I am building a simple full wave rectifier with smoothing capacitor. I am using the following transformer to drop the input voltage:

in: 120V 65W
out: 12V 4170mA AC

The rectifier bridge is 50V 25A (I also have 400V 8A). The power supply is going to be used for 18W guitar amplifier based on this schematic:

schematic

Since I am still a noob when it comes to electronics, I have have no idea what value the smoothing capacitor should have.

Anybody can help me out?

Thanks.

 

C = 0.7 *i /ΔE /F = 0.024F or 24,000uf

Assuming 1V peak to peak ripple and 120hZ ripple freq

 

Read this
http://www.electro-tech-online.com/...-how-calculate-value-smoothing-capacitor.html

Or use this equation

C = Q/V = ( I * t )/V

where
I - load current
t - discharge time 1/50Hz (20ms) for half wave rectifier and 1/100Hz (10ms) full wave rectifier or 1/60Hz and 1/120Hz for countries with 60Hz mains.
V - capacitor ripple voltage

 

Thanks for replies. So if my math is correct, according to Jony130's equation I am getting:

C = (2A * 1/120Hz) / 1V = 0.016F

Is that correct? The amp draws 2A max to produce 18W.

 

Actually, the capacitor may be reduced by the 0.7 factor--jony130's method assumes instant charging--in my method, it is assumed that the transformer supplies the current for 30% of the cycle and the capacitor supplying current for the remaining 70%.

 

Thanks again for the replies. I know now how to calculate the smoothing capacitor value! The problem is that the price of 24,000uF cap is somewhere close to $35. I may be better reusing one of my switching power supplies that I have in my project box.

 

You don´t need a 24000uF cap, you can use for example 5 4700uF caps in parallel. For 35V they should cost like $5 total.

 

But you don't need such big capacitor, simply buy 6800uF.

 

In agreement with Jony130, the value is not all that critical--I would add that it does no harm to go larger as well.

 

I have few 2200uF caps and maybe one 3300uF. I can probably add up to over 10,000uF with what I have in my parts bin.