Why not using large capacitor to filter out all ripples after the full bridge rectifier

Thread Starter

Chaotic Breeze

Joined May 3, 2019
4
Sorry for the naive question, but I'm just wondering why can't we kill all the ripples by using a capacitor with a large capacitance instead of other more sophisticated methods ?
 

mvas

Joined Jun 19, 2017
539
1) Large farad capacitors are more expensive.
2) Large farad capacitors are physically larger
3) Large farad capacitors act more like a short circuit, especially on power-up
4) Some low ripple value is acceptable

The ROI ( return on investment ) for ever larger farad capacitors is smaller and smaller
There are always trade-offs in electrical engineering.
 
Last edited:

DickCappels

Joined Aug 21, 2008
10,187
Besides, just using a huge capacitor does eluminate all if the ripple, and it does very little if anything for regulation, which is sometimes
Important.
 

crutschow

Joined Mar 14, 2008
34,468
why can't we kill all the ripples by using a capacitor with a large capacitance instead of other more sophisticated methods ?
Those sophisticated methods do more than just "kill all the ripples", such as regulating the voltage, limiting the current, and providing short-circuit protection.
 
Last edited:

WBahn

Joined Mar 31, 2012
30,077
Sorry for the naive question, but I'm just wondering why can't we kill all the ripples by using a capacitor with a large capacitance instead of other more sophisticated methods ?
Because, all other factors aside, it would require an infinite capacitance to kill all the ripple. As soon as the capacitor supplies a single electron of charge to the current, the voltage across it changes and, hence, you have ripple. The more current the load needs, the more ripple you have for the same capacitor.

So you have to decide how much ripple is acceptable based on the heaviest load current you are designing for and determine a minimum capacitance that will achieve that result. But if that capacitance is too large, then you run into other problems. You peak currents go up and the circuit supplying the current to recharge the capacitor may not be able to deliver it. Also, if you have a short then that capacitor might be able to deliver destructive levels of current into it. Also, the average voltage supplied by the capacitor will be a function of the load current, meaning that it is not well-regulated. Depending on which of these is important enough to not just be lived with, you need to add additional circuitry to deal with it.
 

Thread Starter

Chaotic Breeze

Joined May 3, 2019
4
Many thanks for all of the responses, I have a better idea about the role of regulation now.

On further consideration, this sounds an awful lot like a homework question. Is it?
It's not a HW, that would be awful xD. I am working on building my own power supply for testing breadboard circuits in general.
 

crutschow

Joined Mar 14, 2008
34,468
I am working on building my own power supply for testing breadboard circuits in general.
The old standby, the LM317 adjustable regulator is often used for that purpose.
It requires only two resistors to set the output voltage (one can be a pot to make it adjustable), and is short-circuit and over-temperature protected, making it nearly bullet-proof.
It will output up to about 1.5A, which is sufficient for most electronic circuits, unless they are driving high power loads.
 

Glenn Holland

Joined Dec 26, 2014
703
On a related note, capacitors are somewhat undesirable in very large DC power supplies that are used mainly for motor loads.

In mass transit, DC traction power for rail cars is supplied by a Wye-Delta transformer connected to a 12 pulse rectifier. This reduces the ripple to an acceptable level while eliminating the needed for smoothing capacitors. Interesting to note that some ripple on a DC traction power supply is desirable as a method of determining whether the overhead wire is energized by the power substation or by regenerated power from the rail cars themselves.

A 12 pulse rectifier will have a 360 and 720 Hz. ripple which is detected by the input to the motor controller and that will enable regenerative braking. However, if the ripple is absent, regeneration will be disabled.
 
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