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please explain the components of a power supply

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SGC622

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Could someone explain the components of this circuit to me and their purposes and if you increase values to any given component what it would do, and say you wanted to get a higher amperage out of this circuit on the output side. how would you do that increase the component amp ratings? thank you ahead of time for any help thank you
 
Transformer adjusts the AC voltage up or down so the RMS of the AC is closer to the DC output you want (it's easier to change AC than DC so you might as well do as much as you can while it is still AC).

Diodes rectify the AC so it is unipolar and the capacitor smooths it out to get a DC with some ripple in it (since the capacitor isn't perfect).
 
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how do you know how many farads the capacitors need to be? and are there any specifics to what size the diodes need to be because i built a fullwave bridge rectifier before and i got 12vdc out of it but i tried to run a cpu fan for a test and it did not work what did i do wrong?
 
THere are calculations but it's a bit complicated. Just make it big. (like 1000uF) or bigger like 4700uF, or even bigger. THe bigger the load, the bigger the capacitor.

THe diodes just need to be faster than the AC input frequency. THe same thing goes for the capacitors. THey have to be of a type good enough to handle the input frequency (larger capacitors at higher frequencies end up looking like open circuits and that does not help). That's about the only reason other than cost why you don't want a capacitor that is too big.

Did you have a capacitor at the end of your full wave rectifier? If you didn't that would explain why your fan wouldn't run. It'd be almost like repeatedly connecting and disconnecting the fan from a battery at 120Hz.
 
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i did not i only had the typical configuration for a rectifier with the capacitor across it as the i assume "bridge" making it a full wave bridge rectifier correct? 120? it seems its doubling the ac frequency true? and if so how? thank you for your posts im learning alot
 
Look at a sine wave. It has two humps per cycle. One is positive, one is negative. THe rectifier makes all the humps positive. What's the frequency now?
 
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ok i get the frequency part and the rest of the circuit but what else confuses me is what the capacitors on the ac side of this circuit are for?
 
Probably because the mains voltage is noisy and the capacitors are there to help quiet some of the high frequency noise so more of a pure sine wave goes into the transformer. The capacitors are probably small so that they look like a high impedance to the 60Hz AC so it does not go through them as much as it goes through the transformer. But for higher frequencies, capacitors look like low impedance or short-circuits so they travel through them instead.
 
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how do you know how many farads the capacitors need to be?
First you have to decide how much ripple you can tolerate on the output. You can then approximate the smoothing capacitor size with this formula:

C = I / (Hz * Vrip)
Where C is capacitance in Farads, I is the current in amps, Hz is the frequency in Hertz and Vrip is the ripple Voltage in volts. Remember that a fullwave rectifier produces 120Hz pulses when the line frequency is 60Hz, so use 120Hz and not 60 in the formula for a bridge rectifier.
 
2 questions
(1)how do i know how much ripple the output device can handle?
(2)Say my output device is 12v 3amps i know to make capacitors big aswell as the diodes but is there a formula to get me close to where i would have to be my concern is would it be less efficient if i used say a higher uf cap rather than a lower?
 
2 questions
(1)how do i know how much ripple the output device can handle?
It depends on the device. Something like a DC motor can handle plenty of ripple, where as an audio amplifier needs very clean DC to avoid hum in the speakers. Usually if you require very low ripple you follow the rectifier/cap with a regulator to reduce the ripple to very low levels. You can also use a LC PI network, but this is seldom used in linear supplies due to size, weight and cost. In switching supplies it is more practical due to the high frequencies used.
(2)Say my output device is 12v 3amps i know to make capacitors big as well as the diodes but is there a formula to get me close to where i would have to be my concern is would it be less efficient if i used say a higher uf cap rather than a lower?
One of the problems with using a large capacitor is the surge current when the circuit is turned on. another is size and cost as the cap gets bigger.
It is basically a compromise between allowable ripple and the largest capacitor that is practical.
What exactly are you trying to build or learn?
 
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im just trying to learn all the factors in what would affect a load i could hook up to this aswell as help me to understand my failed attempt at building a fullwave bridge rectifier and why it didnt work, just to be clear the capacitor that we are talking about that decreases the ripple is the one on the end right before the load that is connected in parallel with the load correct?
 
The I²T rating of the diodes has to be higher with larger filter caps.
 
just to be clear the capacitor that we are talking about that decreases the ripple is the one on the end right before the load that is connected in parallel with the load correct?
Yes, it is the one labeled C1 in your diagram.
 
one more question and i think ill understand this or any power supply
when you make a full wave bridged rectifier say you make the classic diamond shape out of the diodes what is the capacitor that goes across the diamond for? and thats the circuit i built that didnt work(refer to previous replys above). would it then still need the capacitor labelled (C1) on my diagram.
 
what is the capacitor that goes across the diamond for?
For filtering the DC, if you need it filtered.

vac rms 120
vac pk 169.68
vdc halfwave rect 53.95824
vdc full wave rect 107.91648
vdc half rms value 84.84
vdc full rms 120
 
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