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Beginner "reed relay" question

I would like to know why my output voltage drops from 4.5V to 1.3V when I connect a reed relay (N/O type) from this voltage to ground. Further, I also need to know how to overcome this drop. I have a power supply of 9V in the system.

9 Volts Vcc | | 0 or 5volts -----|------------------| 0 or 5volts -----| MULTIPLEXER | Output goes from 4.5V down 0 or 5volts -----| 4:16 | to 1.3 volts when relay is 0 or 5volts -----|------------------| connected. |||||||||||||||| | ----------------- N/O | | --->|------------------| | REED RELAY | Need relay to make --->|------------------| these line common | | with each other. | ----------------- 3Volts __|__ /////

I have a 5 volt reed relay from radio shack. And, I have also ordered a 3 volt reed relay from digikey. Perhaps, there is a better way to do this though. All I need to do is make the Nomally open contact common with the 3 Volt contact as shown above. However, I need to use the output from the multiplexer to do this. Later, this will be controlled with a PIC chip (5 volts). But, for now, I just need these two lines to become common when a particular pin from the multiplexer is high.

Thanks for any help you give, David

Reply to
dwfox1977
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Do This:

In words, take any NPN transistor, like a PN2222 or 2N3904, put a

4.7K resistor from the mux output to its base, its emitter to ground, its collector to the reed relay coil, and +Vcc to the other side of the reed relay coil.

Good Luck! Rich

Reply to
Rich Grise 

--
Your voltage is dropping because the relay coil resistance is too low,
but why on earth are you using a 9V supply?  What kind of a mux are
you using, a 4514?
Reply to
John Fields

Across the coil, such that it's reverse-biased in normal operation (cathode to +Vcc).

Sorry for the oversight.

Thanks! Rich

Reply to
Rich Grise

Thanks for the help everyone. I will definitly try this one out. I am using a 74HC238N Mux. For the power supply, I am using 9V to power a preamp, a decoder, 2 Mux's, and 2 transistors. Later there will be a PIC chip, chipcorder (check it out), and others to go with this list. But, the transistor method seems to be what I need to switch the relay on. Thanks, David

Reply to
dwfox1977 

--
And the diode???
Reply to
John Fields

I concur with Joerg here. But that would be fairly simple; since their supply current is so minuscule, a 5.1V zener and maybe 470 ohms would make a dandy little shunt regulator. This is off the top of my head - you could probably get away with much more resistance; I've never had any luck at all trying to design Zener regulators.

It would have worked the way David had it at the beginning - HC is notoriously powerful in the current-source department, but still,

20 mA is probably not enough to properly drive even a reed relay, unless Watson Name wants to spend the next three weeks winding the silly thing. ;-P

Thanks! Rich

(P.S.: Joerg, is that pronounced, "Yerg?" I've been saying, "Jorg," like "George" but with a hard final G. But when I checked for spelling, I found I've even been spelling it wrong! =:-O FWIW, my name is pronounced, "Rich," as in, speaking to a bar bimbo, "Hi. I'm Rich." ;-) My last name "Grise" has a long "i" and silent "e" - it rhymes with "nice", not the French city, but as in "nice guy." It also rhymes with "lice," so make your own decision in that respect. ;-) But it is _NOT_ "grease." Maybe "Greiss." Thanks!)

Reply to
Rich Grise

Awsome!!! It works perfect. This is exactly what I was looking for, thanks. I do want to know if this will drain my battery though? I would like to have the battery I use to last for at least a year or two. The system would only be used around 5-6 times in the time period. And, the only part of the system I would be worried about draining the battery is this part with the transistor, if it is a possible problem. I may be speaking out of ignorance of the transistor. But, I am learning still here. The transistor does not seem to be getting hot though.

Reply to
dwfox1977

What Mux are you using? I used to use TTL compatable reed relays from Pickering UK., normally connected from Vcc to the Mux output. You could try using a 74HC0( insert favorite IC) buffer driven by the Mux

martin

Reply to
martin griffith

Hello David,

Be careful. A HC238 doesn't like more than 6V.

Regards, Joerg

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Reply to
Joerg

If you use RLY-635 from AllElectronics, it should work directly from the mux. I'm assuming 1) the mux can source 10 mA, 2) and that the .5 amp contact rating of the relay is more than adequate for your needs and 3) that your existing reed relay draws more than 10 mA. You want a relay that draws as little current as possible to maximize battery life, and this one draws only 10 mA and is cheap $.65).

If your mux can't source 10 mA, you can still use your transistor circuit, and replace the existing relay with RLY-635 and a

470 ohm resistor in series with the new relay. Whatever you end up doing, don't forget to add a diode across the relay coil "backwards" (with the arrow pointing toward +).

Ed

Reply to
ehsjr

The '238 is not a "MUX"- it is a decoder. And you can't run the 74HC family off 9V Vcc, it should be 5V or less.

and -

Generally the relay voltage refers to the voltage applied to its coil and not the contacts. So you would want to use a 5V or 9V relay in your circuit.

Reply to
Fred Bloggs

Yeah, I have a 5V regulator to handle the voltage situation. Thanks, ehsjr, for the information on the diode across the coil. I will try this out. I greatly appreciate everyones help through this.

Reply to
dwfox1977

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