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Registered Member #621
Joined: Sun Apr 01 2007, 12:37AM
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Posts: 119
In a small DC power supply consisting of an AC step down transformer and a full wave rectifier bridge and a smoothing capacitor bank, if the power draw from it is about 3-4 amps, would a 6,000uF bank of caps be sufficient to smooth very well?
The reason I ask is I will be pulling just over 3 amps, the transformers and diodes I'll buy will be rated for greater for safety (like 5 amp transformers and 6 amp diodes), but I have read many conflicting ideas on the amount of smoothign capacitance I need.
One source quotes "at least 1000 uF per amp of current draw" so I figured 3,000uF would cover 3 amps, but double it for safety to 6,000uF.
I purchased the capacitors (12 500uF that I will parallel) and they are MUCH smaller physically than I expected! So if I must I could probably easily fit 24-36 of them for 12,000 or 18,000 uF, but would that be overkill?
Is my 6,000uF sufficient for a nice smooth DC under 3 amp draw?
Registered Member #142
Joined: Sat Feb 11 2006, 01:19PM
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Posts: 102
The formula you want is CV=It capacitance times voltage equals current times time here voltage is actually change in voltage (voltage drop in 1/120 sec, if you use a full bridge rectifier on a 60 Hz transformer) so C = .006, I =4, and t = 1/120 the ripple on your caps is 5 or 6 volts. A lot. Add more caps.
Registered Member #621
Joined: Sun Apr 01 2007, 12:37AM
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Posts: 119
Thanks, shoot I just thought of something, will my voltage regulators on the final output remove the ripple? I am kind of thinking they may if the caps voltage is at least equal to what I am pulling plus the ripple?
What is the normal ripple on a decent wall wort transformer or a good power supply, like whats considered acceptable?
Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140
"What is considered acceptable?" depends on the requirements of your circuit!
A typical linear ic regulator has about 60 to 80 dB ripple rejection, that's 1,000 to 10,000 in voltage terms so if the raw dc has 6V ripple, then the regulated output would have 6mV to 600uV ripple you need to decide if that's acceptable.
Most importantly you need to ensure that the minimum voltage to the regulator is greater than (output voltage) + (dropout voltage)
Registered Member #142
Joined: Sat Feb 11 2006, 01:19PM
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Posts: 102
Your transformer ouput will have a peak voltage at no load. Put a cap on it and measure that voltage. Then subtract the ripple you calculated, to find the minimum voltage going into your regulator. That still has to be at least 2 or 2.5 volts above the output of the typical linear 78xx regulator, if I remember correctly, but you can refer to the datasheet. So you have enough numbers to figure it out now.
Registered Member #32
Joined: Sat Feb 04 2006, 08:58AM
Location: Australia
Posts: 549
Sulaiman wrote ...
"What is considered acceptable?" depends on the requirements of your circuit!
Exactly. If it's providing power to a regular lamp, a small amount of ripple doesn't matter. If you're powering an analog circuit with high-gain parts (like most audio circuits), ripple = pain.
If you look at old circuits, they have huge filtering caps but in modern times IC regulators are so cheap and effective it isn't worth it. You use caps to make the output basically smooth and then use a regulator to chop the ripple out.
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Smoothing capacitors in a Small DC Power supply
