Full-wave Rectifier with Smoothing Capacitor

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xxxyyyba

Joined Aug 7, 2012
289
And here is how I found Fourier series for our voltage signal:
\(v_x(t)=C_0+\sum_{n=1}^{\infty}[A_n\cos{nw_0t}+B_n\sin{nw_0t}]\), where \(C_0=\frac{1}{T}\int_{\tau}^{\tau+T}f(t)dt=\frac{1}{T}V_p_k\left ( \int_{0}^{T/2}\sin{w_0t}dt+\int_{T/2}^{T} (-\sin{w_0t})dt\right )=...=\frac{2V_p_k}{\pi}\),\(A_n=\frac{2}{T}\int_{\tau}^{\tau+T}f(t)\cos{nw_0t}dt=\frac{2}{T}V_p_k\left ( \int_{0}^{T/2}\sin{w_0t}*\cos{nw_0t}dt+\int_{T/2}^{T} (-\sin{w_0t})*\cos{nw_0t}dt\right )=...=2V_p_k*\frac{(-1)^{n+1}-1}{\pi(n-1)(n+1)}\) and \(B_n=0\)
 
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t_n_k

Joined Mar 6, 2009
5,455
You perhaps missed MrAl's important point.

"What this does not model is the relatively high impedance of the rectifier bridge when the cap has a higher voltage than the input sine wave. To do that you would have to solve for the start and end of the conduction time and only integrate from the start to end when computing the Fourier series. If you are not interested in too much accuracy this is easy to estimate."
 

WBahn

Joined Mar 31, 2012
29,979
Hi MrAL,
I found Fourier series for full wave rectified sine. I got \(v_x(t)=\frac{2V_p_k}{\pi}+2V_p_k\sum_{n=2}^{\infty}\frac{(-1)^{n+1}-1}{\pi(n-1)(n+1)}\cos(nw_0t)\).
Here is plot of series for period T=0.02s, Vpk=12V:

And now I should analyse circuit below, right?

View attachment 78585
Only if you believe it to be adequate to model the AC source and the bridge as an ideal voltage source having voltage Vx. For some circuits that might be fine. But there is effective series resistance and as long as that resistance is ALWAYS much less than the impedance in the rest of the circuit, you will be in the ball park. But in this case the impedance is nonlinear, meaning that it changes with voltage and/or current. When a diode is conducting it has low impedance and when it is not it has very high impedance -- much, much higher than the other impedances in the circuit.
 

MrChips

Joined Oct 2, 2009
30,714

MrAl

Joined Jun 17, 2014
11,389
This is harder than I thought :)
Hi,

He he, well then it's time to roll up your sleeves and dig deeper into it :)

You are certainly on the right track, and it doesnt hurt to do it this way because after all it is a good exercise. The only difference is when we use the full wave rectified sine as the source voltage we have to consider it an ideal source, which has zero output impedance and that is no matter what the load is (linear or non linear).
In other words, what we are modeling here is a full wave bridge rectifier circuit with a power op amp on the output, and the power op amp feeds the output filter. There's no harm in doing this but just keep in mind that the output of the filter will not be the typical output you find in most circuit applications.
If we split your solution into the DC and AC parts, we can see right away that the DC component that you computed:
Vdc=2*Vp/pi

comes out to 24/pi=7.64 volts DC which might be what we see with most rectifier circuits with load, but it would be more correct for a circuit with the bridge feeding a power op amp and we look at the output of the power op amp, and you can then use that to feed the filter. It's kind of a good preparation for the real thing that comes next, which is solving for the start and end times of conduction and integrating over those periods.

