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

#### Zooxanthellae

Joined Sep 11, 2022
7

#### Papabravo

Joined Feb 24, 2006
20,609
You need to show your best attempt at a solution. This is NOT homework done for you.

#### Zooxanthellae

Joined Sep 11, 2022
7
You need to show your best attempt at a solution. This is NOT homework done for you.
I read that the formula for it is (Vr(rms)/ Vdc) * 100%. So I tried to get the Vdc, I tried dividing the Vrms by 10 or by using the formula. Vdc = ( 1-(1/2fRlC)). But I think I'm wrong. I'm sorry, I'm quite new to electronics,

#### WBahn

Joined Mar 31, 2012
29,519
First, draw a rough sketch of what you expect the general shape of the voltage at the output to be and what the current coming out of the rectifier diodes will be. Then start working things out in terms of when and where the current is flowing into and out of the capacitor. That will tell you what is happening to the voltage on that node.

#### MrChips

Joined Oct 2, 2009
29,853
In my electronics class we were given this assignment without any guidance on how to do it. It turns out that the math is not easy. Hence one has to resort to using approximations.

#### MrChips

Joined Oct 2, 2009
29,853
As WBahn stated, draw a picture of what you expect to see for the output voltage.
Then follow these steps. Try to derive from first principles any formula you use. Don't take any formula for granted.

You should use algebraic variables for the following calculations:

1) Calculate the rms AC voltage after the transformer.
2) Calculate the peak AC voltage after the transformer.
3) Determine the peak DC voltage at the load (Output).
4) Calculate the peak current through the load resistor RL.
5) Calculate the cycle frequency and period of the ripple voltage.

Make the first approximation assumption that the ripple voltage is a saw-tooth waveform.

6) Calculate the charge per cycle delivered to the load.

Let us see if you can continue from here on your own.

#### Zooxanthellae

Joined Sep 11, 2022
7
As WBahn stated, draw a picture of what you expect to see for the output voltage.
Then follow these steps. Try to derive from first principles any formula you use. Don't take any formula for granted.

You should use algebraic variables for the following calculations:

1) Calculate the rms AC voltage after the transformer.
2) Calculate the peak AC voltage after the transformer.
3) Determine the peak DC voltage at the load (Output).
4) Calculate the peak current through the load resistor RL.
5) Calculate the cycle frequency and period of the ripple voltage.

Make the first approximation assumption that the ripple voltage is a saw-tooth waveform.

6) Calculate the charge per cycle delivered to the load.

Let us see if you can continue from here on your own.
Thank you very much for your help. I'll do my best on answering.

#### WBahn

Joined Mar 31, 2012
29,519
Start with simple approximations, such as assuming that the ripple is small enough such that the current in the load resistor is a constant, that the diodes have a fixed forward voltage regardless of the current and that the transformer can deliver unlimited current without any voltage drop. From that, figure out what the ripple on the capacitor voltage will be. If that turns out to be small, then those assumptions are probably not too unreasonable. But if it turns out to be a significant portion of the average capacitor voltage, then you need to discard some of those assumptions and refine your results.

#### MrAl

Joined Jun 17, 2014
10,914

Hi there,

First let me just say that the other suggestions in this thread were very good you should consider them seriously. I just wanted to add a little more info here, and one question.

First, are you supposed to find an EXACT answer to this or is an approximation of some kind good enough?
I ask because although there are several approximations for doing this there is an exact solution to this which is a little more difficult but not all that much, and that is because the capacitor is connected directly to the output of the diode bridge. We do often still accept a constant diode voltage drop, and it gets more complicated if we have to use an actual spice diode, but it would be considered very simple if the scale was FROM this bridge rectifier circuit TO a bridge rectifier with cap ESR and other stuff.
So i guess the question really is, what is acceptable for an answer?
You might also mention what year you are in school that might help members answer this question too, and maybe what you have studied in the past.

I'll save the exact solution (which isnt all that difficult) for later after you've thought about all this and tried to answer yourself.
Good luck

#### Zooxanthellae

Joined Sep 11, 2022
7
Hi there,

First let me just say that the other suggestions in this thread were very good you should consider them seriously. I just wanted to add a little more info here, and one question.

First, are you supposed to find an EXACT answer to this or is an approximation of some kind good enough?
I ask because although there are several approximations for doing this there is an exact solution to this which is a little more difficult but not all that much, and that is because the capacitor is connected directly to the output of the diode bridge. We do often still accept a constant diode voltage drop, and it gets more complicated if we have to use an actual spice diode, but it would be considered very simple if the scale was FROM this bridge rectifier circuit TO a bridge rectifier with cap ESR and other stuff.
So i guess the question really is, what is acceptable for an answer?
You might also mention what year you are in school that might help members answer this question too, and maybe what you have studied in the past.

