Question about relay voltage...

trixdropd · July 14, 2009, 09:26:45 PM

I have been running 5v relays for some time now. http://www.mouser.com/Search/ProductDetail.aspx?qs=sGAEpiMZZMs3UE%252bXNiFaVDVuZJnAgzXjZnSFmYYYj4s%3d
I run the standard 9v dc to them and in parallel I run a led with a 4.7k resister. I've never had any issues.
On my last order. I accidently ordered these: http://www.mouser.com/Search/ProductDetail.aspx?qs=sGAEpiMZZMs3UE%252bXNiFaVDVuZJnAgzXji3GfCWHA/kU%3d
On the build I'm doing today (a relay powered true bypass looper/ signal router) I am using 3 relays and they seem to be running super hot. I figure this is due to the extra 4 volts. Should I throw a resister to lower the voltage to them and if so what would you recommend, or should I throw a 5v regulator on there?
here is the datasheet since mousers link is bad: http://www.omron.com/ecb/products/pdf/en-g5v2.pdf1 (I see now they require 6v max.. I don't know why I've never had an issue...)

Thanks!!

trixdropd · July 14, 2009, 09:32:50 PM

Upon reading the datsheet I see the 1st relays i was buying can support 9v max, where the low sensitivity can only handle 6. This answers the why question.

I would love opinions about resister vs regulator to lower my voltage...

doitle · July 14, 2009, 09:37:20 PM

Do a KVL to determine the resistor and put it inline. I don't see any reason to run it above it's rated voltage when it's such a simple fix.

The datasheet says the current through the coil is 100mA. Since that current will be the same going through your resistor your KVL would look like this:

-9 + R*100mA + 5V

R*100mA+5V=9V
R*100mA=4V
R=40 Ohms

Power handling should be at least half watt as 100mA and 4V drop accross it gives .4 Watts dissipation.

Closest standard value 10% resistor would be a 39 Ohm and that's what I'd use.

trixdropd · July 14, 2009, 09:49:19 PM

thanks you mike!! I'll have to hit radio shack, I don't have any resisters that low...

What does kvl mean? I would have had no idea on that math formula...

doitle · July 14, 2009, 10:39:07 PM

It just means to use the Kirchoff Voltage Law says that the sum of the voltages around a loop are 0. You just start at a point (I started at the 9V supply) and make a loop through the circuit till you come back to your starting point. Out of the power supply, through the resistor, through the coil of the relay, to the negative terminal of the 9V supply (ground). If you were to draw your 9V supply it would have a + and a - terminal. This is called... What is it... passive sign convention? Anyhow whenever you "come out of" a + terminal that voltage is negative and when you enter a + terminal the voltage is positive. Since we came out of the + supply terminal it is:
-9V
Then we add the voltage dropped across the resistor. This will always be positive as (barring negative resistances) resistors always dissipate energy and never generate it (bring it into the circuit). Ohms law tells us the voltage dropped is R*I and since the resistor and the coil are in series they will pass the same amount of current, your target: 100mA.
so -9V + R*100mA
Now we just add in the coil. It will drop 5V accross it since it wants 100mA and has a resistance of 50 Ohms. Once again Ohm's Law verifies this: V = 100mA * 50 Ohms.
-9V + R * 100mA + 5V = 0

Solve for R and you get 40 Ohms.

I hope that explained it a bit. If you have any more questions just ask them here and I or someone more knowledgeable than me will answer them I'm sure.

trixdropd · July 14, 2009, 11:06:12 PM

Thanks for the schooling Mike. That helped my understanding a bit. I am getting ready to make a run for an E.E. degree. I have A lot to learn...

doitle · July 14, 2009, 11:09:10 PM

Quote from: trixdropd on July 14, 2009, 11:06:12 PM
Thanks for the schooling Mike. That helped my understanding a bit. I am getting ready to make a run for an E.E. degree. I have A lot to learn...

I'm nearing the end of that run. Will be starting my senior year here pretty soon.

Also... You have a lot of FUN to be had!

At least I've been having a pretty darn good time with it. It sure makes it easier to learn when the stuff is so interesting. I hope you enjoy it.

jrem · July 15, 2009, 02:47:22 AM

both of those data sheets are showing 5vdc coils, one has 72ma and the other 100ma current. I can't imagine this is what you meant, sounds like your original relays were 9vdc with the 4.7k resistor as a current limiter. You could do a two resistor voltage drop to get the 5vdc from the 9, then figure out your current limiter, that might also solve your problems. Or drop in a voltage regulator.

trixdropd · July 15, 2009, 06:43:02 AM

I slapped a 5v regulator on there. it has 3 pins, in out and ground. with ground connected it doesn't work right, without it overheats.. any ideas?

trixdropd · July 15, 2009, 08:57:34 PM

Quote from: doitle on July 14, 2009, 09:37:20 PM
Do a KVL to determine the resistor and put it inline. I don't see any reason to run it above it's rated voltage when it's such a simple fix.

