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This is my first instructable. I am an engineer working with electronic road signage. Recently a customer came to us looking for a sign which warned road users of dangerous conditions caused by high wind speeds at a bridge site. The sign and subsequent control system were not a problem, however the customer returned to us with the concern that the AC powered sign may suffer power failure during a storm and in turn not warn drivers about dangerous conditions. After some searching I found that uninteruptable power supplies (UPS) are not cheap. This instructable will also work for emergency lighting or with a large enough battery and inverter could keep a fridge or freezer up and running in the event of power failure.

But lets look at what we have to deal with:

  • AC supply
  • AC Load
  • DC Battery
  • AC - DC Battery Charger
  • DC - AC Power Inverter
  • A means of automatically switching over in the event of a power failure.

PLEASE NOTE THIS INSTRUCTABLE DEALS WITH HIGH VOLTAGES, THESE POSE A SERIOUS RISK TO YOUR HEALTH SO PLEASE BE CAREFUL WHILE YOU WORK AND DON'T WORK WITH MAINS POWER UNLESS YOU KNOW WHAT YOU ARE DOING.

Step 1: Equipment

For this setup I have used some basic tools and the following equipment:

  • a 300W 12VDC to 230VAC Power Inverter
  • a 230VAC to 12VDC Battery Charger
  • a 100W Light Bulb to simulate the load
  • a 22Ah* 12VDC sealed lead acid battery (it is important that the battery be sealed, rechargeable battery, ALKALINE BATTERIES ARE NOT SUITABLE FOR RECHARGING, and automotive batteries will not withstand this kind of charging cycle for long)
  • a dual throw, triple pole relay (I could have used a single throw, dual pole for this task but this is what I had

*This instructable will use a 22Ah battery, however my final installation will use 3 of these batteries, you can use any size you wish as long as your battery charger can handle the load. the more batteries you have the more run time you will have in the event of a power failure. To calculate this look for the wattage rating on your appliance and divide this by your supply voltage e.g. *Edited* 50W @ 230V = 0.21A, on the low voltage side this translates to 50W @ 12V = 4.16A for a 22Ah battery this allows for 5.2 hours max run time. Now there are some other factors to consider:

  • The inverter will consume some power so refer to the manual and make the subtraction from the run time
  • The battery will not provide full power until it dies so I'd say to be safe assume the battery is dead from 40% so this cuts the run time for the above scenario to about 2.5-3 hours.

Step 2: Wiring the AC Side

The relay used in this project is a 230VAC coil pulling three sets of contact switches, a wiring diagram is supplied on the side of the relay block.

The idea behind using this kind of relay is to supply the load with power from the mains supply while mains are available, the unit is also designed to maintain the charge in the battery during the good times. In the event of a mains power failure, the coil demagnetizes and the contacts drop into the natural position, this switches in the supply from the battery to the inverter and the inverter to the load.

As per the wiring diagram connect the mains supply to the coil on the relay (A1 & A2), also hook this directly to the battery charger (we want the battery charging whenever there is a supply available). I am really using the relay block in reverse: instead of switching the supply between 2 loads, I am switching 2 supplies to 1 load.

Loop the power to the normally open (NO) side of contacter 1 in this case pin 9, connect the positive output from the inverter to the normally closed (NC) side of contacter 1 (Pin 8) and connect the input of contacter 1 (Pin 11) to the positive of the load.

Step 3: Wiring the DC Side

Now we have a working circuit, we can see from the last image that when the mains power is turned on there is supply to the load. However, if the mains power is removed the load turns off, no good...

So we look to the DC side of the circuit to take up the slack.

We need to use the second contacter on the relay to switch the battery between charging and supply so we wire the positive terminal of the battery to the input of contacter 2 (Pin 6) of the relay block. From here we wire the positive output of the charger to the NO (Pin 7) and the positive of the inverter to the NC (Pin 5). This means that when there is mains available and the coil is energized the charger will be connected to the battery and charging will take place, but when the power drops out and the coil de-energizes the battery will be connected to the inverter and the battery will be in supply. You can connect the negative of the battery, the charger output and the inverter together. MAKE SURE ALL AC NEGATIVES ARE CONNECTED SEPARATELY FROM DC NEGATIVES!!!

Now lets insert the plug to the inverter and we are ready to test.

Step 4: The Moment of Truth

Switch on the mains power and watch the load come on, ensure that the power switch on the inverter is off and kill the mains, the load should die. Power back on the mains and again the load springs to life, this time turn the power on the inverter to the on position and kill the mains. I called this instructable the almost uninteruptable power supply as there is a second or 2 during the switch over where the power is lost, in this time the relay is dropping back to the natural position and the inverter is coming up. This means that is you want to keep your PC running this probably is no good but if it's for something less mission critical and you can allow for the 2 second power loss this is the solution for you. Check out the video of the system in action and let me know what you think...

If you want to be rid of the relays and have 100% uptime you can hook this up as an active ups. The charger both charges the batteries and powers the inverter. The load only connects to the inverter. This means you will need a larger charger that can supply the entire load with some left over to charge the batteries but there would be no time when the load did not receive power. You would also want something to prevent overcharging but you really need that either way it is hooked up.
Nice idea, maybe for the next one. As for the overchage protection, this charge has a battery level monitor and switches out when the battery is full, one of the advantages of off the self I suppose.
Just curious, why not use a UPS? For a client, I imagine off the shelf is quicker and easier for them to fix.
The client came in the aftermath looking for this as an addition this was the cheap way out. Also all this equipment must be housed inside the sign which has only 100mm depth.
A couple of edits pointed out by the readers. The relay is in fact triple pole dual throw, not single throw. Also the battery is sealed lead acid not nimh. Sorry as I said this is my first instructable, I appreciate the feedback and I hope the idea is helpful.
<p>Congrats on your first... Just trying to help for safety sake of people that will use this. ;-) </p>
<p>Very nicely done. Interesting to see how this all went together. </p>
Hi. 500W @ 12V is more than 41A. so your battery will not last 30 minutes at that discharge rate and with inverter losses. You need to calculate the current at the relevant voltage level. Also if you do run that kind of current you need serious wire diameter on the low voltage side.
Thanks for the comments, I'm not actually running 500W in my final project. I just arbitrarily selected a wattage for the example. Also yes high voltage is dangerous so please be careful.
<p>Yes I realize your 500W is just an example, but lots of people makes this error, you cannot use the high voltage side current as the same as the low voltage side. W(in) = W (out); V(in).A(in) = V(out).A(out) in perfect situation. So if your voltages are 230V(in) and 12V(out) your currents will be the inverse, low(in) and high(out). If you monitor your battery discharge current when the inverter is running you will see how high it is. ;-) </p>
That could be a sealed lead acid battery but I doubt if it is NiMh. Also your circuit shows a flyback diode. You cannot use that on the AC side. you will short out half a waveform of your AC mains.
Apologies, the circuit diagram was drawn using a free online package, I hadn't noticed the diode in the block I pulled in, this is not present in the build.
<p>No worries, just keep in mind that people will try and replicate your instructable and if they put the diode there it will go up in smoke...</p>
Hi there... Please be very careful, you are working with high voltages and high currents here. Check your calculations and wire diameter required.
I agree with teratron. something's wrong with your computation.
Very interesting design!
Great instructible! <br>-one thing though: that relay is a triple-pole, double-throw.

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Bio: I am an automation engineer but I will give anything a go. I don't know if you call if pessimism or just being an ... More »
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