AC/DC Rectification. VDC = 1.4 x VAC?

Thread Starter

TheLaw

Joined Sep 2, 2010
228
Okay,

This is a super quick question. I haven't build a power supply in ages. Need one for my amp.

I think I remember hearing that VDC is equivalent to 1.4V x the VAC? Or roughly around there. But if my bridge rectifer has a voltage drop of around 0.9V, how does that play in?

So 12VAC transformer + rectifying bridge with 0.9VDROP = X-voltage DC?

What would be the VDC then? I need atleast 14.5-15VDC.

Thanks.
 

Kermit2

Joined Feb 5, 2010
4,162
if you are correct - then 12 volts times 1.414 is 16.9. Minus the diode drop(call it 1.3 volts because its bridged). Leaves 15.6 volts.

Would seem you are golden. :)

Since you are going to 'build' this thing...do you have a multi meter?

Also I seem to remember a bridge rectified circuit uses a different multiplication factor than a full wave rectified circuit uses, when calculating DC for a power supply. Check out the two types using google or wikipedia
 

Thread Starter

TheLaw

Joined Sep 2, 2010
228
if you are correct - then 12 volts times 1.414 is 16.9. Minus the diode drop(call it 1.3 volts because its bridged). Leaves 15.6 volts.

Would seem you are golden. :)

Since you are going to 'build' this thing...do you have a multi meter?

Also I seem to remember a bridge rectified circuit uses a different multiplication factor than a full wave rectified circuit uses, when calculating DC for a power supply. Check out the two types using google or wikipedia
Yes I have a multimeter.

Thanks for verification.
 

Kermit2

Joined Feb 5, 2010
4,162
Ya know. A little more reading on my part would have revealed the fact my probing for shop equipment was not called for.


My apologies for my seeming rudeness.
 

t_n_k

Joined Mar 6, 2009
5,455
Presumably you intend to add some reasonable capacitance in parallel with the DC output to provide some filtering of the DC load. Your amp may not work very well on full-wave rectified AC.

Also, even a simple power supply design requires some knowledge of the expected load current.
 

Thread Starter

TheLaw

Joined Sep 2, 2010
228
Presumably you intend to add some reasonable capacitance in parallel with the DC output to provide some filtering of the DC load. Your amp may not work very well on full-wave rectified AC.

Also, even a simple power supply design requires some knowledge of the expected load current.
Yes of course. I'm going to use a 12V regulator with 2200uF on the primary side + 100nF film + 220uF on the secondary and another 100nF fim.

This is a headphone amp by the way. Don't need much output 100mA is about peak power draw.

By the way, I've heard that putting ceramics on each IC input reduces noise. So a few on the rectifing bridge and the regulator? How do you determine their value? And is it like inline with each lead?
 

Thread Starter

TheLaw

Joined Sep 2, 2010
228
It is a handy article, but I kind of knew that much so far. I'm talking about the noise filtering around ICs. =D
 
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