The datasheet says the current through the coil is 100mA. Since that current will be the same going through your resistor your KVL would look like this:

-9 + R*100mA + 5V

R*100mA+5V=9V
R*100mA=4V
R=40 Ohms

Power handling should be at least half watt as 100mA and 4V drop accross it gives .4 Watts dissipation.

Closest standard value 10% resistor would be a 39 Ohm and that's what I'd use.

I couldn't get half watt, so I bought 1/4 watt. They feel like there gonna burst into flames... Can I parralel them to get half watt?

doitle · July 15, 2009, 09:05:53 PM

You could series two 1/4 watts of 1/2 the desired resistance value. Then since only half of the power is dissipated in each they would be dissipating .2W which is less than the .25W limit. They would need to be 1/4 Watt 22 Ohm resistors though since that'd be the closest 10% standard value.

trixdropd · July 15, 2009, 09:17:20 PM

Quote from: doitle on July 15, 2009, 09:05:53 PM
You could series two 1/4 watts of 1/2 the desired resistance value. Then since only half of the power is dissipated in each they would be dissipating .2W which is less than the .25W limit. They would need to be 1/4 Watt 22 Ohm resistors though since that'd be the closest 10% standard value.

Alright, I can try that.While I got you here, I have 2 more questions...I'm running 3 relays in this build. Can i use 2 22 ohm resisters in series to supply all 3 or does each one need it's own resisters?

2nd, I'm using a 7805 regulator as well in my tests on my breadboard. I have 9vdc going in, 5vdc coming out. i have a 10uo cap between input and ground, and another between output and ground. No matter what, even with a pretty decent size heatsink this thing is cooking too... do 3 relays pull too much current for the 7805?

doitle · July 15, 2009, 09:45:21 PM

Are you sure you are using a 7805 and not a 78L05? The 78L05 can only output 100mA of current.

I'm afraid I can't be too much help beyond that though, I've only ever used a Vreg once and it worked without getting too hot. It was for a 8051 development kit I designed.

I will say though that it is probably going to get hot and it's a question of whether or not it is too hot. On the Emac SDK we use at school the Vreg on that is a 7805 and it is too hot to touch but they've been running just fine for probably 8 years now.

Hopefully someone with more experience with the Vregs can help out here.

trixdropd · July 16, 2009, 02:21:12 PM

Quote from: doitle on July 15, 2009, 09:45:21 PM
Are you sure you are using a 7805 and not a 78L05? The 78L05 can only output 100mA of current.

I'm afraid I can't be too much help beyond that though, I've only ever used a Vreg once and it worked without getting too hot. It was for a 8051 development kit I designed.

I will say though that it is probably going to get hot and it's a question of whether or not it is too hot. On the Emac SDK we use at school the Vreg on that is a 7805 and it is too hot to touch but they've been running just fine for probably 8 years now.

Hopefully someone with more experience with the Vregs can help out here.

thanks again.. it is indeed a 7805. The comment about the 7805 at school helps me... I have 3 different relay types and tested them all on the breadboard. The 7805 gets real hot with the omron g5v-2 but not with the others.

Thanks again...

Jeremy

trixdropd · July 16, 2009, 04:29:56 PM

added note:

One strange thing that may or may not be of value; The g5v-2 relay registers a beep with a continuity tesy when probing the two relay coil conductors. This is in and out of a circuit. none of my other relays do that. Could that be causing a short to the 7805?

doitle · July 16, 2009, 05:04:19 PM

The coil should be continuous. If it wans't no current would flow and cause the magnetic field that flips the relay contactors. What models are the other relays you are using? The coil should show continuous on all of them unless they are some extremely high resistance coil.

trixdropd · July 16, 2009, 06:09:25 PM

Quote from: doitle on July 16, 2009, 05:04:19 PM
The coil should be continuous. If it wans't no current would flow and cause the magnetic field that flips the relay contactors. What models are the other relays you are using? The coil should show continuous on all of them unless they are some extremely high resistance coil.

ok, I wanted to make sure that wasn't causing my problem... The other relays that don't show continuity in the coil are the Omron g5v-2-hi and some tyco axicom v23079-a2001-b301's. They have a higher resistance than the g5v-2.

Thanks again, I think I'm cool now...

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Question about relay voltage...