DC power supply from AC through bridge rectifier

[vick5821]

because as the current increases, the capacitor struggles to maintain the average DC voltage. It results in a voltage drop, mainly caused by the increase in ripple voltage.

So as you again increase the value of the capacitance to overcome the increase in ripple voltage, it also has the effect of holding the average voltage up.

This is why guys with really hi power stereo systems in cars use HUGE capacitors across the 12V car battery. it holds the average voltage up during the hi peaks of output of the amplifier ( which of course coincides in peaks of current demand)

Dave

Why it is so ? Any detail explanation ?

Thank you

 

[jackorocko]

think of a glass of water. Capacitors store current, like a glass holds water. As the load consumes more current, the capacitor is emptied per say quicker. Like a glass of water that is being poured faster.

capacitors dump stored current to try and maintain the voltage across the capacitor. The faster the current is dumped the quicker the voltage across the capacitor drops.

 

[(*steve*)]

Vick5821, have you covered the equation i = C dv/dt ?

If so, do you understand it at all? If you're not overly mathematically inclined I understand, but if you are it should bring some clarity.

 

[khankll]

nice discussion learned few bits from it..
let me throw in my question though i made a seperate thread and waiting for responses...

what should be the voltage rating of output capacitor? say my output is 12volts after all rectifications etc..
and what should be the value of capacitor capicatance ?
how to get ideal value of capicatance ?
and what if that value is not used and somehigher value is used or some low value is used..
ho muh can one goo high and low from ideal value ?

i found this eqiuation from previous page
Δ V = I / 2fC

where V is ripple voltage (mV), I is DC load current (mA), f is frequency of AC supply and C is smoothing capacitor value (F). ..
sayif i want max ripple of 100mV,i=350mA,f=60hz,
c=9.2 mF ...or 7291 uF.. ?
is this correct ?
and fore voltage rating say my output is12 volts i should just make sure my cap o/p volts are >12 volts...?

 

[poor mystic]

Power supplies are difficult to model.
Try making a model yourself, and see what you think. There's so much to consider!
For any given capacitance, with some transformer, and some DC load, there will be 120 'inrushes' of current every second (in the USA) or 100 per second elsewhere.
The size of the maximum current in the rush depends on the value of the cap, the impedance of the transformer, and the load. If the inrushes are great (large capacitor, high current draw, big transformer) you need to use big rectifying diodes, to handle the currents.
Sometimes it is much much easier to try something and measure the results, and make adjustments to the design based on what you find.
For the voltage rating of the capacitor, things are easier. Use a cap whose voltage rating is no more than twice the highest expected voltage, which means in your case a 20V to 40V capacitor. Higher voltage capacitors might not work so well in your low-voltage setting. My 40V is pretty arbitrary.
Use big fat high-rated diodes for everything. They're cheap, time isn't. If you want an economical design to share around you can come back and make measurements later.
Don't use an under-rated capacitor.

 

[gorgon]

i found this eqiuation from previous page
Δ V = I / 2fC

where V is ripple voltage (mV), I is DC load current (mA), f is frequency of AC supply and C is smoothing capacitor value (F). ..
sayif i want max ripple of 100mV,i=350mA,f=60hz,
c=9.2 mF ...or 7291 uF.. ?
is this correct ?
and fore voltage rating say my output is12 volts i should just make sure my cap o/p volts are >12 volts...?

The easiest way to get a correct result is to use the basic units when computing the formula. As far as I can see, you are a bit off in yours results. I think you will be surprised, due to the low ripple of 100mV.

For the voltage value of the capacitors, use 20 or 25V, for a 12V DC.

