Assignment Help

DamienH

Hello,
I've been given an assignment to do by Friday and I'm just trying to get my head around it.

We've got to design a remote control system for control of a light. We are also required to design a seperate power supply for the reciever. The transmitter has one button, when pressed the light should change state. There should also be no greater than a 100mv P2P ripple from the regulators.

There should be two regulated supplies one for the light ( rated 12V +/- 2%, 10W ) and one for the receiver system.

We can assume that the transformer can be represented as a Thevinin equivalent ac generator of 22volts peak amplitude with output resistance of 0.9ohm.

I've decided to use the HT12A and HT12D as the transmitter and receiver. We were told that we don't have to worry about the IR diode bit of it. Both of these circuits can be pulled straight from the data sheets. My main problem is with the Power Supply.

Never had to do something like this before. I've tried sitting down with a few people from my class but the solutions that they've come up with seem a bit too simple to work. I've drawn my circuit ( with my paint skillz ) in the first file below. Now I've done a good bit of head-scratching on this and I'm pretty sure it's okay.

As regards values for the C's and R's in the diagram I'm slightly stuck.

My first question is whether or not I need a resistor before the regulators to limit current. From the data sheet it says they are rated for P<15W. My rectified voltage will be approx 20.6Volts, with current being something like 17A ( <---- V/R Think it's correct ) which is obviously going to blow the IC sky high. Would a resistor somewhere in the KOhm range do the job? This could be put after the regulator's couldn't it in order to control the current flowing into the regs

As regards the capacitor in front of the regulators I found the formula shown in cap.jpeg.

These are the main questions I can think of now and I greatly appreciate any help that anyone can give me!

Thanks a lot,
Damien

DublinDilbert

Hi,

Firstly your supply.jpg looks pretty good....

If your bulb is 10W 12V, the regulator feeding the bulb must supply 10/12 A ( P=VI). hence that regulator must supply 0.83A to the bulb.

You'll need to calculate the power dissipation in the 12V regulator, its going to be large, if the peak output from the rectifier is 20.3V, then the regulator must dissipate 0.83x(20.3-12) = 6.9W, which is a big amount of heat, you'll need to calculate the size of the heat sink for this.

You'll need to spec regulators to get the required tolerance for the applications, eg +/-2% on the 12V supply.

Michael Collins

I was going to question your 17 A part, but DublinDilbert has worked out what it should be for you. Also just remember that 10% of 12 V gives you 1.2 V so you will need a larger capacitor than that equation gives if you want < 100 mV ripple.

DamienH

Thanks a lot for answering all these questions for me.

The only thing that stumps me is that the regulators are rated at less than or equal to 15W? Since the voltage is 20.3 at it's peak will this not destroy the IC?

As regards the ripple, from the data sheet of the 7805, it says that regulation is within 100mv p2p when Io= 250ma -> 750ma. Am I right in thinking that by sticking a resistor of the correct value after the 7805 I can limit it's output current? This should take care of the ripple without having to use the cap...

Thanks for the power dissipation aswell, forgot about that

Michael Collins

Sorry I misread the question a little, so ignore my previous post!

As regards the ripple, from the data sheet of the 7805, it says that regulation is within 100mv p2p when Io= 250ma -> 750ma. Am I right in thinking that by sticking a resistor of the correct value after the 7805 I can limit it's output current? This should take care of the ripple without having to use the cap...

If you only required up to 750 mA from your regulator that would be fine, but as DublinDilbert showed, the current output for a 10W bulb at 12 V is actually 0.83 A -> 830 mA. You can't just stick a resistor at the ouput because your specification requires up to 0.83 A to be made available - a resistor would reduce that.

The absolute minimum voltage you are allowed at the output is 12 V - 2% = 12 V - 240 mV. So your tolerance needs to be +/- 240 mV. So if an output current of 750 mA gives a voltage tolerance of 100 mV, will an output current of 830 mA give you less than 240 mV? It's hard to say, but it will probably be OK - the only way you'll know for sure is when you build it!

The only thing that stumps me is that the regulators are rated at less than or equal to 15W? Since the voltage is 20.6 at it's peak will this not destroy the IC?

If the 20.6 volts was across the regulator then you would be correct. It isn't though. The voltage across the regulator is 20.6 - 12, so the max dissapated power is (8.6 V)*(0.83 A) = 7.138 W, which is under the 15 W limit. As DublinDilbert says though, it is still a reasonably large amount of power to dissapate, you will need a significant heat sink.

DamienH

That's a bunch man, good old boards to the rescue. Better than the lecturer

DublinDilbert

DamienH wrote: »

The only thing that stumps me is that the regulators are rated at less than or equal to 15W? Since the voltage is 20.3 at it's peak will this not destroy the IC?

These are 2 separate things...

The max input voltage of the regulator should be greater than 20.3V, most linear regulators will meet this requirement, but check in the data sheet.

Yes you'll need to choose a regulator which can dissipate the power. You'll need to pick a heat sink too, these are rated in degrees C / watt dissipated. You'll want to keep the temp of the regulator below say 80C. Your regulator will dissipate 6.9W, as calculated above... So if you pick a heat sink which has a rating of 10 degrees C / watt, it will go to 69 degrees above ambient..