How to wire a Reed Relay

Let's forget about your experiments, other than to say it is obvious you don't know anything about electricity... The old adage that " a little bit of knowledge is a dangerous thing" applies in your case, but don't take this too personally. It is usually a good idea to ask the questions BEFORE you start to blow expensive items up due to lack of knowledge.

To begin with;

The relay in question is a DIL 5V, 200 ohm coil polarised relay with a single make contact (plus diode for user configuration).

You must apply +5V to terminal 6 and -VE to terminal 13 in order to energise the relay.

The diode is for user configuration. In the usual arrangement the diode is connected across the coil (insert strap between 13 and 9) to SUPPRESS the back emf from the coil during release. The disadvantage of this method is that the release time of the relay is increased considerably but if this isn't a problem then connect it like this.

The diode can also be wired in series with the coil by connecting +VE to terminal 9 instead of 6 (leave 6 open or no connection). The diode will now BLOCK the coil back emf during release to prevent damage to the relay driver and will have the advantage of minimising the release time of the relay. A small voltage will be dropped across the diode (0.7V) but since the relay will operate reliably down to 3.5V this shouldn't be a problem.

You will need to check that your relay driver TTL output can source or sink at least 25mA depending on your arrangement. What TTL driver are you using?

Well, "John Smith", out of a total of your first four posts to this newsgroup, three were insults and one was essentially parroting my post, which you obviously read.

Now, had you paid any attention at all, you would have noticed that I wrote "What do you mean by TTL?" , _not_ "What is TTL?".

Also, had you paid any attention at all, you might have noticed that the OP was trying to look less ignorant by using acronyms like 'TTL' and 'CMOS' which, to him, might have widely different meanings than to someone with a working familiarity with the term. Also, had you been paying attention, you might have noticed that the 200 ohm relay coil would draw 25mA from a 5V supply. Interestingly, the OP stated that he wanted to use a logic _high_ to drive the relay, and had you been paying attention to that and been familiar with TTL yourself, you would have realized that standard bipolar TTL _can't_ supply that kind of high drive. Even using it for low side drive, which it was designed for, will only get you 20mA from a Schottky.

So, with all that in mind, maybe it's dawned on you that my question to the OP was designed to find out what he knew about TTL so that if he posted back other drive recommendations could be offered.

Finally, had you been paying attention, you would have noticed that he was using a 5V supply to try to drive the relay, and that a 5V output certainly is _not_ TTL.

In conclusion, I think it would be a good thing for you to learn to keep a civil tongue in your head before you get yourself into any more trouble.

--- No, you stupid f*ck, +5 goes to pin 6, and pins 9 and 13 go to ground or to a low side driver. TTL can't supply the 25mA needed to drive the high side of the relay, plus, your'e trying to drive the coil in series with the diode, when what's supposed to happen, if you use the diode, is that it goes in _parallel_ with the relay coil to soak up the spike when the driver turns off.

Like this for low drive:

And like this for high:

GND--[R]---B PNP C | +--------+ | | [DIODE] [COIL] |K | +--------+ |

-- John Fields Professional Circuit Designer

Ooops...

Should be:

....and not all TTL is 0-5V, and not all 0-5V "TTL" is, in fact, TTL. ;-)

Hi Ross,

Thank you very much for your answer and time

What does VE stands for?

release

or

If relay driver TTL output current is the output current at the terminal of the TTL chip, im getting 735 mVolts and 735mA . The TTL driver is a 74F00 I got those values setting the current prove sensitivity to 1 V/A (Unfortunatelly I don't know what is the meaning of V/A and neither if I actually I'm measuring the real current, sorry about that)

As for now, I'm testing the relay as follows:

Terminal 1 connected to 330 Ohm Resistence and resistence to Power Supply's GND Terminal 6 connected to Power Supply's +5 Volts Terminal 7 connected to a LED and LED to Power Supply's +5 Volts Terminal 9 and 13 connected to Power Supply Ground Terminal 14 and 8 are not connected (Seem to be the same as 1 and 7 respective)

Terminal 2 has two "states":

a) Connected to Power Supply's +5 Volts -> The LED emits light b) Disconnected from Power Supply's +5 Volts -> The LED doesn't emit light and the relay makes click.

If the test connection is ok, the following step would be to wire:

- output of the 74F00 to Reed Relay's Terminal 1

- input of a 74HC to Reed Relay's Terminal 7

- output of an other 74F00 to Reed Relay's Terminal 2

As much as I can assume, the terminal 2 of the reed relay chip is the one that controls the "open" and "close" states between terminals 1-7 and respective 14-8. Is this assumtion right?

So, do you have any new suggestions?

Thank you very much and sorry about my electronics/electricity languaje

Best Regards

Without trying to prempt John's reply, you can use IC based TTL or discrete bipolar TTL. John was probably asking to find out if the OP knew the difference or indeed if he knew there was other than IC type TTL.

"professional" and not understanding TTL???? Get real.

Comments interleaved

You blew up the diode. Both sides of the coil were at +5V

? Not possible. You still haven't energized the coil, so the contact should still be open.

No, they are on opposite sides of the contact switch.

As Q1. One side of the switch is connected to pins 1 and 14, and the other side of the switch to pins 7 and 8.

You should be able to, but you now need your own back EMF prevention diode.

Wire (a new one, with an intact diode) up as follows Pin 9 to +5V (i.e. to your TTL output - but make sure the output can handle the current!) Pin 13 to GND This will energize the coil, closing the contacts.

:)

(Not having a go at you John, it just made me smile! :) )

Shorthand abbreviations for; POSITIVE = +VE NEGATIVE = -VE

Comes from symbol + (plus) which has traditionally been used to symbolise POSITIVE (pole of battery), and - (minus) for NEGATIVE (pole of battery) when related to direct current working.

correct

correct

correct

correct

correct, 1,14 are commoned and 7,8 are commoned

This is NOT correct. Terminal 2 is simply connected to a single wrap turn of copper foil (open circuit at ends) around the complete relay internally. The purpose of this is to minimise the effect of external electrostatic fields inducing stray currents into the coil or the reed element. Terminal 2 is not electrically connected internally to the coil itself or the contact, and if your tests are correct then the indication is that there has been an insulation breakdown between the shield and the other terminals internally. Use your ohmmeter to test whether terminal 2 is electrically connected to any other terminal of the relay. It should not show anything other than open circuit. Make sure the relay is disconnected from your circuit. In operation terminal 2 would normally be connected to ground (GND).

????? 74HC

A 74F00 gate cannot drive a relay directly due to maximum current limit of 20mA.

I suggest you do some reading on driving relays with logic before going further

Incorrect, see above.

Hi Ross,

Aga> On 31 Mar 2005 11:18:46 -0800, "Lathe_Biosas"

I was told than an Analogue Switch could help do the job. That is, switch between to lines: one that connects CMOS with CMOS technology and the second one that connects a 74F04 chip to the a CMOS chip. The Analogue Switch would be driven by a 74F04.

Best Regards

Hi John,

Thank you for your answers.

It was a nice feeling when copy pasting your proposed circuit and changing to Courier, the circuit was then very clear.

I think I will change the reed relay and place instead an Analogue Switch. I will now try to find information on the Analogue Switch

Best Regards