Hello forum,

I have a project where I want a pump, and a florescent bulb to go off for set times. (pump for 30s and bulb for 5 mins)

I have decided to go the route of 556 timer to accomplish this.
what I did was make a standard monostable 556 timer and put a relay at the output of the timer. The relay on the pump side is plugged into the wall. Great it works. (I'm just doing the pump for now, later I want the florescent bulb to go off via relay as well)

My issue is, I'm currently using this RadioShack electronics learning kit (the 556 circuit is on there) and the pump is plugged into the wall.
The learning kit is powered by couple of AA batteries. I want this project to be stand alone and run on the wall socket only.

I'm sure this is discussed somewhere but I don't really understand electronics too well and I couldn't understand this issue (yes I used the search function) unless explained for my own unique case.

Thank you so so much in advance.

-Kaz

 

So you need a power supply adaptor , ideally 6 to 12volts, and make sure the relays are the same coil voltage, and the contacts are rated at AC230v 5A should be ok.


post a circuit so we can assist you further.

 

What is the rating of the power going to the pump? Is it a wall wart (large black box that plugs into wall) or just a cord? If the former, what is the OUTPUT voltage and current rating? It will be printed on the wall wart and say something like 9VDC, 1000mA or 8VAC, 850mA.

If the power going to the pump is DC and within the specs of your 556 IC, you should be able to use it to power your circuit too. If it is much higher or the pump is driven by AC voltage, you'd be best off with a wall wart for your circuit, meaning you'll need two outlets - one for the pump and one for the circuit. If two AA batteries are powering your circuit now, you'll look for a wall wart rated at 3VDC with a current rating of 100mA or more. RadioShack should sell these.

 

Thank you for the response guys.

I have attached a circuit diagram.

The pump does not have an adapter. it is just a cord that comes out from the pump.

I understand I may have to get a wall adapter (wall to 9VDC or something) to power the timing circuit,
but this whole contraption would have 2 wall plugs. (In the end it would have 3 because remember, I essentialy have 2 of the attached circuits one for the bulb one for the pump).

I know I can complete this project with 2 or 3 wall plugs but I feel it doesn't look good and is sort of unprofessional (is it just me?).
I just think ok most of our commercial products have a microcontroller or IC in them and they have some sort of light, pump, or motor that the Ic can't source. yet you dont see TV's or microwaves with multiple wall cords comming out the back.

I hope I have explained the problem at hand slightly better this time.

P.S here is the specs on the pump:
UL listed pump has a voltage of 110V-115V, a frequency of 60Hz, a wattage of 2.19W and a flow rate of 45 US Gph.

Thanks again everyone.

 

I just think ok most of our commercial products have a microcontroller or IC in them and they have some sort of light, pump, or motor that the Ic can't source. yet you dont see TV's or microwaves with multiple wall cords comming out the back.

That is because they have a step-down transformer and voltage regulator (another power supply) built into them. You can either:
1) Use multiple cords and wallwarts.
2) Build your own power supply and put it inside your project enclosure.
3) Add an outlet box inside your project enclosure and plug in a wall wart.

(In the end it would have 3 because remember, I essentialy have 2 of the attached circuits one for the bulb one for the pump).

Are you using different times for each or does the pump and light come on at the same time for the same amount of time? If the latter, you can use one circuit and control both relays. If there isn't enough power out of the 555 to do so, you can add a transistor or MOSFET to increase the power going to both relay coils. If the former, you only need one wallwart to power both circuits. Not sure what your relay coils are drawing, but a 500mA or higher 9VDC wallwart would probably work fine.

Hope this helps.

 

-elech_mech

Thanks, yeah I was afraid I was going to have to do #3) .
That was the original plan I had but I want to make this thing small as possible (about the size of those old school electric pencil sharpeners).
The adapters and the outlet box would become bulky.....I will also look into what you suggested in #2)

I wanted to:

1) press the switch
2) both the light and the pump go off
3) the pump to turn off after 30s
4)the light to turn off after 5 mins after the switch was pressed.

Thank you for your help!

 

That was the original plan I had but I want to make this thing small as possible (about the size of those old school electric pencil sharpeners).

In this case, I'd suggest using one 9VDC wallwart since you can use it to power both circuits. Can you find out the maximum current draw of the circuit (both relays on)? This will help determine how much current you need which will impact the cost but mostly the size of the wallwart. I think you can get away with a 9VDC, 200mA wallwart and these are reasonably small. Then you can solder the wallwart contacts to some wires and connect those to the AC cord powering the pump. Just be sure to cover the contacts with heat shrink. This would allow you to stick with one cord and keep your enclosure small and hook up simple. Alternately, you could pick up a 120VAC to 12.6VDC transformer and add a bridge rectifier to the output. However, this is likely going to take up more space than a simple wallwart.

For either case, be sure to add a electrolytic (say 47uF rated 25VDC or higher) and a cermanic (0.1uF, 50VDC) capacitor to the output of the regulator. Also add a 1uF electrolytic and a 0.1uF ceramic cap to each 555 IC across Vdd and GND as close to the IC as possible.