Relay

Thread Starter

aaad2

Joined May 30, 2012
50
Do not get hung up on the pin numbers. You will find that when you order the relay you need or if you look at a datasheet of a relay, the pin numbers are really different than what you think they are.

Just look at the contacts as NO - C - NC, which is Normally Open - Common - Normally Closed. If you can identify them that way you will be in better shape.
found this
 

BSomer

Joined Dec 28, 2011
434
It is kind of hard to make out on that image but I think the following is correct...

Pins 1 & 16 = Coil
Pins 13, 11, 9 = Contact set 1 Common, Normally Closed, Normally Open
Pins 4, 6, 8 = Contact set 2 Common, Normally Closed, Normally Open

Once you have the relay in your hands you can check it with a multimeter. Place the meter in a ohms setting and check the resistance between the pins.

No power applied to coil.
13 -> 11 & 4 -> 6 = very low reading
13 -> 9 & 4 -> 8 = open or O.L.
9 -> 11 & 8 -> 6 = open or O.L.

Power applied to coil.
13 -> 11 & 4 -> 6 = open or O.L.
13 -> 9 & 4 -> 8 = very low reading
9 -> 11 & 8 -> 6 = open or O.L.
 

Thread Starter

aaad2

Joined May 30, 2012
50
It is kind of hard to make out on that image but I think the following is correct...

Pins 1 & 16 = Coil
Pins 13, 11, 9 = Contact set 1 Common, Normally Closed, Normally Open
Pins 4, 6, 8 = Contact set 2 Common, Normally Closed, Normally Open

Once you have the relay in your hands you can check it with a multimeter. Place the meter in a ohms setting and check the resistance between the pins.

No power applied to coil.
13 -> 11 & 4 -> 6 = very low reading
13 -> 9 & 4 -> 8 = open or O.L.
9 -> 11 & 8 -> 6 = open or O.L.

Power applied to coil.
13 -> 11 & 4 -> 6 = open or O.L.
13 -> 9 & 4 -> 8 = very low reading
9 -> 11 & 8 -> 6 = open or O.L.
then if everything is correct and works, the lights should all work :)
also, how many LEDs do you think I can add onto the relay at pin 8 and 9?
 

Thread Starter

aaad2

Joined May 30, 2012
50
Since the contacts on that relay can handle up 2 amps @ 24 VDC and figuring in aproximately 10 - 15 mA for each LED. Quite a few. I would suggest reading through Bill_Marsden's blog posts about current limiting and LEDs in series or parallel. There is a lot of good information there.

http://forum.allaboutcircuits.com/showthread.php?t=19075
Hi,

This is what my PCB should look like, please can you check to see if everything is ok.

thanks in advanced.

 

BSomer

Joined Dec 28, 2011
434
I can see all sorts of issues with this layout...

1) The connections to your relay are incorrect. The three top right pins are one set of contacts according to the datasheet of the relay you showed us you were getting for this project. The three lower left pins are the other set of contacts. V+ should connect to pins 4 & 13. The LEDs should connect to pin(s) 9 and/or 8.

2) There are no current limiting resistors connected to the LEDs.

3) R2 - 6 are all connected in series. Why? You should only need one 2k7Ω resistor from the motion detector pin 2 to the base of transistor 1.

4) The capacitor pads you are using in the layout are for surface mount capacitors. The capacitors in the schematic you provided are 100uF 25V capacitors. I am unaware of any SMD capacitors of that rating that will fit the pad dimensions you have in the layout.

There may be other problems with the layout as well but the ones I mention are the most glaring issues.
 

Thread Starter

aaad2

Joined May 30, 2012
50
I can see all sorts of issues with this layout...

1) The connections to your relay are incorrect. The three top right pins are one set of contacts according to the datasheet of the relay you showed us you were getting for this project. The three lower left pins are the other set of contacts. V+ should connect to pins 4 & 13. The LEDs should connect to pin(s) 9 and/or 8.

