I have a project to make a cell phone charger for my beginner's electronics class. Can someone please suggest which value capictors i should use for the filtration of initial 220 v 50 hz ac (of course will step it down and then rectify it) and the output is 9v 0.6 a. thank you.
9V I've not seen in any cell chargers. There maybe.
The most chargers of Nokia and all I've seen 5.xV output. I assume they maybe using a cap of something 1uF or more rated at 250V. However I'm not sure. Note that wrong output may lead to cellphone battery blasting and injuries. The best idea is to open a charger and see the component values inside. Not anything interesting project
This is an educational assignment so it doesn't have to be practical. You could use a 12V mains transformer to convert the 220VAC to 12VAC, a bridge rectifier and smoothing capacitor, to convert it to DC and a regulator such as the LM317 to reduce the voltage to 9V.
A 2200µF smoothing capacitor should be big enough to reduce the ripple before the regulator to an acceptable level, the circuit for the LM317 is on the datasheet which can be found using Google and the resistor values can be calculated using the utility in the tools section of this site.
If the link provide a good way of calculating the smoothing capacitor, for an unregulated power supply at 9 volts, 0.6 Amp. 50 Hz, a 6800µF (theorical 6667µF) capacitor should be used to obtain an acceptable ripple near 10%.
The ripple only needs to be low enough so the valley voltage never falls below the output voltage plus the drop-out, in this case about 11.5V.
The peak voltage is 17, minus 2V rectifier loss gives 15V, minus 11.5 is 3.5V of ripple which will require a 1700µF capacitor, 2200µF is the nearest standard value.
I was giving the answer for what I understand to be the original poster question: The smoothing capacitor needed for a power supply outputting 9V, 0.6 Amp. (50Hz for the calculations).