TOK

 

[vick5821]

hi Vick

your scope is showing you the peak voltage out of the the transformer, where as the transformer is usually rated by its RMS voltage. That is.... if you transformer has an output of 10VAC... that is its RMS voltage. Its peak voltage is the RMS value x 1.414

your DC voltage out of the rectifier is the RMS value minus the voltage drop in the diodes ( ~ 1.7V for 2 diodes). You need to remember that the diodes are only conducting current in brief bursts. And when you add a capacitor to the output, the current through the diode charges up the capacitor, which then slowly discharges through the load till the next burst of current from one of the diodeson the other 1/2 cycle.
So the capacitor is holding up the voltage during the time that the diodes are not conduction. If you have only a small capacitor, then the voltage is going to drop more ( cuz it holds less charge) than a larger capacitor.
this is where your ripple comes from....
The rectifier diodes will charge up the filter capacitor, to the peak DC value, and between non conducting cycles of the diodes, it will discharge into the load resistor. This creates the sawtooth waveform known more commonly as ripple voltage. The value of the ripple voltage is dependant on load current, power supply frequency and capacitor value. Approximate ripple voltage is calculated using:

Δ V = I / 2fC

where V is ripple voltage (mV), I is DC load current (mA), f is frequency of AC supply and C is smoothing capacitor value (F).


have a ponder on that.... I need to go out for some hours

cheers
Dave

Rectifier diode ?? Not the output voltage /rectified voltage ?

Thank you

 

[khankll]

Rectifier diode ?? Not the output voltage /rectified voltage ?

Thank you

he meant to say that the voltage accross rectifier diode..

normally people say it in the way as he mentioned ..
for more ifo read what a diode is.. its not a source ...

 

[khankll]

The easiest way to get a correct result is to use the basic units when computing the formula. As far as I can see, you are a bit off in yours results. I think you will be surprised, due to the low ripple of 100mV.

For the voltage value of the capacitors, use 20 or 25V, for a 12V DC.

TOK

so for such a setup whats the normal ripple size and m i using the correct equation correctly ? and its my 'unrealistic ripple' voltage choice thats giving me strange cap value ?
thanks

 

[gorgon]

so for such a setup whats the normal ripple size and m i using the correct equation correctly ? and its my 'unrealistic ripple' voltage choice thats giving me strange cap value ?
thanks

No, that is not the point. The ripple voltage you have specified is not unrealistic, it's just the reason that the capacitor needs to be rather big.

The main point in what I said was that you should use the basic units of the different values, that should make the computation easier to get a correct answer. Use just Volt, Ampere, Hz and Farad, not all the prefixes. Like 0.1V, 0.35A and so on.

TOK

 

[khankll]

No, that is not the point. The ripple voltage you have specified is not unrealistic, it's just the reason that the capacitor needs to be rather big.

The main point in what I said was that you should use the basic units of the different values, that should make the computation easier to get a correct answer. Use just Volt, Ampere, Hz and Farad, not all the prefixes. Like 0.1V, 0.35A and so on.

TOK

still it comes quite a huge value .. .. 29.16 mF .. aint it humongous ?
all i have noted is just 470uF.. wonder y such a low value.. ?
this thing always confuses me..

 

[(*steve*)]

still it comes quite a huge value .. .. 29.16 mF .. aint it humongous ?
all i have noted is just 470uF.. wonder y such a low value.. ?
this thing always confuses me..

Well, if you reduce ripple solely using a filter capacitor, you need very large capacitors.

You can improve things further using inductors.

Or you can use a regulator.

Another option is to increase the frequency...

 

[gorgon]

still it comes quite a huge value .. .. 29.16 mF .. aint it humongous ?
all i have noted is just 470uF.. wonder y such a low value.. ?
this thing always confuses me..

If you use a 4700uF capacitor, you'll have 0.62V ripple, maybe you forgot a '0'?

TOK

 

[vick5821]

How is consider engineering work ?

How would be the title of the graph be ? And the label of the axis ?

For example, I want to draw a graph of sine wave input and draw another graph for sine wave output from a transformer. So what shoud be my engineering way of title ?

Thank you

 

[gorgon]

The name depends on what you want to show with your graph.

A sinus wave graph is a representation of an alternating voltage over time, so the labels should be simple.

TOK

 

[vick5821]

Any example ?