I'm pretty new to making circuits and I ran across this issue . I've got an SPDT 5vdc micro relay that I have a 9V battery with a resistor bringing it down to 4.5v on the magnetic side of the relay, but with the resistor it doesn't work if i remove the resistor it works fine . I'm sure its something stupid, like i said i'm new to making circuits . any help would be great!

 

What relay? Just reduce the value of the resistor until it works and you really need alkaline batteries.

Have a circuit we could look at? You may be asking too much from the battery or your driving circuit is dropping too much voltage.

 

Measure the resistance of the relay coil. Calculate the coil current using I = E/R. E would be the 5 volt rating of the relay coil.
Lets assume your alkaline battery is 9.4 volts. Then the resistor you need must drop 9.4-5 volts at the current (I) you previously calculated.
Using R = E/I, with E being 9.4-5 = 4.4 volts, you get R = 4.4/I.
The larger I is, the shorter your battery will last.

 

I tend to use the 4.5V rated ones when dealing with logic, Goldmine almost gives them away. Not the smallest relay in the world but very tiny: http://www.goldmine-elec-products.com/prodinfo.asp?number=G17319 and I've had extremely good luck with them. 30 mA pull in isn't bad, hold current seems to be less and if you need to drive more than one a simple 2N3904/3906 works well.

They will drive off of simple 5V logic as you can see by the specs. Seems like I'm low on this part so I best order more.

 

I've tried every resistor in my arsenal including the largest (10,000,000 Ω) and the smallest (100 Ω) and everything in between but i get nothing . soon as i take the resistor out of the loop and let the full 9v flow it works fine. even three AAA batteries at 3.6v with no resistors the circuit works. with the resistor in place with the 9v brought down to 3.8v i get nothing. is there an amperage problem? i can't get my meter to read the current for some reason but i suspect that has something to do with it.

 

If the battery voltage dips that much, it sounds like the resistor is in parallel with the battery. It should be in series with it.

 

Use an LM7805 to drop voltage down to 5 VDC for the relay coil. Also, put a rectifier diode across the relay coil to protect your circuitry from the kickback current.

 

Hello,

What type of 9 Volts battery are you using?
The small 9 Volt block batteries are very weak.

What is the coil resistance of the small relays?

Bertus

 

the coil resistance of my relay is 56Ω (10%) the pickup and drop out is 3.5/0.25vdc the nominal coil current is 89.mA the coil voltage is 5vdc maximum of 6.5vdc and its rated 1A at 120ac/24vdc with pc contacts.

 

56 ohms ia darn low.

 

I managed to make it work with a 7505 voltage regulator and some capacitors but I'm still curious what the resistor has against the relay... if any one could shine some light, that would be great... thanks,

 

Could you please put up a circuit?

Try out http://www.diptrace.com and get the student or non-profit edition. Even the free one is fine.

 

I've tried every resistor in my arsenal including the largest (10,000,000 Ω) and the smallest (100 Ω) and everything in between but i get nothing . soon as i take the resistor out of the loop and let the full 9v flow it works fine. even three AAA batteries at 3.6v with no resistors the circuit works. with the resistor in place with the 9v brought down to 3.8v i get nothing. is there an amperage problem? i can't get my meter to read the current for some reason but i suspect that has something to do with it.

I should think that the resistor has nothing "against" the relay, but you may not have used an appropriate value. If the smallest resistor you have is 100Ω, this may explain why the relay did not work from a 9V battery with a series resistor added.

Your relay has a 56Ω coil, therefore with a 100Ω coil added the coil voltage woild have been 9V*56/(56+100) = 3.23V

The voltage would have been lower with any higher resistance, and of course the battery voltage may have been less than 9V.

Your relay passes about 90mA. To drop 4V would require a resistor of 4/.09 or 44Ω, which is much less than the 100 ohms you used.

 

 

I didn't put it in there but i've got the relay activating a christmas light

 

This is a very bad circuit idea. Don't do it! If your relay pulls in, the NO contact shorts the battery. The relay then drops out as the battery voltage collapses, and the cycle repeats....

This will probably ruin the relay contacts. The current could be large enough to cause a fire hazard, especially if you used a large battery. The battery would run down fast, and could even burst.

 

maybe i made the diagram wrong cuz i've tried it and with out the resister it works just fine

 

With a small weak battery, it may work, for a while. It is still a bad idea, and will probably quickly wear out the relay and the battery.

Do you not understand why it is bad to connect a switch directly across a battery? This is very important for your safety.

 

Adjuster is right. It is a very bad idea since you are shorting out the battery. Why don't you use a LM555 to drive that relay instead?

 

no no no. lol made the diagram to show mostly the resistor in relation to the relay there is more to the circuit then just the relay completing the circuit. the formula you showed to me earlier made sence after i did the math. i was leaveing out the resistance of coil in mine.