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design ac to dc converter

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Deexith Hasan

Advanced Member level 4
Oct 24, 2014
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using transformer input is 220 o\p 12 volt rms. bridge drop is 1.4v [full wave rectifier] so o\p is 10.6 volt...
how to find the filter capacitor needed if load is 1k.

10.6 volt rms after filtering i should get 10.6*sq.root (2)=14.9 volt but getting 8.98v..... y???

i need to measure ac voltage so using this setup


the 8.98 is with or without cpacitor?

with big capacitor you will get peak voltage of about 14.9V,
without capacitor you get average voltage of rectified sine. (depending of measurement setup)
If you have a scope then look at the waveform with and without cpacitor.


with cap got 8.98 getting transient error in proteus...

bridge rectifier drop s 1.4 volt in ac rms or ac peak....??

1)12v ac rms - 1.4v ac rms drop=10.6v rms....10.6v rms*sq.root (2)=14.9 volt dc

2)12v ac rms*sq.root (2)=16.9v dc - 1.4v dc drop=15.5v dc

which is crt 1 or 2??

Number 2 is correct BUT:
1) The transformer is 12VAC only when it has its rated load current. With less current then its output voltage will be higher.
2) The voltage drop of a bridge rectifier is 1.4V only when the load current is low. With a higher current it is 2V or more because it conducts a very high peak currents for short durations.

I think your Proteus simulation software flunked electronics or math or both in school.

A simple question, but a difficult ..

If you have a bridge fectifier with resistive load..
Imagine a sine waveform, then mirror the lower negative half to be a positive.
Then shift everything down by 1.4V. Cut all negative values at the zero line.
Now you see two hills per fullwave, but with a small horizontal connection at zero line.

You see that the peak is really shifted down by 1.4V, but the exact average value is difficult to calculate.
It drops more than 1.4V/sqrt(2).

Because the peak value drops 1.4V you will see this when you connect a capacitor as load.
@klaust 1.4v drop is on ac rms [in my case transformer secondary is 12 v] or on 12*sqrt(2) ??

The bridge rectifier conducts for a moment only on the peaks of the sinewave. Then its current is MUCH higher than the average load current so its voltage drop is higher than only 1.4V.
The voltage drop is on the 16.97V DC peak voltage , not on the 12V AC RMS voltage.

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