So it's up to you how you want to proceed, using the power op amp as buffer or solving for T1 and T2 the start and end times of conduction.
Once you have T1 and T2 you can then proceed in almost the same way except you'll be integrating from T1 to T2 over each half cycle instead of from 0 to T/2 and T/2 to T. The An will be different and possibly some Bn might come in too but you can solve that quite easily with the integration.
To solve for T1 and T2 i suggest you use an estimate to start with based on some simple principles, you can always get more detailed later. For example, use the half cycle period minus the peak sine time to next sine amplitude time period knowing the filter RL*C time constant. So the end would be at one peak, and the start would be at the coincidence of the falling RC wave and the next sine. You can even estimate the RC falling wave as a straight line assuming the time period is short. That's a good place to start.

There is one circuit application i can think of that would perhaps use the simpler way with the op amp, and that is a circuit that needs to calculate the average voltage of a sine or rectified sine.

So you dont have to give up just yet :)
 
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t_n_k

Joined Mar 6, 2009
5,455
......
In other words, what we are modeling here is a full wave bridge rectifier circuit with a power op amp on the output, and the power op amp feeds the output filter. There's no harm in doing this but just keep in mind that the output of the filter will not be the typical output you find in most circuit applications.
......
I'm not convinced this will produce the same waveform as the "original" circuit.
The power amplifier presumably drives a component of the full-wave rectified output into the filter. That is to say, part sinusoidal voltage pulses concurrent with the T1 to T2 conduction interval. What happens to the filter capacitor voltage during the "off" condition - when the power amplifier output is zero volts? Doesn't the capacitor discharge back into the amplifier output terminal? Unless something blocks that current / charge reversal ..... like a diode.

Edit:
Decided to try some simulations. Hopefully the attachments indicate my meaning:

Rectifier Simulation Options.jpg
Rectifier Waveforms.jpg
 
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MrAl

Joined Jun 17, 2014
11,389
I'm not convinced this will produce the same waveform as the "original" circuit.
The power amplifier presumably drives a component of the full-wave rectified output into the filter. That is to say, part sinusoidal voltage pulses concurrent with the T1 to T2 conduction interval. What happens to the filter capacitor voltage during the "off" condition - when the power amplifier output is zero volts? Doesn't the capacitor discharge back into the amplifier output terminal? Unless something blocks that current / charge reversal ..... like a diode.

Edit:
Decided to try some simulations. Hopefully the attachments indicate my meaning:

View attachment 78637
View attachment 78636
Hi,

I thought i made it clear that it would not be the same with the op amp so im not sure i understand what you are saying. The op amp is just there to make the math more exact when we dont consider the diodes to ever stop conducting. That's just to make the first analysis easier so we dont have to solve for T1 and T2, and is mostly just an exercise in doing circuits this way.

Oh did you perhaps mean that we had to use a current generator instead of a voltage source when we do it the second way? Maybe we should stick to the first method then, which is the method with the full wave rectification and op amp feeding a low pass filter.
For this the ratio of Rs to RL is small so the DC output is almost the same, but the filter cap is selected to attenuate all the AC components, especially the 2nd harmonic.
What you need to do in your simulations is show the component values for the resistors and capacitors of the filter. We need a pretty big cap to see this work, on the order of 150000uf using a 0.1 ohm resistor for Rs, or 15000uf for 1 ohm.


Lets look at the two cases:
1. rectifiers only: the cap gets charged only for part of the time and discharged through the load. The DC voltage might come out to the same as the average DC.
2. rectifiers with power op amp:The DC voltage is the average DC voltage.

So in case 1 we have to assume the load is enough to discharge the cap so the DC voltage is not very near the peak.
In case 2 we dont have to assume this because the op amp draws the voltage down and the DC voltage ends up being the average DC.

I corrected my previous post too where i mistakenly typed that 7.6v is not typical, but with 12v peak it may be typical because it is the average DC voltage of the full wave rectified sine. But i was in no way trying to say that we get the exact same result with the op amp, especially for example at no load. What it looks like now however is that we get a better result with the op amp then when solving for T1 and T2, unless we can use a current generator, or solve for the wave between the peak and the coincidence of the next sine and also for the coincidence to the next peak and use both these times to generate the Fourier coefficients.
 
