hi,
im new to electronics ive made a 5v power supply but got a problem with my bridge rectifier im measuring 11.3ac input but im getting 12.3 dc out of it
am i right in thinking the output will be half of the input so i should be getting about 6v?
thanks
Did you add a capacitor at the DC output of the bridge rectifier ?
If so, this is how it is calculated:
The top of the sine wave is 1.4 * the measured AC voltage.
The diodes have a voltage drop of perhaps 0.7V each.
The output is ( 11.3 * 1.4 ) - ( 2 * 0.7 ) = 14.42 V
Since you are measuring 12.3 V, pehaps that is with a load.
yes i have a 1000uf cap after it which feeds into a 7805.
Krodal isn't wrong, but there is the option without a capacitor.
Peak voltage after the bridge is 14,42V. A multimeter measures the rms-value, the root mean square. This is, in essence, the equivalent constant DC-voltage over the measuring period (display refreshes about every second, I would guess).
After the bridge you don't have a constant voltage, it is pulsating at double your mains frequency.
Imagine a sine wave, it has peaks alternatingly on the positive side (up) and the negative (down) side. A bridge rectifier, to put it simply, flips the negative side up and lowers the peak voltage (as explained by Krodal).
If you have your multimeter set to V DC, it will measure not the peak voltage, but the equivalent DC-voltage.
Imagine your pulsating voltage after the rectifier. There are peaks and there are valleys. Now imagine taking some voltage from the peaks and filling the valleys. Sooner or later you won't have peaks and valleys, but little hills, and if you keep going you will have a flat line. THAT is what your meter is displaying for you. It will be higher than your AC-rms-voltage and lower than your DC-peak-voltage.
This much to how you got more voltage. Now to what you could try.
This way you cut your voltage in half (or more), because after each peak you have along pause.
Now try the whole thing with a capacitor AFTER the diode. The voltage you measure will rise. The bigger the capacitor, the higher the voltage.
However, if you attach your load (or, for tests, any load), your will see the voltage drop again.
So you will need to stabilize it, but you should find enough info on the web. Or else, just ask again 
Greets
Tom
OK, so I took too long to reply 
Anyway, it should work, if you don't draw too much current. The 7805 has to "burn" the excess power (example: 12V in - 5V out = 7V; 7V x 1A = 7W) and will get warm as you draw power, and warmer, the more power you draw. If it's too much, use a heat sink or your regulator will die. And watch your max. current for the 7805.
Greets
Tom
thanks very much for the help, that really helped a lot, my load is 225ma so well within my 7805 max load with a heat sink fitted, still gets really hot though
Pillager:
OK, so I took too long to replyAnyway, it should work, if you don't draw too much current. The 7805 has to "burn" the excess power (example: 12V in - 5V out = 7V; 7V x 1A = 7W) and will get warm as you draw power, and warmer, the more power you draw. If it's too much, use a heat sink or your regulator will die. And watch your max. current for the 7805.
Greets
Tom
Is the die for drama, or are you not aware that a 7805 like most all 3 pin linear voltage regulators has internal self protection circuitry to automatically shutdown if device temperature or maximum output current exceeds it's design limit values? It's usually a datasheet front page bullet point feature of such devices.
Lefty
yes i read some where they have a thermal overload, tested one by shorting the + to gnd, very cleaver, is there any need to use a fuse in a circuit if thats the case
thanks all
welovedc10:
yes i read some where they have a thermal overload, tested one by shorting the + to gnd, very cleaver, is there any need to use a fuse in a circuit if thats the casethanks all
Not really, a fuse would be used to protect other things upstream if needed to be protected, but the internal regulator's automatic shutdown can be allowed to continue indefinitely if you so desire and will self recover if the output short circuit is removed, which a standard 'go and blow' type fuse can't recover from.
Lefty