# Abusing the MAX232

• posted

=20

What would be the difference, Rick ?

The circuit was a simple [PSU] -> breadboard for ease of testing -> [relays= ]=20

... where each relay has a common connection to 9v, and an individual conne= ction to ground. The relays only switch at 9v (well, about 8.2v actually, b= ut let's use 9v since that's what they say on them). It seemed to make sens= e to me that that was the current required, since the measured resistance w= as 163 ohms per relay in isolation, and=20

9 =3D 163 * I =3D> I=3D 9/163 =3D ~0.055A 8 * 0.055A =3D 0.44A

... and the bench PSU was reading 0.4A (it's only got one significant digit= after the decimal point).=20

So (1) there's nothing else going on in the circuit, and (2) the relays req= uire 9v (or just under, to be pedantic). I can buy that the hold-in current= would be less than the activation current, but I'm not getting why the 440= mA isn't a good approximation to the activation current.=20

For what I want, it's entirely possible that the system could come out of a= power-failure mode and want to switch on all the relays immediately, so I = have to be able to supply the maximal activation current.

Cheers Simon. (Always willing to ask stupid questions in order to learn :)

• posted

If I could use 5v relays, I'd be there like a shot. The problem is that I want to plug in mains power-cords, don't really want to make my own PCB that handles the mains, and the chauvet enclosure is (a) reasonably cheap and (b) built relatively well.

Cheers Simon

• posted

connection to ground. The relays only switch at 9v (well, about 8.2v actually, but let's use 9v since that's what they say on them). It seemed to make sense to me that that was the current required, since the measured resistance was 163 ohms per relay in isolation, and

after the decimal point).

require 9v (or just under, to be pedantic). I can buy that the hold-in current would be less than the activation current, but I'm not getting why the 440mA isn't a good approximation to the activation current.

power-failure mode and want to switch on all the relays immediately, so I have to be able to supply the maximal activation current.

Yes, asking questions is exactly how you will learn... assuming you get answers from people who actually know... ;)

First I want to point out that the activation voltage or current is a property of the relay and is independent of your design. The fact that you are using 9 volts and 55 mA does not make that the activation voltage and current.

The pull in voltage or current is the minimum voltage or current that is guaranteed to pull a relay to the active position. You are powering the relays with 9 volts, sort of, but what if they will pull in with only 7 volts? Then your circuit only has to provide 7 volts rather than 9.

Likewise the hold voltage or current is the minimum current required to maintain them in the active state. If the voltage or current drops below this value they might release. If this voltage is 5 volts, then your circuit only has to maintain the activation voltage for a short period (from a cap would do the job) and the converter circuit only has to provide this voltage to keep the relays activated.

Another spec is the drop out voltage or current. A voltage or current below this value is guaranteed to cause the relay to return to the inactive state. You have to lower your current or voltage below this value to make sure the relay releases.

If power failure or other issues are really a design requirement, then you can design the circuit so that the relays don't all switch at the same time even after a power failure.

Rick

• posted

=20

Right, but as I said above, it takes 8.2v to *just* switch the relay (it so= unds like a more ... tentative... switch than the positive *click* at 9v), = and they are marked as 9v relays. I was therefore assuming that 9v (or to b= e pedantic, just under) was the activation voltage, and the activation curr= ent would be whatever the circuit pulled from the power supply at that poin= t.

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Yep - like I say, it takes 8.2v to trigger the relay. It definitely doesn't= work at 5v :)

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I guess I'd have to activate the relay and then dial down the PSU to see wh= at voltage the relay deactivated at, in order to find the drop-out voltage,= then. Presumably the hold voltage lies somewhere just above the drop-out v= oltage. I could try that and see what I get.

Yeah, I thought of that, and it's under microcontroller control, so it'd be= easy to do. I'd just have to allow time for the voltage-reserve capacitor = to charge up again.

It does seem a little less robust as a scheme, though. What if I put anothe= r 75' of cable between the relay pack and the controller (this is a possibi= lity, the controller is mainly based in the house, but it's possible I migh= t want to be controlling things in the shed, which is about 100' away). I d= oubt I'd have any issues with a 1A 9v wall-wart...

Cheers Simon

• posted

Well for future reference consider Pickering 101 Series, unless you are switching heavy loads. Max 8mA coild current 5V or 3V3

One of my designs has banks of 12 relays being switched in and out (total of 432 relays required in design only 12 active at most) All 7 boards are powered by USB device running at 200mA limit on port. Includes LED feedback and simple error detection communications.

