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AC to DC conversion

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alok1982

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Can anyone share the formula to convert AC voltage to DC voltage across the rectifer?
Consider input voltage is 230Vac/ 50Hz I am applying through Full bridge rectifier what will be the DC voltage at output?
 

Willen

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Here is a simple calculation but should we have to multiply 1.414 also if we don't have high value smoothing capacitor with bridge rectifier while calculating average DC output? Anyone please!
 
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Reloadron

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230 VAC 50 Hz RMS Single Phase would be 230 VDC less the forward voltage drop of the diodes. In a full wave bridge 2 of the 4 diodes are conducting at any given time. If we assume 0.7 forward voltage drop for a Si Diode then we have 2 * 0.7 = 1.4 volts. That gives us 230 volts - 1.4 volts so we get 228.6 VDC.

Consider input voltage is 230Vac/ 50Hz I am applying through Full bridge rectifier what will be the DC voltage at output?

Since you did not mention a filter capacitor I did not include a filter capacitor.

Now if we have a filter capacitor the 1.414 mentioned by Willen figures into things. The filter capacitor will want to charge to Vpk (Peak). The peak value of a sine wave is the RMS value * 1.414.

Ron
 

cowboybob

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Here is a simple calculation but should we have to multiply 1.414 also if we don't have high value smoothing capacitor with bridge rectifier while calculating average DC output? Anyone please!
A smoothing cap does not (cannot) "add" voltage as you are suggesting.

Mulitplying the RMS voltage (as indicated by, say, a normal VOM) by 1.414 ia necessary to give us the peak voltage level available to the rectifying circuit and, ultimately, the maximum resulting DC value as per the formula of the example.
 
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Willen

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*Double post removed!
 
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Willen

Well-Known Member
The terms 'average DC out' or 'maximum DC out' confused me.

Main confusion is- can't we simply calculate like this way including Filter caps-

230V*1.414= 325V-(0.7*2)= 323.6V DC peak out. Then in which condition we need the formula- 2Vp/pi?
 
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cowboybob

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Do not confuse AC and DC voltage levels and, tangentially, the ability to do work.

AC "peak to peak" is the voltage differential between the maximum positive and negative swings. Once DC (either from a single rectifer or a bridge), only the positive (and/or negative) excursion(s) remain (zero to peak).
 
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NorthGuy

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A smoothing cap does not (cannot) "add" voltage as you are suggesting.

Mulitplying the RMS voltage (as indicated by, say, a normal VOM) by 1.414 ia necessary to give us the peak voltage level available to the rectifying circuit and, ultimately, the maximum resulting DC value as per the formula of the example.

If there's no load, then smoothing cap charges to the peak voltage and stays that way. So, it does "add" voltage by keeping voltage steady between peaks.

If there's a very small load and very big cap, the voltage still stays at the peak all the time.

If load increases or cap gets smaller, the cap starts discharging between peaks, so the voltage gets lower between peaks.

If there'a a very big load and a very small cap, the voltage drops almast all the way to zero between peaks, so the voltage approaches the RMS voltage minus diode drops, as it would be without the cap. Someone has used too big a load or too small a cap.
 

Willen

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I am still confused when we use 2vp/pi... :)

Sorry KeepItSmile, just now I've been far from computer and using java cell phone just for few days (not able to run .pdf).
 

KeepItSimpleStupid

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OK.

The average value of a function is the integral of the function divided by it's length. (Calculus stuff)

So, the definite integral of Sin(t) from 0 to PI is equal to 2. (1/2 of a cycle)
https://www.wolframalpha.com/input/?i=inegral sin(x) from 0 to PI

The sin(t) function has a max of 1, corresponding to Vp. in v(t)=Vp*sin(t)

Our full sin(t) goes from 0 to 2*PI, but let's invert (from 180 to 360 deg), so the integral of abs(sin(t)) from 0 to 2* PI = 4 for the rectified sin function.

So, the average is 4/2*PI for the sin wave and therefor 2*Vm/PI for the sine wave having Vmaximum of Vm.

Sorry, for the round-about way of getting there. My Calculus squeaks.

The real average value of a sine wave is zero, but the average of a rectified sine wave is 2*Vm/PI

I could have got there easier by saying a rectified sine wave has a period of PI and the definate integral of from 0 to PI of 2. Therefore, the average is 2/PI when Vm =1. When Vm <>1, it's 2Vm/PI.

The capacitor filter basically charges the cap to the peak value and if the cap isn't big enough or the input current is lower than required, it starts to decay.

The wierdness with average, also determines the max secondary draw. It's less than the RMS current rating of the secondary.
 
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cowboybob

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If there's no load, then smoothing cap charges to the peak voltage and stays that way. So, it does "add" voltage by keeping voltage steady between peaks...
Since you quoted me in your post, I should point out that I was responding to Willen's post (#2) wherein he seemed to posit that a cap could "add" to the peak voltage, thereby increasing that peak value which, of course, it cannot.

Although, I could have been wrong in my interpretation of his query.
 

KeepItSimpleStupid

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Willen:

I finally looked at the post you referenced: https://www.industrial-electronics..../5_Four-Diode_Full-Wave_Bridge_Rectifier.html and it's a perfect example of how the Internet get's confusing. It's perfectly correct, EXCEPT it might not be used for much.
It could provide the basis of:

1) A power fail detector
2) a 120/240 auto line switch
3) The front-end to a average responding, RMS reading meter

The link, sort of, misleads you into the formula for finding out what transformer you need for a rectified AC source, which it does not.

A little more goes into capacitor selection. See https://www.irjes.com/Papers/vol2-issue6/Version-1/E02064249.pdf In general, you do not want the ripple factor of a capacitor to be exceeded and you select the cap based on those two parameters. In a pinch, you can use 1000 uF/amp as a "rule of thumb".
 
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