Determine the ripple factor for the filtered bridge rectifier with a load as indicated in the figure below

ericgibbs

Joined Jan 29, 2010
16,808
hi Al,
It is just a diode symbol artefact, I have corrected it.
This plot shows the Ripple for stepping Rser.
Note the change in ripple content. [Vrip]EG57_ 38.pngEG57_ 37.png
E
 

WBahn

Joined Mar 31, 2012
27,925
Hi,

Not sure what you mean because i specified a diode drop of 0v meaning that even 2*Vd=0 volts, or did i not make that clear enough maybe?
To switch to the diode voltage drop version with Vd=0.7v the total drop would be 1.4v and that would be dealt with in a slightly different manner. It would at least have to subtract from the input sine voltage by the amount of 2 diode drops.
I always start with a diode drop of 0v because that establishes a 'base' version of the calculation after which we can start to impose other deviations and reformulate somewhat. The 0v version is like a 'reference' version that can show the basic idea about how the calculations are developed without being too complicated.

Is that what you meant? If not, please elaborate if i overlooked something i'd like to know what it is.
That's right, you did spec a diode drop of 0 V. My bad. My default is to always start with a drop of 0.7 V unless I'm going for a real quick estimate. In this case, the diode drop of 1.4 V has a noticeable impact on the answer because it lowers the peak output voltage by nearly 10%, so in my calculations I used a sine function with the corresponding peak value to determine the conduction angle (in order to find the current when recharging the cap) and I had that stuck in my head.
 

MrAl

Joined Jun 17, 2014
9,633
That's right, you did spec a diode drop of 0 V. My bad. My default is to always start with a drop of 0.7 V unless I'm going for a real quick estimate. In this case, the diode drop of 1.4 V has a noticeable impact on the answer because it lowers the peak output voltage by nearly 10%, so in my calculations I used a sine function with the corresponding peak value to determine the conduction angle (in order to find the current when recharging the cap) and I had that stuck in my head.
Oh no problem at all, i use 0.7 a lot too when explaining things because often people expect to see some kind of drop at least.
I might go back over it with the 0.7v drop but unfortunately the results will not match up with the simulations exactly because that would be more ideal then the spice model.
 

Thread Starter

Zooxanthellae

Joined Sep 11, 2022
7
Still waiting to see your best attempt at a solution. We really can't do much more for you until we see that.
307849337_552420949990989_2170825754371674587_n.jpg308023597_1160455721492856_2451024854061212020_n.jpg
I tried two approaches but I got wrong answers. I don't really know which formulas to use. Also, were studying a book by Boylestad and Nashelsky called "Electronic devices and circuit theory" if that information helps I guess.
 

MrChips

Joined Oct 2, 2009
27,702
Why are you trying random formulae?
How can ripple voltage be 130.90V? Does that make any sense?

Take one step at a time.

Input AC rms voltage =
Secondary AC rms voltage =
Secondary peak DC voltage =
Rectified DC voltage =
Peak DC current =
Coulombs/s =
Voltage decrease =
% ripple =

Edit: See post #6.
 

WBahn

Joined Mar 31, 2012
27,925
I tried two approaches but I got wrong answers. I don't really know which formulas to use. Also, were studying a book by Boylestad and Nashelsky called "Electronic devices and circuit theory" if that information helps I guess.
The problem is that you are just wanting to through numbers at equations instead of analyzing the circuit. As a result, you are using (or rather misusing) a slew of equations with little rhyme or reason.

Draw the waveform you would expect to see at the output of the bridge if the capacitor were removed. Identify the period of the waveform and also what the peak voltage would be.

Then add to that the waveform that you expect to see at that same point once the capacitor is put back in. Identify the ripple voltage on that waveform. You don't need to have a number for it, just show where on the waveform it would be measured.

Do just that. There is no point going any further until you have that.
 
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