# Calculate capacitance across full wave bridge rectifier?

I've added some pics(not much of help but.....)

OK,so I need an AC to DC converter.I used a 9 Volts transformer to convert 220VAC to 9VAC,then rectified it to 9VDC using a full wave bridge rectifier(IN4007).I need to know the value of capacitance which is to be connected to the 9 volt DC output(I heard somewhere that its got by multiplying the forward current*5 then dividing it by input voltage to the rectifier and the result is obtained in Farads........well I'm not sure if this is right and I need clarification).I've currently used a 1000uf capacitor and my multimeter shows the dc output from the bridge as 14 volts which is really confusing because before I used the capacitor it showed 9 volts.I really need a way to deal with capacitance.Pls do help me.

I've added some pics(not much of help but.....)

I've added some pics(not much of help but.....)

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The no-load voltage appears to increase when you add the capacitor as the capacitor charges to the peak output voltage, and stays charged to that level as there is nowhere for the charge to go.

You'll have to use a REGULATOR to give you 5V.

Think about it. What happens if you connect the other end of your 470k to the -ve of your supply ? You'll see 0V, and 7/470,000 Amps flows.

What happens now, if we put a load that is say 100Ohms, instead of 0 ohms( a short) ? V= 100/(470,000+100) x 7 = 100/470,100 = 0.001 V !!

What happens now, if we leave the resistor OPEN circuit ? V=7, I=0.

So the bigger the load = the smaller the resistance,, the LESS output voltage you get

What you have done is put a voltmeter (nominal input resistance ~1,000,000 ohms across the supply, now, v=7*1000000/(1000000+470000) and v= roughgly 5 !!!

Your buzzer's resistance is more likely to be LESS than 1000Ohms. Measure it, put the numbers in the equation

v = 7 x buzzer resistance / (buzzer resistance+470,000) ....and measure v