How can I reduce AC ripple current in a power supply?

Hello. I have a 40 amp, full wave bridge rectifier DC power supply that I would like use for radio equipment. At the moment this is not possible because there is too much AC ripple on the DC side. I know that using some form of capacitor you can make a conditioner, but I am not technical enough to know what capacitor to use.

Basically I'm asking what I need to do to cut the AC ripple.

Thank you very much :) 

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The attached image was pulled from the Wikipedia article titled "Power Supply",

http://en.wikipedia.org/wiki/Power_supply#DC_power...

And this image is, I think, the circuit you are describing. The output voltage is the voltage on that capacitor, and it is essentially DC plus a smaller AC ripple voltage.

The magnitude of the ripple voltage, call it Vpp, depends on the size of the capacitor C, the current I drawn by the load, and the frequency f of the AC supply (e.g. 50 or 60 s^-1).

For a full wave rectifier,

Vpp = I/(2*f*C)

Basically the ripple gets worse with larger current draw I, and the only way to compensate for that is to use a bigger capacitor. There's another Wikipedia page that formula comes from, here:

http://en.wikipedia.org/wiki/Ripple_%28electrical%...

Anyway this trick of using a capacitor, can only do so much. It can reduce ripple to a reasonable size, e.g ~ 1 volt peak-to-peak, but it can't eliminate it completely.

So usually the next trick is to feed the capacitor-filtered DC to a linear voltage regulator IC, like the LM317, or LM7812, or whatever.

And this arrangement: transformer, plus bridge rectifier, plus big filter capacitor, plus linear voltage regulator, comprises your traditional old-school linear, regulated DC supply.

Of course the new-school way to do this (I mean new-school at the time of this writing) is to just use a switchmode DC supply.

http://en.wikipedia.org/wiki/Switch-mode_power_sup...

Switchmode power supplies are more complicated, pretty much too complicated for an amateur to build. However, they are lighter weight and cheaper to mass produce, and work over a wider range of input AC voltage. Consequently most new DC adapters, also called "wall wart", "power brick", etc, are some kind of switchmode supply, and there plenty of these on the used market, like in thrift stores, or closets, or dumpsters.

So a used switchmode power brick might now be best, and least expensive, way to power your..., whatever your equipment is.

ACtoDCpowersupply.png
-max-2 years ago

Some bigass electrolytic capacitors (;

https://www.google.com/webhp?sourceid=chrome-insta...

You will often see these huge caps used in old large linear power supplies. If you need more regulation than that, you can then attach the output to a linear voltage regulator, and since you need 40A, the pass transistor elements will need to be in either the TO3 package, or bare minimum a modern TO-247 or similar, and you will need quite a few pass elements and a BIG heatsink.

OR, for that many amps, it will probably be more practical to use a large switch mode power supply, That would be more difficult to design, it is probably easier to buy one premade.

I am guessing OP does not truly need 40 A of load current for his radio whatever, although I could be wrong about that.

Guessing "40 amp" modifies "bridge rectifier". I mean, that formula for estimating ripple current, Vpp=I/(2*f*C), gives some big numbers for C when I= 40 A, truly some big-ass caps, like you say.

E.g. 1 volt ripple, 60 Hz, gives C = 40/(2*60*1) = 0.333 F
0.1 volt ripple, 60 Hz, gives C = 40/(2*60*0.1) = 3.333 F

ckgarside (author)  Jack A Lopez2 years ago
Realistically I'm going to be drawing 15 amps peak for transmit and about 1.5 idle.

Yeah. Wow! So the current draw really is that big. Then I guess you need a big capacitor to quiet the ripple. The same formula applies,

1 volt ripple, 60 Hz, gives C = 15/(2*60*1) = 0.125 F
0.1 volt ripple, 60 Hz, gives C = 15/(2*60*0.1) = 1.25 F

So for the size of capacitor needed, it is the same ballpark. I guess. I don't really know how small you need the ripple to be.

BTW, if you're going to be using a voltage regulator also, it will defintitely be dissipating some heat.

How many watts of heat to sink? Well, to calculate that, I multiply the drop across the regulator (the minimum is around 3 volts) times the max current (15 amperes) which gives around 45 watts, minimum, which implies kind of a big heat sink.

A switchmode supply, if you can find one, should waste less power.

-max- ckgarside2 years ago

Yeah, then just get some giant capacitors. The more the better! There are calculators online to help you calculate the amount of ripple at given frequencies (120Hz, or 2*60Hz in this case) and amounts of capacitance and current draw. Jund one and plug in numbers until you see an acceptable amount of ripple. Note: You can remove all ripple unless you have a regulated output. Alternatively you can use that formula that jack mentioned up there ^^^.

seandogue2 years ago

generally speaking, in modern electronics, we start with a "bulk" cap to knock down the ripple (generally high capacitance, w/appropriate V rating well above the max of the peak input voltage to the cap), then pass it through a voltage and/or current regulator prior to consumption by the load(s)

Common regulators one encounters in the wild are the National Semiconductor LM78xx series voltage regulators, along with the ever relevant LM317, which is often used both for voltage and also *current regulation. ("LM" is a distinction of Nat Semi, so MC78xx may be the label, or any one of many other prefix tags. the 78 and 317 topographies are widely licensed, if indeed if licensing is even still an issue after ~4 or 5 decades.)

Note: There are a wide variety of alternate, suitable regulators that are far more efficient and/or full featured than those I mention above, but they're reliable standard that most of us start with.