Normal/Alternative DC uninterruptable power supply with battery back lighting circuit

Hi guys.

I need some help designing a Normal/Alternate DC uninterruptable power supply with battery back-up to illuminate a 14v lighting circuit.

I will be using Locktronics to build the circuit and it needs to be as simple as possible.

SPECS: Normal and Alternative supplies to be rated @ 14v
Battery back-up is to be 12v
supply available lamps are to be red LED's
supply in use lamp to be red for N/A supply and green for battery supply

I'm not really to sure where to start as my electronics knowledge is very basic.

I have relays, caps, transistors, resistors, led's and diodes in the toolbox I have been given so I'm guessing I can get a solution using just these items.

I have searched for basic circuit diagrams that fit my needs and that I can easily understand but I have so far come up with nothing.

Any help or assistance that can be offered would be greatly appreciated.

what is this some kind of homework? Who thinks you will build what you want this these parts?

When I first read this I was thinking you'd need a boost converter at the very least. Maybe not depends on what your needs are exactly. You will need to spell out exactly what your load is gonna be if you want more definitive answers.

Yes it's homework. We use Locktronics to build circuits on

The load is a 14V lighting circuit.

I'll write out my brief:

To design and build and test a Normal/Alternate DC uninterruptable power supply with battery back-up to illuminate a 14V lighting circuit.

Objectives
When Both supplies are available, normal supply is to be automatically selected

To automatically switch to secondary in the event of a total mains failure OR an approximate 20% fall in the mains supply

In the event of the secondary also failing or an approx 20% loss in secondary supply, the circuit should automatically switch to battery back-up.

Whilst on battery back-up, with the restoration of normal OR alternate supply, these must automatically select with the priority as above

(then there is some guff about indication LED's)

Equip required

Loctronics components supplied in tool box (I may be able to ask for other basic components)
Dual DC power Supply
Relays
12V battery box
Digital volt meter

My box contains the items listed and unfortunately I don't know what a boost convertor is.

I really feel a little out of my league with this and really don't know where to start.

I was thinking of using transistors as switches or just have the relays switch between the powered paths of the circuit but my electronics knowledge isn't really that good.

Well haven't you studied the basics of electronics? This seems to me to be an end year project or something?

You can use the transistors as switches, but you need to figure out the values of resistors you will use to get it to turn on when the voltage is 20% below normal. You should probably already have learned how to do this or you wouldn't be asked to handle this problem.

I know some basics about potential divider biasing and I have some vague ideas about how it could all work.

I'm actually in a 4 man team for this project and am seen as very much the weakest link so I thought I'd try to learn how to do it and take some pressure off the others for a change. But so far I haven't been able to come up with a viable working circuit. I have very rudimentary knowledge of electronics (the electronics module I did was 3 weeks long whereas the top set people all did much longer and have a much better understanding)

Anyway, enough excuses from me, I'll keep plugging away at it I guess and hope someone can point me in the right direction.
Thank you for your reply

Your on the right track. Now a NPN transistor needs 0.7V before it will start to turn on. So you need to figure out the values for your divider to put at the very least 0.7V on the base of the transistor when VCC is 20% below normal. Is that what you had in mind? Where are you struggling to understand?

We aren't suppose to do your homework for you, so if you want help you will have to at least try and we will help you get there.

I thank you for replying. I don't want anyone to do my homework for me, I'd just like some guidance. I am prepared to play catchup and try and learn what I need. I'm just not to sure where to start.


Anyway, I know how to set up PD biasing to activate the transistor and allow it to "switch" on using r2 over r1 + r2 x Vin. Largest resistor at the top etc.
And that allows my current to flow to my load, which is the lighting circuit.

I am thinking along the lines of using a relay to switch from that power supply (am I right thinking that the relay would close the circuit if there was power on that connector, and if I use the right value relay if it drops 20% it would switch to a second position thus activating the second power supply. We learned next to nothing about relays. Just how they work and how to draw them.)

Or is there another way in which I could get the supplies to switch if there is a drop in power ?

What connector? what type of relay?

A relay will either open or close a circuit (mechanically, like a switch) if power is applied to the coil. Vice versa when power is taken away. NC = normally closed NO = Normally open. When the relay is not powered the state of the 'switch' will be either NC or NO. When powered its state changes. Does that make a it a little clearer?

That's where I'm stuck. I need the relay to drop out (or the supply to switch over) when I get a 20% drop in Vout from circuit 1. And then again if circuit 2 drops out I need it to switch to battery backup.

In my head I think of a relay with a power rating that if it drops below will switch over and the 2nd circuit will be made. I don't know if this exists though so I'm trying to come up with a simple way of the circuit changing over in event of voltage drop in the respective "active" circuit.
I also need it to switch back to "primary" (or the next level up) if the power i restored.

Infact, That's why I think I need the transistor, so that the current will flow, but the circuit will be shut if there is a drop in voltage and I don't get the .7 at the base of the transistor.I'd have the relay on the output of that circuit
and if I lose that 0.7 to the transistor the relay would switch to the other position.
I see the whole 20% drop thing as inconsequential, as I'd want the transistor to shut if I had any drop voltage and secondary woud kick in.

It's quite helpful writing this to get it out of my head.
Saying that, am I making any sense and would what I AM SAYING MAKE ANY SENSE

Your board in the picture gives you a good hint how to use the parts together. How one can control the other.

A npn transitor needs a positive voltage on its base 0.7V greater then its emitters voltage for its collector-to-emitter junction to conduct. a pnp transistor needs a negative voltage of 0.7V less then its emitter voltage for the C-E junction to conduct. You can use this with a voltage divider circuit to control the relay. Does that help?