got a broken circuit

[GeneralHulio]

I am using these circuits to create an automated sprinkler system, but I'll keep this brief. I'm running a DC solanoid valve and a relay switch using an Arduino from my computer. The Ardiono cannot power either device directly, so it switches the device indirectly via a NPN transistor. The 9V batteries can power both devices sufficiently. The diode is there neutralize any kickback in the coils (both the valve and the relay have coils so I'm using the same circuit design for both of them). I had the relay and valve turning on and off every 5 seconds to test them but then they stop doing this after a minute. I replaced the diode and that solved the problem for about a minute but then it stops working again (I think current is building up and gettting trapped somewhere but I don't want to remove the diodes). Is there any electronics whizz who can help me solve this riddle and stop the electical buildup or whatever it is causeing this?

I got the schematics from here using their design but I think there is something wrong and I can't figure it out . http://www.instructables.com/id/Control ... h-arduino/

Here are the design schematics in this picture

[pebe]

The 'Instructables' website shows two PP3 batteries. Is the battery you are using a PP3? If so, it's pushing it a bit to get enough current to drive a solenoid, and it may be that the high current being used is draining the battery and lowering its voltage. Then the rest period is refreshing it for a while.

What type of battery is it - alkaline or rechargeable? Can you measure its voltage?
Do you know the coil resistances of the solenoid and the relay?

[GeneralHulio]

Well my solanoid coil resistance is 5.3 oHms
The resistance on the relay is 157 oHms

The batteries are both PP3 and Alkaline. I'm getting voltage readings of 9.05 and 8.56 volts. The odd thing is that they measure at 7 volts when integrated into the circuit with the relay because I also tested them while they were connected. The other thing is that when I tried to drive the relay it would half click but not properly, like something was stopping it (like maybe you are right and I needed a higher voltage). However, when I unplugged the relay from the circuit and put a battery directly against it, it worked properly again. The problems only happen when it's all wired up into the circuit and not when I test components separately.

Also, I'm confident to run my solanoid on 18 volts but not the relay ( the relay is a 12 volt coil while the solenoid is 12-24 volts, so i am not sure if the relay can handle an extra 6 volts). I bought durable parts that should be able to handle the voltages but I feel like if I put it up to 18 volts I might risk blowing up something on the circuit.

[pebe]

It looks like your problem is due to a number of things. Your relay is designed for 12V minimum, so with a coil resistance of 5.3 ohms, it needs a minimum current of just over 2A to operate it. Components will often work at voltages below their designed limits, so it is not surprising that it will work on a new 9V battery.

But in your circuit, the solenoid is fed from a TIP120. That is a ‘Darlington pair’ transistor and will always have a voltage drop between emitter and collector of at least 1.4V. So with a new battery, and allowing for that voltage loss, the coil will only be fed with 7.6V, and the solenoid just manages to work. Then when the battery voltage drops to 7V, the solenoid only gets 5.6V and that is well below the level at which it will operate.

The obvious answer is to increase the battery voltage. That said, a PP3 is not intended to supply such a high current (50-100mA would be about its norm). If you can run it from the mains then a 12V 2A switch mode wall adapter would be the ideal. If you have to run it on batteries, then a rechargeable Li-ion battery would be far better. Individual cells like the 18650 give a nominal output of 3.7v (just over 4V when fully charged) so you could make a battery of 3 cells to give you 11.1V nominal (12.6V max). They need a proper LI-ion charger and you would need ‘protected’ cells to prevent damage if you discharged them too low. The cells can be supplied with capacities of up to 4Ah and you can buy cell holders for them. These are all available quite cheaply on ebay, like these:
http://www.ebay.co.uk/itm/2x-UltraFire- ... 485c06c780

http://www.ebay.co.uk/itm/2xUltrafire-4 ... 258c0aaa74

Or if you want to bid:
http://www.ebay.co.uk/itm/2pc-Ultrafire ... 4ad8872ae2
The costs would soon be justified compared with PP3 replacements.

If you wanted to further refine the circuit, you could replace the TIP120 with a FET to save the ‘lost’ 1.4V. Let me know if you want more details.

[yy482]

check the power 9v make sure its ok,or using a switch dc power for test.
Try to add a resister about 10k to the B and E