What does 40VA rating for transformer mean

transformer

I am working on a circuit for a home make home security system. I bought the following wall transformer to power my circuit. I was going to use with a 5VDC Linear voltage regulator to get 5 volts for my digital circuitry.

http://www.mpja.com/prodinfo.asp?number=12635+PA

I couldn't find any documentation on this transformer. When I plug it into the wall and measure the voltage at the output I get zero volts. I tried attaching it to a couple different loads (with fairly small resistance) but I still couldn't measure any voltage at the output. I realize this is probably because the transformer is not regulated and thus requires a load for the output to be 12V but I am still not sure if I can use this transformer with my design. Does this thing only output 12VDC if it has the exact right load? Or since I will be connecting this to a linear voltage regulator will that handle it all for me? The total current draw of my circuit will typically be less than 200mA unless the alarm is sounding, and then it will go up some.

Also, I was planning on using the 12V from the transformer to power the horn(siren) directly and switch it on with a power Mosfet. The 12 Volts will not be regulated so do I need to add a big resistor in series with the horn so that it will have the proper 12 Volts across it?

I guess I should have just spent a little more and got a regulated supply!

Best Answer

Those 12V are the effective AC voltage. To get the peak voltage you need to multiply by the square root of 2, which results in about 17V.

40VA means that at an effective voltage of 12V it can handle an effective current of 40VA / 12V = 3.3A, which should be enough (though you neglected to tell us how much current the horn needs).

You need a rectifier and a smoothing capacitor in front of the linear regulator. The rectifier will reduce the voltage by about 1.4V, so at the input of the linear rectifier you will have a little more than 15V, with some ripple depending on the capacitor and the current drawn.

At 200mA and 10V difference between input and output your linear regulator will need to get rid of 2W of power, so it will need some cooling.

As for the horn, it depends on the specification of the horn, how precisely it needs those 12V. If it can tolerate some ripple you could use five 1N4004 diodes in series (5 * 0.7V = 3.5V, 15.5V - 3.5V = 12V) to lower the voltage. This is better than a resistor, because the voltage across a resistor depends much more on the load than for a diode.