I'm currently working on my project that teacher gave.I need to convert my 15VAC input to 5VDC output.

EDIT : I am so sorry. I am not experienced user. I should give you more information about where i am stuck at.

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    \$\begingroup\$ What else do you know about the requirements? \$\endgroup\$ – Spehro Pefhany May 14 '14 at 19:51
  • \$\begingroup\$ I am going to use bridge type diode to make my voltage all positive.Then i need to filter it to make it seem like DC.After that i should use 7805 IC to make my voltage fixed 5VDC.But i do not know which values i should use as my capacitors. @SpehroPefhany \$\endgroup\$ – Pyro May 14 '14 at 20:01
  • \$\begingroup\$ I am not a lazy student to let you make all work for me.I just want your help.Is it bad? \$\endgroup\$ – Pyro May 14 '14 at 20:03
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    \$\begingroup\$ It's best to let us know everything you know about the problem, and show that you've made a serious effort to solve the problem and explain where you are stuck. If you do that, you'll be sure to find friendly help. If you ask someone to do your homework for free right now, you'll not find a warm welcome. It's more typing for you, but that's the way it works. \$\endgroup\$ – Spehro Pefhany May 14 '14 at 20:11

The main concern you have is to calculate the filter capacitor value. The recommended output bypass capacitor value you can get from the regulator datasheet.

If you use a full-wave bridge, you can approximate the capacitor as being charged only at the peaks of the AC waveform, which is twice whatever the mains frequency is in your country.

So you will need to calculate the capacitor value based on the allowable "droop" in the voltage when the capacitor is not being charged. So if your mains is 50Hz, it will be charged every 10msec, and will discharge with the current going into the regulator.

You can calculate this easily from the definition of the behavior of a capacitor with constant current flowing (borrowed from this website).

enter image description here

If you ignore the curvy bit and just assume it's charged right at the peak, it will be a bit conservative (which is good).

In a real design, you would have to consider line voltage variations, capacitor tolerance and aging, capacitor ripple current rating, and dropout voltage of the regulator under all conditions.

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