Security System Power Saver

   This circuit is used for automating a day/night scheduled security system which, best of all, could only cost cost you $0.25 a YEAR to run. (All electricity prices based on a 2009 national average cost found here .) If you have a security system circuit you would like to run for a long time, then you probably don't want to pay more than you need to pay nor do you want to have to switch it on and off manually in order to save money and have it run at nighttime. This circuit takes care of both those problems. The circuit can be changed around to run on a daytime or nighttime schedule and is low cost to run along with the other circuit. (I'll explain how to calculate these prices of run time in the last step).
   Materials differ depending on which of the two circuits you plan to build, but here are the components used by both circuits:

1x 555 timer
3x resistors (1x 470 ohm; 2x 470K)
1x Photoresistor
1x LED
1x non-Zener diode
various jumper wires

   For the 5v relay circuit, you will need these in addition to the components listed above:

1x 5v relay
1x 12v regulator
at least 2 high rated capacitors with voltages ratings of 20 volts or above)

   For the 12v relay circuit, you will need these in addition to the components listed above:

1x LM741 op amp
1x NPN type transistor
1x 12v relay
2x 65K resistors

Step 1: Building the 5 Volt relay circuit

   Here is a series of mini steps to help build the 5 volt relay circuit. Keep in mind the higher the values of capacitors you use the better off you'll be in the long run.

Step 1. Plug the 555 timer into board.

Step 2. Apply power and ground wires to timer chip.

Step 3. Plug 5v relay (I'm using the reed switch style) into the board, applying 555 timer output to relay power and ground to ground.

Step 4. Connect pin 2 and pin 6 of the 555 timer with a jumper wire.

Step 5. Connect 470K resistor from power to pin 2.

Step 6. Connect 470K resistors from power to pin 4.

Step 7. Connect photoresistor from pin 4 to pin 6.

Step 8. Connect pin 6 to ground with jumper wire.

Step 9. Connect a non-Zener diode so that the cathode connects to the power side of the relay and the anode connects to the ground side of the relay. (This is used to keep sparking from happening when the relay turns on and off)

Step 10. Connect 470 ohm resistor from power to an open end of the switching contact. This step differs depending both on relay rating AND what you plan on switching with the relay.

Step 11. Connect LED to other open contact and ground it. This step differs depending both on relay rating AND what you plan on switching with the relay.

Step 12. Plug 12v regulator into board.

Step 13. Connect the ground of the regulator to ground and the output pin to power on the board. (Your wall source power is connected to the power INPUT on the REGULATOR.)

Step 14. Connect a capacitor from output to ground on the regulator.(Remember: Your capacitors must be rated to handle your source voltage.)(Remember: Your capacitors must be rated to handle your source voltage.)



Step 15. Connect one or more capacitors to the power side of the relay, remember the higher the rating (to an extent) the better! (I'll explain why in the last instructable step)

Step 16. If you plan on using a wall supply, DO NOT JUST GO BY WHAT THE WALL WART IS RATED!!! (As evidenced by two of the photos above) Even I didn't believe that the wall wart was immune to differences in its rating, so use a multimeter to check for the ACTUAL voltage and amperage.

Comments

[LegendofPedro]

Cunning use of a 555, but you could use an op-amp as a Schmitt trigger instead. Or a setup with a thyristor.

And you realise in one of your circuit diagrams, you've left the non-inverting input of your op-amp floating, right?

[MacDynamo]

I'm glad you caught that floating pin. I was actually unsure about using a pull-up/pull-down resistor on the inverting input because I wasn't sure if that would affect the output regardless of whether the non-inverting input receives power or not. If you happen to know, tell me and I'll fix it as soon as I can.

[LegendofPedro]

Your op-amp is running as an open loop inverting amplifier. The gain of the amplifier is unimportant as it is going to be very high; you're essentially just flipping the waveform (as you put a signal on the inverting input).

Since an op-amp is a differential amplifier and the output of the 555 will always be between GND and +Vcc I would assume you want to tie the non-inverting input to GND. It's bad form to have a floating input since it's supposed to be amplifying a 'difference'.

[surfingump]

Well done and thank you for posting!

[aslanidis]

very nice job!!!

[jensenr30]

amazing! i love this!