2) There are no current limiting resistors connected to the LEDs.

3) R2 - 6 are all connected in series. Why? You should only need one 2k7Ω resistor from the motion detector pin 2 to the base of transistor 1.

4) The capacitor pads you are using in the layout are for surface mount capacitors. The capacitors in the schematic you provided are 100uF 25V capacitors. I am unaware of any SMD capacitors of that rating that will fit the pad dimensions you have in the layout.

There may be other problems with the layout as well but the ones I mention are the most glaring issues.
Should be all correct now...


reference:
 

Thread Starter

aaad2

Joined May 30, 2012
50
Also, I orginally used R2-6 is because I dont have any 2K7 ohms resistors so i thought I might use two 2200ohms resistors and 3 100ohms resistors.

For the capacitors I cannot find the normal type of capacitors so now I have just replaced them with pads.

And once again thanks for the help :)
 

BSomer

Joined Dec 28, 2011
434
The first one will work if you want more than one LED in series and is what I was referring to in my previous post.

The second will work if you are just doing a few LEDs, though the power dissipation of the resistors will go up because it will have a higher voltage across it.
 

Thread Starter

aaad2

Joined May 30, 2012
50
The first one will work if you want more than one LED in series and is what I was referring to in my previous post.

The second will work if you are just doing a few LEDs, though the power dissipation of the resistors will go up because it will have a higher voltage across it.
Thanks, Im still waiting for my PIR sensor to arrive then I can get started on this project. :)

also what can I use the Pins 6 and 11 for? more LEDs?
 

BSomer

Joined Dec 28, 2011
434
Pins 6 & 11 are the normally closed contacts. This means that while the relay is not turned on those contacts are closed, and when the relay is powered on the contacts open. You can put LEDs on those contacts but they will work oposite of the other LEDs.
 

Thread Starter

aaad2

Joined May 30, 2012
50
Pins 6 & 11 are the normally closed contacts. This means that while the relay is not turned on those contacts are closed, and when the relay is powered on the contacts open. You can put LEDs on those contacts but they will work oposite of the other LEDs.
arhh, so I can add one LED(such as a small red LED) to indicate that there is still power. :)

also I have bought some solar panels to charge the batteries in the circuit.
so far I have connected 4 solar panels in series and they produce 1.5v and 5.5mA(i think it is mA - I put the dial on 200m--> which is 0.2Amps) [the solar panels might produce more voltage and current as I have only tested them on the window sill.]

I am thinking of connecting some more solar panels but not sure how much more I need, the battery is made of 8 1.2v 800mAh batteries so a total of 9.6v.

I would like to use just a single diode to do this job but scared that it will overcharge.

the circuit is designed to be switched on not all the time, only sometimes when someone goes to the back garden which therefore switches on the LEDs.

So I was thinking about 10 solar panels altogether? and just let it slowly charge away... because it will slowly charge all day which wont really produce that much current or voltage...

what would you say?

THE PICTURE BELOW IS FOUND ON ANOTHER WEBSITE.


AND THANKS IN ADVANCED... YOU HAVE REALLY HELPED ME THROUGH THIS :)
 

BSomer

Joined Dec 28, 2011
434
The diode is there to protect the solar panel(s) from reverse current. For instance if there is no voltage being produced because it is dark the battery could possibly start putting voltage back into the panel, which is a bad thing. You will want to size the diode above the maximum possible reverse current and voltage.

The set up you show there has no overcharge protection. You would have to monitor the battery voltage and then disconnect the solar panel when the battery voltage is where is should be.

I must admit though that I do not have too much experience with solar panels and charging batteries. I recently just started playing around with that stuff and bought a premade charger with a solar panel.
 

Thread Starter

aaad2

Joined May 30, 2012
50
The diode is there to protect the solar panel(s) from reverse current. For instance if there is no voltage being produced because it is dark the battery could possibly start putting voltage back into the panel, which is a bad thing. You will want to size the diode above the maximum possible reverse current and voltage.