....

```--
Paul Carpenter          | paul@pcserviceselectronics.co.uk
PC Services```
• posted

I'm well aware that there are easier relays to use. What I want is a ground= ed metal box with 8 mains-sockets each of which is controllable by a relay.= I don't really want to get into the manufacturing of said box because unti= l you get to pretty high volume, custom enclosures are expensive. I don't h= ave that high volume...

The Chauvet SR-8 is ~\$40, is metal (and properly grounded), has the 8 contr= ollable sockets, and someone else has the liability if it goes on fire. If = there's an equivalent product that uses friendlier relays, I'm all ears, bu= t relays on their own aren't my problem, it's the package as a whole that's= the issue.

Simon.

• posted

nded metal box with 8 mains-sockets each of which is controllable by a rela= y. I don't really want to get into the manufacturing of said box because un= til you get to pretty high volume, custom enclosures are expensive. I don't= have that high volume...

trollable sockets, and someone else has the liability if it goes on fire. I= f there's an equivalent product that uses friendlier relays, I'm all ears, = but relays on their own aren't my problem, it's the package as a whole that= 's the issue.

considered opening the box and changing the relays? often relays come in various voltages in the same footprint

-Lasse

• posted

Since he mentioned liability, I suspect he's worried about modifying the existing hardware as that may incur additional liability on his part.

Simon. (A different one. :-))

```--
Simon Clubley, clubley@remove_me.eisner.decus.org-Earth.UFP
Microsoft: Bringing you 1980s technology to a 21st century world```
• posted

Yup. Replacing them would actually be feasible, but in terms of liability I= might as well paint a target on sensitive parts and hand out hob-nailed bo= ots :) Even if the device was going to fail anyway, once it had been modifi= ed they'd disclaim any liability, and rightly so.

Cheers, Simon.

• posted

sounds like a more ... tentative... switch than the positive *click* at 9v), and they are marked as 9v relays. I was therefore assuming that 9v (or to be pedantic, just under) was the activation voltage, and the activation current would be whatever the circuit pulled from the power supply at that point.

work at 5v :)

voltage the relay deactivated at, in order to find the drop-out voltage, then. Presumably the hold voltage lies somewhere just above the drop-out voltage. I could try that and see what I get.

That's one way. I wouldn't make a measurement like this and rely on it being accurate over various conditions, temperature and aging are two that come to mind. I also wouldn't trust the pull-in measurement either. I have been assuming you could get your hands on the data sheet for the relay in question.

easy to do. I'd just have to allow time for the voltage-reserve capacitor to charge up again.

75' of cable between the relay pack and the controller (this is a possibility, the controller is mainly based in the house, but it's possible I might want to be controlling things in the shed, which is about 100' away). I doubt I'd have any issues with a 1A 9v wall-wart...

If you use 28 gauge wire, that might be a problem, 6.5 ohms per 100 ft, but 22 gauge is under 2 ohms per 100 feet. Did you ever really say what your setup is? You have been talking about a controller, is this connected by a 100 foot cable to the PC? That won't work as USB has a limit of 5 meters between repeaters (about 16 feet). So I assume the controller is next to the PC and the cable to the relays is a bundle of wires that you are running, no? So why can't you put the wall wart next to the controller and PC and run 20 or 22 gauge to the relays? Actually, the cable to the relays can be more like 26 gauge if you use multiple returns for the relays.

Actually, I think I'm not sure anymore what you are asking. This whole thing seems rather clunky. I think I would run a single pair and put power and signal on the same pair of 20 gauge. Then I would put my controller next to the relays with a driver from the PC. I suppose this would require two units, one at each end, but both can be the same design, with different jumpers for driving vs. receiving power on the cable.

Rick

• posted

It's still simple enough. There's no PC, there's an ARM device running linux with a couple of USB outputs, but it looks like:

[arm device] [controller]
• posted

tiistai, 28. elokuuta 2012 2.11.39 UTC+3 Simon kirjoitti:

being bus-powered.=20

end of a 25' wire (DB9DP9). This part is fixed because it's an off-th= e shelf part.

und) in voltage-doubling configuration. That worked fine until I actually w= anted to use the relays [grin]. The 7660 could produce 9v with no load, but= as soon as I connected just one of the relay pins, the voltage dropped to = ~4v and the relay failed to switch.

at it's supposed to be used for, but all I really want is +10v from the USB= 's +5v that will drive 8 relays... I'm supposing that the RS232 drivers wou= ld probably be specced to drive the voltage down a long serial wire anyway,= so maybe they'll do better ?

er (and basically make it not bus-powered) and then there's no problem, but= if there's a good way of avoiding that I'd like to know about it :)

Another alternative is to forget the mechanical relays and use solid-state = relays (ie. opto-couplers).

- Calvin

• posted

with a couple of USB outputs, but it looks like:

Ok, that's what I didn't understand. The wall wart wouldn't be in the shed with the relays, it would be by the PC so not really a big deal, just a small one. Ok. I got it now.

Rick

• posted

metal box with 8 mains-sockets each of which is controllable by a relay. I don't really want to get into the manufacturing of said box because until you get to pretty high volume, custom enclosures are expensive. I don't have that high volume...

controllable sockets, and someone else has the liability if it goes on fire. If there's an equivalent product that uses friendlier relays, I'm all ears, but relays on their own aren't my problem, it's the package as a whole that's the issue.

Then yeah, the wall-wart solution is your best one. Simple, easy, and gets you done with the project. By the way, that little Chauvet relay pack just went into my "One of these days, I'll find a project that uses a..." box.

```--
Rob Gaddi, Highland Technology -- www.highlandtechnology.com
Email address domain is currently out of order.  See above to fix.```
• posted

Hmm - re-reading that it came over as a bit snarky. No offense meant, I was just trying to explain what I wanted.

Cheers, Simon

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