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

#### WBahn

Joined Mar 31, 2012
27,855
We are missing some important context about the problem, and thus aren't in a position to firmly understand what the TS means when he says that the 3.9% figure is "the exact" solution. He could mean nothing more than that it is the solution that was provided, with no actual claim that it is somehow exact. If it were truly meant to reflect an "exact" solution, I would expect at least one more sig fig to be given. We also don't know how detailed of an analysis was expected and what factors were to be considered and what was to be ignored. Does the fact that specific diode type was given mean that part of the exercise was to consider things like the actual (typical) forward diode drop during the brief time that they are conducting a rather substantial current? It would help if the TS would show some work, but that doesn't appear likely to happen.

Doing the normal, simplified analysis I get a ripple of 3.79% if I define the ripple as the voltage droop divided by the peak voltage. If, instead, I define the ripple as the voltage droop divided by the average voltage I get 3.86%, which rounds to the given answer. I would expect that if I divide by the rms voltage I would get a slightly smaller, but very close, value.

#### MrAl

Joined Jun 17, 2014
9,623
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
View attachment 276481

Hello again Eric,

Ok thanks for that information. I will use to to compare notes. My exact calculation came out to 3.4074 percent but the precision there can be extended to any number of digits. That is also with Vd=0, which is no voltage drop for the diode during conduction. We could change this to some constant voltage too like 0.7v and get a different result.

I did not do a completely 'practical' problem because i always believed that the practical trickles down from the theoretical so if we can know the theoretical we can always find a way to get to the practical, but we cant go the other way around. I do however still believe that the practical aspects have to be investigated as well as the purely theoretical.

With that in mind, i did an exact calculation using the strategy shown in the attachment.
I found it particularly interesting that there were certain facts about this circuit that come about from knowing that the capacitor voltage is forced absolutely during the charge times, as if driven directly from the AC source itself (when Vd=0 or less by 2*Vd if Vd is a constant non zero voltage). That means that except for the very first wave front, all the remaining cycles will look and measure exactly the same with no variance no matter how long we wait.
Also, the key to solving this exactly using this strategy is that the current though R and C (in parallel) is zero when the exponential part of the response departs from the sinusoidal part of the response. That gives us an equation that can be solved for t2 (and similar time points) directly to any precision desired. Once we have that we can solve for t4 using Eq2 and Eq3 and that's about all we need to complete.
It's an academic solution but very interesting and informative.

So the first hint is to solve for t2, then knowing that, v2. I wont give too much else away until the OP tries whatever he wants to do first.

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#### Pyrex

Joined Feb 16, 2022
24
To evaluate the given circuit i use a simplified formulas, nevertheless, the result is acceptable for practical use ...
First of all, checking if ripple is small enough and following formulas can be applicated: C*Rl=0.22 sec>>8.33 msec (ripple period). Ripple is small enough.
Output voltage is close to the peak voltage minus diode bridge drop, assuming the transformer impedance is quite low :
(12-(0.7*2))*sqrt2= 15V
So, current is : 15V/Rl= 15V/220 Ohm=68 mA
One more useful formula to find the required capacity: C=(I*t)/Ur, where t is 8.33 msec, Ur is ripple voltage. So, Ur=(I*t)/C=(0.068*0.00833)/0.001=0.57V
Ripple factor is : Ur/15V=0.57/15=0.038=3.8%

#### WBahn

Joined Mar 31, 2012
27,855
To evaluate the given circuit i use a simplified formulas, nevertheless, the result is acceptable for practical use ...
First of all, checking if ripple is small enough and following formulas can be applicated: C*Rl=0.22 sec>>8.33 msec (ripple period). Ripple is small enough.
Output voltage is close to the peak voltage minus diode bridge drop, assuming the transformer impedance is quite low :
(12-(0.7*2))*sqrt2= 15V
So, current is : 15V/Rl= 15V/220 Ohm=68 mA
One more useful formula to find the required capacity: C=(I*t)/Ur, where t is 8.33 msec, Ur is ripple voltage. So, Ur=(I*t)/C=(0.068*0.00833)/0.001=0.57V
Ripple factor is : Ur/15V=0.57/15=0.038=3.8%
(12-(0.7*2))*sqrt2= 15V
What is the basis for claiming that the voltage drop of the diodes gets multiplied by the square root of two?

Assuming very low impedance of the transformer, the RMS voltage of the transformer output is

Vac_rms = (Vac_in / 10)

The peak voltage is therefore

Vac_pk = √2·Vac_rms

Then the peak voltage across the capacitor has two diode drops due to the full-wave bridge rectifier.

