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Calculate capacitance across full wave bridge rectifier?

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.....)

Picture of Calculate capacitance across full wave bridge rectifier?
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1000uF should be fine for smoothing the output. If the transformer is designed to put out 9V, the output should drop back down to about 9V once you connect a load accross is.

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.
Adarsh_tronix (author)  The Skinnerz1 year ago
Thanks friend that solves a problem;but the calculation part......any help?
Sorry, I've not come accross that calculation before, but if that's what gave you the 1000uF value, it sounds about right. The exact capacitance needed is not particularly important, but bigger is generally better as the voltage accross the capacitor will wary more for lower values.
One of the very first projects I ever did when I left university, 28 years ago, was writing an accurate simulation for a linear PSU at switch on. LTspice will do the same job now. Its quite amazing some of the things that happen, like the HUGE diode currents that flow, even when the output is quite low current.
Adarsh_tronix (author) 1 year ago
1)And does these diodes eat power?because though I gave a 9 VAC input I am getting only 7 VDC.2)Also I used a 470 kOHM resistor to bring down this 7 volt to 5 and miraculously no device seems to run on this 5 volt.I tried a buzzer which worked perfectly on the 7 VDC but not on the 5 VDC.I use a 3v lithium battery to power the buzzer and it worked....so its no question of low volts...got something?
Yes, diodes drop ~0.6 volts EACH..
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

Adarsh_tronix (author)  steveastrouk1 year ago
To be honest I am very confused with resistance and capacitance but I think you have given me enough to understand the problem with my circuit.Thanks!
Resistors waster energy: capacitors store energy.
Adarsh_tronix (author) 1 year ago
You know what?....you are right bigger's better and steveastrouk thank you for setting me on track!!next time I wouldn't have to guess..
If you REALLY want the maths, start here