The set up you show there has no overcharge protection. You would have to monitor the battery voltage and then disconnect the solar panel when the battery voltage is where is should be.

I must admit though that I do not have too much experience with solar panels and charging batteries. I recently just started playing around with that stuff and bought a premade charger with a solar panel.
Also, I was thinking of adding a LDR in the circuit so that the LED will only switch on when there is a motion WHEN IT IS DARK'. this allows me to go to the garden during daytime and not cause the LEDs to turn on.

this is the circuit I have found, this mean that when the environment is dark, the LEDs will switch on but I am not too sure where I can add this circuit to my current one.



Should I just add it to where my LEDs are? pin 8 and 9? this means I have to make two seperate parts to this:
pin 8 to ground and
pin 9 to ground

and I dont know of these components will work with a 9.6v(approx.) etc...
 

Thread Starter

aaad2

Joined May 30, 2012
50
The diode is there to protect the solar panel(s) from reverse current. For instance if there is no voltage being produced because it is dark the battery could possibly start putting voltage back into the panel, which is a bad thing. You will want to size the diode above the maximum possible reverse current and voltage.

The set up you show there has no overcharge protection. You would have to monitor the battery voltage and then disconnect the solar panel when the battery voltage is where is should be.

I must admit though that I do not have too much experience with solar panels and charging batteries. I recently just started playing around with that stuff and bought a premade charger with a solar panel.


there is something wrong, when I set up the circuit above, the batteries starts to become warm!!!! arhh!!!!

please can you check my diagram?
I have confirmed that all the components are connected in the right way, i somehow have the feeling that it is the relay? i dont know... please help :)
 

BSomer

Joined Dec 28, 2011
434
The board layout looks correct according to the schematic. If the battery is getting hot there is a possibility of a short or too much current draw.

Check the voltage input without the battery connected with a multimeter set to ohms and see what the resistance is. If it is very low you may have something connected wrong.

Can you post a clear picture of your set up?

Can you check to see what the current is for the whole circuit when powered up with your battery?

Do you have a bench power supply with current overload protection?
 

Thread Starter

aaad2

Joined May 30, 2012
50
The board layout looks correct according to the schematic. If the battery is getting hot there is a possibility of a short or too much current draw.

Check the voltage input without the battery connected with a multimeter set to ohms and see what the resistance is. If it is very low you may have something connected wrong.

Can you post a clear picture of your set up?

Can you check to see what the current is for the whole circuit when powered up with your battery?

Do you have a bench power supply with current overload protection?
Ok, I connected the multimeter and set to ohms, and connected to the positive and negative ends of my circuit. It does not display any resistance? why is that? I then connected to some of my resistors which the multimeter displays the values for those resistors? why is there no resistance?

Currently, I am doing all this on breadboards so that I can do a proper version on circuit board when the circuit works.




the red lines just means that the connection inside the breadboard goes like that and the top row is the positive and the bottom row is the negative

Few days ago, my battery was about 2.9A but today is only about 1.5A... think I need to charge my battery again...

the total voltage is about 10v-11v

i checked everything to my diagram but dont know why it is not working?
and thanks for the help :)
 

Thread Starter

aaad2

Joined May 30, 2012
50
The board layout looks correct according to the schematic. If the battery is getting hot there is a possibility of a short or too much current draw.

Check the voltage input without the battery connected with a multimeter set to ohms and see what the resistance is. If it is very low you may have something connected wrong.

Can you post a clear picture of your set up?

Can you check to see what the current is for the whole circuit when powered up with your battery?

Do you have a bench power supply with current overload protection?
are you sure the relay is correct?
i thought it would be the other way round...
currently, the positive is connected two the two pins on the 2nd from left side(cant remember pin number). shouldn't be horizontal? connect to the top row 2nd and 3rd pin? confused...
 
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