Vout_pk = Vac_pk - 2·Vd

So the peak voltage on the cap is

Vout_pk = √2·(Vac_in / 10) - 2·Vd

If Vac_in = 120 Vrms and Vd = 0.7 V, then we have

Vout_pk = √2·(120 V / 10) - 2·(0.7 V) = 15.57 V

In terms of the ripple factor, it doesn't matter what the actual peak voltage is because that drops out of the equations leaving you with

ripple_factor = T/(RC)

where T is the period between peaks (in the simplest approximation), which is half of 1/f for a full-wave rectifier.

ripple_factor = 1/(2RCf)

ripple_factor = 1/[2(1000 uF)(220 Ω)(60 Hz)] = 3.79%/cycle

#### WBahn

Joined Mar 31, 2012
27,855
While we don't know what this problem was intended to highlight, be it nothing more than the basic notion of ripple in such a rectifier circuit or the ability to do a more detailed analysis, if only as an exercise in being able to model and perform the analysis, hopefully (at some point) some of the practical aspects are at least discussed. I recall having to plot the current in the capacitor over an entire cycle to show the high currents that the transformer and diodes have to support during the very brief recharging window. A good approximation of that, assuming low transformer output impedance and low capacitor ESR, can be obtained by determining how much charge flows into the load over a half cycle divided by how much time the diodes are conducting. For low voltage droop, the voltage change during this window is reasonably linear, so the current is reasonably constant.

#### MrChips

Joined Oct 2, 2009
27,662
Let us all agree that there is no “exact” solution unless by “exact” we choose that to mean that the calculation is correct based on the model chosen.

My model selected was Vd = 0.7V and that the output waveform is a sawtooth shape. For the later, this assumes that the time to recharge the capacitor is zero and the load current is constant.

#### WBahn

Joined Mar 31, 2012
27,855
Let us all agree that there is no “exact” solution unless by “exact” we choose that to mean that the calculation is correct based on the model chosen.

My model selected was Vd = 0.7V and that the output waveform is a sawtooth shape. For the later, this assumes that the time to recharge the capacitor is zero and the load current is constant.
I don't know that we can all agree to that. For instance, we might all agree to use a model in which Vd = 0.7 V, but assuming zero time to recharge the capacitor and a constant load current is an approximation that, by necessity, results in an answer that is an approximation and, therefore, not exact. An "exact" solution implies no approximations, but it is reasonable to dictate (and specify) the device models used and consider the solution "exact" in light of those models provided no further approximations are made.

One of the common uses for this problem is to show how good a very simple and easy to calculate approximation can be relative to a full-blown, no-approximations calculation, thus dispelling a common belief amongst beginners that such approximations only yield rough, qualitatively useful results.

#### MrChips

Joined Oct 2, 2009
27,662
As someone else in this thread indicated, everything in the real world of electronics is an approximation or gross assumption.

For example, to say that the value of the capacitor is 1000μF is a gross assumption. In other words this assumption is good enough for all intent and purposes.

#### MrAl

Joined Jun 17, 2014
9,623
Let us all agree that there is no “exact” solution unless by “exact” we choose that to mean that the calculation is correct based on the model chosen.

My model selected was Vd = 0.7V and that the output waveform is a sawtooth shape. For the later, this assumes that the time to recharge the capacitor is zero and the load current is constant.
Hi,

Yes that is worth discussing since it keeps coming up.

It became obvious that the OP problem definition of exact is very mild at best.

When i say 'exact' for my solution, i first assume nothing but nothing is unknown, and apply analytic geometry to the 'exact' waveforms that would be produced in a circuit with everything known, and they happen to be fairly simple if we quote the diodes as being 'ideal' in that they have either 0v drop or something like 0.7v drop and that is constant while conducting, and in reverse bias they have infinite resistance so they are open circuit. The transformer and the line both have zero impedance. The capacitor has zero ESR unless we want to get a bit more complex and then we can assign it some constant ESR value. The resistor is spot on balls accurate (quote from My Cousin Vinny) and has no temperature drift. All the connections as well as all the connecting 'wires' have zero resistance as well as no inductance and no skin effect.

So what would be the purpose of all this. It would act as a baseline calculation for a calculation that would be more and more complex depending on what we wish to make 'non ideal'. It can get pretty dang complicated even with just cap ESR, but that's life. A spice model diode also makes the calculation quite a bit more complicated even with a knocked down version.

So the main idea here is that the 'exact' solution is a mathematical solution that can be taken to any level of precision we wish, with the chosen components and their models, and what may be important is that a simulation will never get to a very high precision of accuracy because it would take too long to run the simulation, while the calculations may take an hour if it is that complex, but then the calculation can be repeated with other components of the same type to see what the effect would be, such as changing the cap from 1000uf to 500uf or to 2000uf, etc. This would be used in a study that would be used to make a judgement on what the other effects would cause.
For example, back in the 1980's i worked with a professor from Perdue University and we had to make a chip that would be used to study the effect of line tied synthesized sine wave converters being powered from solar arrays. The study was to find out what the harmonics would look like on the line if there were a lot of these converters tied to the line all at the same time. For that, the chip had to be 'exact' in that it had to produce a set number of harmonics of the right number and amplitude. That just really boiled down to a certain switching pattern, but that's what another form of exact may mean. It has to be exact to some standard so predictions could be made about other things that have to work with it without having to explore a host of other variable effects.

