Hey guys, sorry if this has been asked before, I did searches but couldn't find any thing that remotely covers what I would like to do. I have limited electronics knowledge

http://www.kpsec.freeuk.com/projects/trafficlight.htm

The above link refers to a traffic light simulator that runs off a 9v PP3 battery, I would like to use a 12v SLA battery which has a nominal output voltage of 13.8v, but I doubt I can just connect it up?

So what would I need to add or change to allow this to occur?

I assume just a greater resistor?

 

Hello,

For the 555 and the 4017 the 13.8 Volts is no problem.
You can change the three 470 ohm resistors for the leds to 820 ohms to have about the same current through the leds.

Greetings,
Bertus

 

Seriously that's it ?!

Thanks for your help

 

Hello,

To be sure it works change the 1 μF and the 10 μF capacitors to 25 Volts or higher.
The meantioned 16 Volts will be a little short.

Greetings,
Bertus

 

Second question! Same project.

Can I just connect 12v relays to where the leds would connect if I just exchanged the LED resistors for wire links and fitted diodes across the relays coil or would this be too much load for the IC to handle? And have to use transistors to fire them?

TIA

 

Second question! Same project.

Can I just connect 12v relays to where the leds would connect if I just exchanged the LED resistors for wire links and fitted diodes across the relays coil or would this be too much load for the IC to handle? And have to use transistors to fire them?

TIA

No, You cannot connect the 12V relays directly. You should add a transistor before that as it is too much load for the IC. And also add a freewheel diode across the relay.

 

So will this be how I need to connect using a transistor?

 

Anyone help ?

 

Close, one side of relay coil would connect to the positive power, the other side would connect to the collector of the transistor. The band/cathode of the diode is in the correct position, across coil, towards positive.

As it is shown now, there isn't a power source for the relay, only a switch to ground.

Base through a 100 Ohm resistor, emitter to ground as already shown (without resistor in your diagram).

 

So this is correct? Sorry I hate burning plastic!

 

If I'm reading the relay part right, that should work for a 2N2222 transistor.

I made a mistake on base resistor value, I was thinking 5V. The max base current for a 2n2222 fully on is 15mA, and you have four outputs in parallel to drive it at 12V. Change the base resistor to 220 Ohms, that will protect both the IC and transistor, and and still allow it to operate a 200mA relay coil. If the relay coil needs more drive, you can go to a lower resistance, but keep the base current under 15mA.

Without a current limiting resistor, the base/emitter junction is essentially a short circuit to ground, with a 0.6V drop.

 

Right so I've re done the schematic to incorperate the transistors etc so I can exchange LED for relays, I'm completely a noob when it comes to IC's and transistors, could someone confirm this is going to work or am I going to end up with a door stop?

So is the type of diode across the relays the right one?
Are the previous mentioned 2N222 going to work ok?

Thanks guys,

 

Should work OK. If there are problems driving the two relays on the right, use 100 Ohm resistors instead of 220.

 

The nearest match to those transistors appear to be these type

http://www.maplin.co.uk/module.aspx?moduleno=32927

Are they ok or should I be going for something like this

http://www.maplin.co.uk/Module.aspx?ModuleNo=32952