// design guidance, particularly from Biff.  Of course it all depends on the specs of the DAQ.  I haven't seen any mention of this model, manufacturer, etc.  

In any case, I think the 5V pull-up is a 'hint' from the manufacturer that this is a logic-level I/O, not suited for externally applied voltage more than 7.5 Vdc.  PLEASE consider using one of the industry standard DC Output modules from Crydom, Grayhill, Omron, Opto22 etc. http://tinyurl.com/ckfpvh

These are optically isolated, use 5V control input, and support up to 60V DC on the output.  They can be purchased from catalog outfits like Mouser, Digikey, Newark etc, for less than $15.  Mounting boards are available, or a module with screw terminals with more research.

In GENERAL a sure way to DAMAGE electronic I/O is to apply a voltage more than one diode-drop outside the chip supply rails.  Keep in mind this relationship is true when the DAQ is powered off as well, and Vcc/Vdd = 0 V.  I've seen no comment about sequencing the 20 V last-on and first-off.  This is in addition to all the other warnings.
Johnee2 (Electrical)
25 Jan 09 11:57
Sorry, the tinyURL doesn't work.  

The link was to the Opto 22 Part ODC5.  Everyone seems to use this name, so just GOOGLE "dc output module" for lots of results.
Skogsgurra (Electrical)
25 Jan 09 14:25
The OP mentioned a 0.5 A capability on his output pin. That is hardly a TTL or CMOS chip specification. More like a relay driver or discrete transistor. Hence my comment that he can use an ordinary diode and a 24 V supply.

With what we know so far, a relay coil and a free-wheeling diode is all that is needed.

It would be a lot better if we eventually could learn what board the OP is using. And the real specs.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

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