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DC power supply from AC through bridge rectifier

vick5821

Jan 22, 2012
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What are the circuits combined together to build up a DC power supply from 240V AC supply ?

Need to do step by step. Pls guide me. I will post my work soon.In the mean time, pls suggest to me the ways to do.

Thank you
 

davenn

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What are the circuits combined together to build up a DC power supply from 240V AC supply ?

Need to do step by step. Pls guide me. I will post my work soon.In the mean time, pls suggest to me the ways to do.

Thank you

you have shown all the parts required in the image in your last post

you name each part in your diagram from left to right and what you think it does
and we can help with corrections or suggestions as needed :)

Dave
 

vick5821

Jan 22, 2012
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Is this consider as stpe by step seperate circuits used to build up a AC to DC power suppply ? I wish it is and I get what my lecturer want so that I didnt waste my time on it. No matter how my friends and I do, always wrong. So I hope that I am correct this time. This is the lab sheet for the project :)

Labsheet :
IMG.jpg


Sorry that it is in malay ><

step 1 : Output voltage from rectifier
I manage to get 5.999 from a 6V trasnformer
OutputVoltage-1.png


step 2 : Rectifier using 4 diodes as bridge
Output voltage meausred from multimeter = approximate 4.6 ( 6 - 0.7x2)
Rectifier-1.png


step 3 : smoothing(adding in a 10uF) Capacitor
Output voltage meausred from multimeter = 5.636V ( why increase from 4.6 to 5.636 ??)
Smoothing10uF.png


smoothing(adding in a 1000uF) capacitor
Output voltage meausred from multimeter = 7.073 ( why higher ?? Average voltage ?)
Smoothing1000uF-2.png
 
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vick5821

Jan 22, 2012
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This is what we call seperate circuit used to build up a DC power supply or how ? I need to make sure I get what my teacher want. Or else I am wasting my time to do all this :(
 

davenn

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Hi Vick

I had a wonderful response typed out for you. Unfortunately the site threw a fit for a short while and my response got lost :(

just briefly again....

that looks good the way you have looked at each section (not being your teacher, I cant say exactly what he/she expects from you)

do your simulation, with an oscilloscope across the output of the bridge rectifier and see what it shows you... hopefully it will show what I am expecting, I will await a response from you befor posting a pic

cheers
Dave
 

vick5821

Jan 22, 2012
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Yea..The oscilloscope show the output I expected. Sine wave, then sine wave with negatie side flipped then a DC with ripple :)
 

davenn

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ok excellent
so you can see that altho we call that a DC output from the rectifier, its pretty unstable, varying in pulses from 0V to max volts

ok now stick you 1000uF cap across the output and look at the waveform again
what do you get ? ( again post an image ) :)

Dave
 

vick5821

Jan 22, 2012
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12.png


Waveform with a little bit ripple if I use 1000uF capacitor. How does the capacitor do the smoothing process ?

Below show another smoothing action with 10uF capacitor which shows a poor smoothing effect.
13.png


My question is that, why when smoothing, the voltage will be increased ? After rectified, the voltage will be around 4.6 . But why when smooth with 10uF, the voltage go to 5.5+ ? and when with 1000uF, the voltage rise to 7.0 + ? WHY ?

Thank you
 
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davenn

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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
 

vick5821

Jan 22, 2012
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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

Thats means that I should get 6 x 1,414 = 8.484 V ? after step down from a 6V tranformer ? The 6 is the RMS value ? DC out after rectified = 6 - 1.4 = 4.6V ?(This is the avaerage value?) After it is being smoothed, the average value increases ?
 

vick5821

Jan 22, 2012
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Labsheet :
IMG.jpg


What does my lecturer wants actually ? What do you understand from this labsheets ?
Is there any other circuits ?

My circuit correct ?
 
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vick5821

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6V.png


What is the purpose of having three capacitor there? I wonder since my lecturer gave us 3 capacitors :(
 

davenn

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Thats means that I should get 6 x 1,414 = 8.484 V ? after step down from a 6VAC tranformer ?

yes, your scope will show the peak voltage ( your multimeter if its a true RMS meter will show the RMS Voltage)

The 6VAC is the RMS value ?

Yes, and that is what is always assumed for a AC supply UNLESS it specifically states PEAK or PEAK to PEAK value

DC out after rectified = 6 - 1.4 = 4.6V ? (This is the average value?)

No I dont think its the avg value ( unless its the avg value before a smoothing cap is added)
I have not been able to find a suitable answer to that yet -- I will search further even if just for my own learning :)

After it is being smoothed, the average value increases ?

After smoothing, the average (mean) DC voltage lies midway between the peaks of the ripple voltage. Yes, the avg DC value will increase towards the PEAK DC value as you provide more capacitance, which decreases the P-P value of the ripple voltage.

attachment.php


3 caps ... well no real need for three, a single larger value would work. BUT that being said, there are other reasons when very different values of cap's are used...
say a 2200uF, a 10uF and a 0.1uF. This provides filtering over a wide range of frequencies.

Maybe one of the others will comment about using multiple caps of similar values has something to do with ESR ratings of the caps etc --- not something I'm totally up on

cheers
Dave
 

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vick5821

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hey, how does the load resistance affects the output waveform ?

See below :
13.png

This is produced when I have a 1k ohm load with 10uF capacitor

12.png

This is produced when I have 10k ohm load with 10uF capacitor

Any idea?
Thank you
 

davenn

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hey, how does the load resistance affects the output waveform ?

This is produced when I have 10k ohm load with 10uF capacitor

Any idea?
Thank you

basically, the greater the load (the lower the resistance) there will be more current
flowing so therefore for a given value of smoothing capacitor, the more difficult it will be to maintain that output voltage between peaks and the output DC voltage will drop. The ripple voltage will increase and you will need a larger capacitance to hold up the voltage between charging cycles from the diodes.

You can see that in your sim's where there is a larger current flowing through the 1k resistor than through the 10k resistor. And as a result the ripple is higher when there is only a 1k load across the supply.

cheers
Dave
 

davenn

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on a different subject for a moment.... did you get that friend request from me in facebook ?

D
 

vick5821

Jan 22, 2012
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basically, the greater the load (the lower the resistance) there will be more current
flowing so therefore for a given value of smoothing capacitor, the more difficult it will be to maintain that output voltage between peaks and the output DC voltage will drop. The ripple voltage will increase and you will need a larger capacitance to hold up the voltage between charging cycles from the diodes.

You can see that in your sim's where there is a larger current flowing through the 1k resistor than through the 10k resistor. And as a result the ripple is higher when there is only a 1k load across the supply.

cheers
Dave
I didnt get you.Why more current the output DC voltage will drop ?
 

davenn

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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
 
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