Hi guys,

1st post so be gentle!

im wondering if you guys can help me.

i need to try and make a circuit that is fed from a 12vdc source that will operate a 12vdc relay when it gets either a positive or negative trigger and then stay closed for about 10/15 seconds once the trigger is removed.

i have tried to make a few of the circuits off the site in crocodile clips and it either pops the 555 or dosnt work

my electronic understanding is very limited too.

many many thanks

martyn

 

Well... First, you must limit the input current feeding the circuit to less than the max allowed on the datasheet. If your feeding it 12 volts, and the input current is maxed at 10mA-15mA then 12/ .01 = about a 1 Kohm resister. The chip will source about 200mA max, read the max current on the relay coil. Put a limiting resistor between the output and coil. 12/.02 = 600 ohms. Now you are protecting the chip and the coil. Read Bill Marsden's section on 555 monostables. The time delay formula is...CxRx1.1= Time in sec's. A 500k resistor and a 20uF cap gives you this... 500,000x.00002x1.1 = 11 secs. Long times can be problematic, but I think your well within the specs. Also read up on RC circuits.

http://www.datasheetcatalog.org/datasheet/philips/NE_SA_SE555_C_2.pdf

 

You want a Delay on break relay.

Delay on break means once you cut power, the delay counts down, then the relay opens.

No need to make one. They are pretty cheap.
http://www.airotronics.com/site/timing_delaybreak.php

 

The chip will source about 200mA max, read the max current on the relay coil. Put a limiting resistor between the output and coil. 12/.02 = 600 ohms.

600Ω in series with a relay coil? What a load of ...

 

600Ω in series with a relay coil? What a load of ...

EBLC1388, you seem to be a very argumentative individual. I answered the OP is as simple a manner as possible and as conservative a manner as possible. He is shorting components and will get more out of a simple explanation as opposed to a lesson in electrical pyhsics. Is is obvious that many here possess a higher level of technical knowledge than I, I am trying to contribute in some small way as many here have done for me.

No, I don't understand why you protect the VCC pin of a 7555 down to uA's, but it will source much more than that. I do know the output will burn out a LED though... Here is the link to a randomly selected relay I found. The coil uses 30mA so in this case 400 ohms would work... I was attempting to demonstrate the protection of components whose current draw is less than the max output of the IC. My relay uses 20mA. Are such precautions for the beginner unreasonable?? My humble circuits may not win any prestigious engineering awards, but they usually work.

http://www.hobbyengineering.com/H1325.html

To the OP, Sorry for that... I also forgot to recommend a protective diode in parallel with your relay coil. Those coils generate voltage spikes that can damage things when they operate. See the 555 circuits here.

 

Is is obvious that many here possess a higher level of technical knowledge than I, I am trying to contribute in some small way as many here have done for me.

For a person who don't even know to place a series resistor to prevent the LED from burning out, my advice would be to learn more from the forum posts first than to offer half baked advice.

No, I don't understand why you protect the VCC pin of a 7555 down to uA's, but it will source much more than that.

I never did that or say that and have absolutely no idea of what you are talking about.

I think you should know if you place a series resistor with the same value as the relay coil, it would cut the coil voltage by half. Then it will be 6V for a 12V relay. The relay will not work anymore.

I know you are angry but I don't want to argue with you anymore. Have a nice day.

 

Use transistor to drive your relay and a timer/counter for your delay. If you connect your relay direct to 555 it will of course damage your chip either by kick back of the coil and overloading. simple counter/timer/driver/comparator will do the job what you want. This is just my suggestion.

 

Mr. Chung, My point wasn't that my advice is always correct, or that I never make mistakes.(my LED oversight proved that, and I DO know better)I am also pretty sure the coil current of a relay is self regulating, and the large resistor was probably not needed. My point to the OP was to protect the downstream devices from excessive current as well. My point to you is that you are coming across as rude and arrogant. There are many people here who probably noticed that putting a limiting resistor of that size would cut the voltage, but they did not see it as an opportunity to berate another for a poor answer, but to elighten both as to a better solution and to politely explain my mistake. Putting any resistor on that output will cut the voltage by too much. I now realize that. I learned today. Yes, I AM angry, I strive to help others and to learn from my mistakes, Perhaps you should take some classes in manners.

To the MODS.. I apologise for my outburst, I will try to refrain in the future. Mike.

 

You want a Delay on break relay.

Delay on break means once you cut power, the delay counts down, then the relay opens.

No need to make one. They are pretty cheap.
http://www.airotronics.com/site/timing_delaybreak.php

Hi,

how would i wire my negative or positive trigger into this?

thanks

 

Is your trigger just a pulse or is it a steady LOW or HIGH?


These are set up to trigger on a 0v signal.

SO if you wanted to trigger when a signal goes HIGH, wire a schmitt trigger to it. That will make your high a low to the relay.

If you want to trigger when the signal goes low, then it is a direct wire.

The data sheet shows different applications and wiring diagrams.

 

hi,

just found thats its a pulsed positive.

 

A positive pulse will work fine with this relay: (as long as the pulse is 6 milliseconds or less)
http://www.airotronics.com/site/product-delaybreak_TGM.php

 

the trigger lasts around 1 second

 

Here is what newbies hobbist suggested... See "Driving high current loads"

http://www.talkingelectronics.com/projects/555/Page3-555.html

 

Hey Guys.

this circuit sort of does what i need it to do, it works when a +ve trigger is applied and can operate a relay.

what do you guys think? is there any way i could reduce the parts in this to make it cheaper to build?

thanks

martyn

 

My mistake in my earlier reply becomes even more obvious now. Those 100 ohm resistors in series with the relay coils form a voltage divider. This means they are pulling down your relay voltage to the point where it won't work at all. They probably aren't needed . With other components like transistors and leds, limiting resistors are required.

Sometimes my explanations are vague and sometimes incorrect. I have seen relays being driven by 555's on many circuits online, but the people here like Sgt. Wookie, and Bill Marsden possess a far greater knowledge than I. You might want to specifically request them on a new thread.
Using relays may destroy the chip prematurely.

 

if i dont put a resistor onto output of the IC then it says that it blows the relay.

 

I don't see how, I double checked a reed relay on my breadboard, and the coil current is exactly follows the resistance of the coil. No more, no less. Might be a glitch. I use multisim, and it is flawed because it does not recognise a short condition, It simply calculates mega current flowing. I provided two additional examples of output -relay usage. I suggest you build it on a breadboard, and test out your theories. The components are pretty cheap at Radio shack. A couple of bucks for a 555, and resistors are 99 cents a pack. Caps are a couple of bucks.
http://www.555-timer-circuits.com/driving-a-relay.html
http://www.physics.unlv.edu/~bill/PHYS483/transdemo.pdf

 

 

ok guys, im thinking about fitting this relay on the output side of the 555. will the chip be able to power it ok?

http://cpc.farnell.com/tyco-electronics-schrack/rt314012/relay-pcb-spco-12vdc/dp/SW03368

thanks in advance guys