I'll save the exact solution (which isnt all that difficult) for later after you've thought about all this and tried to answer yourself.
Good luck
Thank you for asking. There's an exact answer that was given to us, which is r = 3.90%. I'm a college sophomore at the moment. I guess we just need to give a solution on how the ripple factor became 3.90%

#### ericgibbs

Joined Jan 29, 2010
18,245
hi Zoo and Al.
Now that you have completed the assignment, I can post what LTspice shows.
My maths show 3.21% , close enough.
E

#### MrAl

Joined Jun 17, 2014
10,914
Thank you for asking. There's an exact answer that was given to us, which is r = 3.90%. I'm a college sophomore at the moment. I guess we just need to give a solution on how the ripple factor became 3.90%
Ok one more little detail. What kind of diode model are you using in this problem, is it a constant voltage diode like one that drops 0.7v in the forward conduction mode, or is it 0v in the forward conduction mode, or some other model?
I ask because often these problems are based on ideal diodes not real diodes with reasonably good spice models, and the diode makes a bigger difference than one might think at first because the peak current in the diodes is often larger than we might think in rectifier circuits with no input inductor.

#### MrChips

Joined Oct 2, 2009
29,853
Using first order approximations, single diode drop of 0.7V, average DC is 15.3V, ripple is 0.6V, percent ripple is 3.9%.

#### Zooxanthellae

Joined Sep 11, 2022
7
Thank you for asking. There's an exact answer that was given to us, which is r = 3.90%. I'm a college sophomore at the momen
Ok one more little detail. What kind of diode model are you using in this problem, is it a constant voltage diode like one that drops 0.7v in the forward conduction mode, or is it 0v in the forward conduction mode, or some other model?
I ask because often these problems are based on ideal diodes not real diodes with reasonably good spice models, and the diode makes a bigger difference than one might think at first because the peak current in the diodes is often larger than we might think in rectifier circuits with no input inductor.
All of the given diodes are 1N4001

#### WBahn

Joined Mar 31, 2012
29,519
All of the given diodes are 1N4001
Still waiting to see your best attempt at a solution. We really can't do much more for you until we see that.

#### MrAl

Joined Jun 17, 2014
10,914
hi Zoo and Al.
Now that you have completed the assignment, I can post what LTspice shows.
My maths show 3.21% , close enough.
E
View attachment 276367
Hi there Eric,

How did you calculate that? I was wondering because i got a different result and would like to compare notes. Thanks.

I think you may have done the same as in the other post with Vpp/Vaverage. That would be reasonable.

Last edited:

#### MrAl

Joined Jun 17, 2014
10,914
Using first order approximations, single diode drop of 0.7V, average DC is 15.3V, ripple is 0.6V, percent ripple is 3.9%.
Oh thanks for that explanation i was wondering if considering a constant voltage drop for the diode was OK for this problem.

#### ericgibbs

Joined Jan 29, 2010
18,245
Hi Al,
The LTspice .meas command [Black Text] posted these results in the LTS Error Log.[Blue Text]
Ref image.
What parameters and results did you get.?
E

#### Jerry-Hat-Trick

Joined Aug 31, 2022
451
I do question why the "exact" answer is 3.9% but in practice it's close enough.

The way that I would solve this problem, near enough for practical purposes, is to recognize that with 60Hz the time between voltage peaks is 8.33mS during (most of) which the capacitor discharges it's peak voltage into the resistor at near enough a linear rate.

C = Q/V translates to dv/dt = I/C = V/RC implies dv/V = dt/RC = 8.33/(220) = 0.378 which is about 3.8%. But I'd expect it to be better than this because the voltage slope reduces slighly with time and it meets the rising voltage again before 8.33mS. Note that the value of V is not important. The tolerance of a 1000uF capacitor is typically being +/- 20%

Frankly, if this is not an acceptable answer, with apologies to Indiana Jones "if you want to be an engineer you need to get out of the library".

#### ericgibbs

Joined Jan 29, 2010
18,245
hi,
The only exact answer to this problem is the one achieved in calculating the numbers.

There is NO exact answer in the practical World, due to the variation in the tolerances of the components used.

We all live in a World of close approximations, which are close enough to get us an acceptable, workable solution.

E