For me though it has been a story of never finding a good, theoretical solution to something we use almost every day, a full wave rectifier, so i set out to find a way to calculate any full wave rectifier with any components. It turned out to be a strategy rather than a complete formula like with Ohm's Law, but it works to whatever accuracy we want it to work.

There is one more work like this online somewhere by a professor that set out to completely analyze a circuit like the one in this thread but i think he included a series R in series with the diode bridge. I guess i was not the only one that got tired of reading about 'approximations' and wanted something more concrete. This doesnt mean i wont use approximations though, i welcome them when they work, and in fact this circuit with the 1000uf cap and 220Ohm resistor can be approximated by assuming that the diode turns 'off' at the peak of each half cycle sine. That's only because the RC time constant is so large the droop is very small. As we let the RC time constant go lower, that approximation becomes more and more inaccurate and may not be useful at all at some point.

Some of these circuits are simply studied for the pure academic value too, and with everybody on the same page everybody should come out with the same result

#### MrChips

Joined Oct 2, 2009
27,662
There is another option of what “exact” means.

To me, using a fullwave rectified power supply circuit as a working example, if the calculation produces a result that is within ±10% (pick your own error) of actual measured values then that is good enough for me.

#### WBahn

Joined Mar 31, 2012
27,855
There is another option of what “exact” means.

To me, using a fullwave rectified power supply circuit as a working example, if the calculation produces a result that is within ±10% (pick your own error) of actual measured values then that is good enough for me.
So "exact" means the same as "good enough"?

#### MrChips

Joined Oct 2, 2009
27,662
Yes. If I got the “exact” answer then that’s good enough for me.
In other words experimental results confirm the model used.
That is how engineering modelling works.

#### WBahn

Joined Mar 31, 2012
27,855
Yes. If I got the “exact” answer then that’s good enough for me.
In other words experimental results confirms the model used.
We'll have to agree to disagree, since I don't consider that every answer that is "good enough" is somehow an "exact" answer. I stick much closer to the meaning of the word "exact" as being "not approximated in any way" -- and note that covers the use of models that are approximations, in which case the implication is that the result is exact with regards to the models used.

Furthermore, in general, the "exact" answer may or may not be good enough. I might determine the exact power dissipated in a device, but that power may be too much to be acceptable.

For a mechanical example to illustrate the distinction, if I were to fill my 3/4 ton truck with a bunch of boxes and not want to exceed the rated load capacity of 1500 lb, I might weigh each box on a scale that I know gives me the correct result within +/-10 lb and take each box's displayed weight and add 50 lb to it. If the result came out to a total of 1357 lb, I would call that a "good enough" value for the total weight of the load, but I would hardly call it the "exact" weight of the load.

#### ericgibbs

Joined Jan 29, 2010
16,751
hi,
For interest, these LTS plots and Text show the variation in Vout etc.
For smoothing components that have a +/10% value tolerance.
E

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#### MrAl

Joined Jun 17, 2014
9,623

#### WBahn

Joined Mar 31, 2012
27,855
With that in mind, i did an exact calculation using the strategy shown in the attachment.
On thing you didn't take into account is that the underlying waveform isn't truly the absolute value of a sinusoid due to the diode drops. While the deviation is small near the peak, it is an approximation.

#### MrAl

Joined Jun 17, 2014
9,623
There is another option of what “exact” means.

To me, using a fullwave rectified power supply circuit as a working example, if the calculation produces a result that is within ±10% (pick your own error) of actual measured values then that is good enough for me.
Well there is a difference to what exact means to the academic world vs the practical engineering project, except when it affects the practical world too. One notable example is with time dilation, which usually does not affect anything so it never has to enter into any formula here on earth. Most every calculation is 'exact' enough without it. With GPS however that's not the case, and time dilation has to be taken into account to make up for the error in location data.

In the academic world though an answer of 3.14 may not be good enough for pi it may be necessary to provide more digits to get the right result. In some cases 3.14 would be good enough, in other cases maybe we would need 3.1415, and in other cases 3.1415926, etc. If we were mathematicians working on a new algorithm for pi however, we'd need to look at billions or trillions of digits.

I could also bring up the model specifications, such as with a diode. In some cases a diode with 0v forward drop is considered 'ideal' while other cause 0.7v is ideal. There are variations here too.

#### ericgibbs

Joined Jan 29, 2010
16,751
hi Al,
Out of interest, I ran a simulation, stepping the Rsource resistance for the Sine wave driver.
Rser = 0, 0.5, 1.0 Ohms
E

#### MrAl

Joined Jun 17, 2014
9,623
On thing you didn't take into account is that the underlying waveform isn't truly the absolute value of a sinusoid due to the diode drops. While the deviation is small near the peak, it is an approximation.
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.

#### MrAl

Joined Jun 17, 2014
9,623
hi Al,
Out of interest, I ran a simulation, stepping the Rsource resistance for the Sine wave driver.
Rser = 0, 0.5, 1.0 Ohms
E
View attachment 276542
Hi Eric,

Oh i was just wondering why D4 was drawn as a Schottky diode, or perhaps drawn as a zener diode while the others seemed more like regular rectifier diodes.