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t_n_k

Joined Mar 6, 2009
5,455
Lets look at the two cases:
1. rectifiers only: the cap gets charged only for part of the time and discharged through the load. The DC voltage might come out to the same as the average DC.
2. rectifiers with power op amp: the cap gets charged for part of the time and discharged part of the time by the op amp itself. The DC voltage is the average DC voltage.
I thought I did - in fact I looked at three cases as per my attachments. Perhaps you could draw a diagram to clarify your meaning.
 

MrAl

Joined Jun 17, 2014
11,389
Hi,

Sorry about that, the diagram showed up a little bit too light on my monitor so i did not see the values in the first op amp circuit. If i did see that i would have assumed you wanted that for all the circuits.

Lets look at the first case where we have a bridge rectifier feeding the op amp, and using ideal diodes for the bridge so there is no drop.
The op amp puts out a perfect absolute value of the sine wave input which we take to be 12*sin(w*t) so the output is abs(that).
Next, we apply the output of the op amp to the filter which is made of 1 ohms for Rs and 20000uf for the cap and some reasonable load for RL.
The output is a wave that is similar to but not exactly the same as the full wave rectification with diodes only.
Note that there are no switches involved in this case.
 

t_n_k

Joined Mar 6, 2009
5,455
OK.
I need to get straight in my head what you are proposing.
This is what I think are the two cases:

Case 1. An single phase diode bridge rectifier with 12V peak sine input feeding a total load including a series resistor Rs = 1 ohm and a parallel load combination comprising a filter capacitor of 20,000 uF and a resistive load RL - let's say RL=100 ohms. The diodes are assumed to have zero forward voltage drop.

Case 2. An ideal full wave rectified AC source is input to a unity gain power amplifier. The full wave source waveform would be same as that seen at the bridge rectifier output terminals in Case 1, if the entire load were removed. The power amplifier output drives the same total load as that in Case 1 - a series resistor of 1 ohm and a parallel load combination of 20,000 uF and 100 ohm resistor.

You are claiming the load voltages will be very similar. This can't possibly be true.
At the very least, the difference in the steady state average load voltage (across RL) would be a point of significant difference. For Case 1, one would obtain an average of about 11.24 V and 7.56 V for Case 2. There would also be a substantial difference in the ripple voltage.

So I must have got you wrong.
 
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MrAl

Joined Jun 17, 2014
11,389
Hello tnk,

You seem to be following this just fine but perhaps a little too verbatim. I realize also that this is not your fault but mine for not being too specific from the start.

We want to look at this in the light of how bridge rectifiers are typically used and what kind of loads they have. I think it would be more rare to see the output loaded very lightly with respect to the capacitor size, so i would suggest that if you want a more direct comparison then you also have to load the output of case 1 with an appropriate load resistance...one that would be more compatible with the capacitor value. Note that 11.24v is about 94 percent of the peak of 12 volts, which i think is a bit rare for a power rectifier circuit. It's not impossible, but i think it would be more typical to see it loaded more again with respect to the capacitor value.
For case 2 we dont want a significant load because we get the expected waveform already after we filter even without a load resistance.

So for case 1 we use a smaller load resistor, and for case 2 we dont have to use any. The two are then more similar and comparable.
I realize that this is starting to sound like string theory where we make up the rules in part by intuition as we face different situations, but that's what works so we can do it and it doesnt hurt to keep in mind that this is mainly for an exercise in Fourier and not intending to be an exact solution, even though the solution in case 2 will still end up looking like a typical bridge rectifier with filter. We note that the difference between the two circuits is mostly the impedance, so we must make a concession somewhere in regard to the impedance to get them to act in a similar fashion. What we see on the output is the average DC of the full wave rectified sine with AC components that resembles ripple although not an exact replica either. The output is DC with somewhat low level even harmonic content due to the nature of the rectified wave and the filtering action that acts on these AC components.

Without doing any plotting, try to figure out which one of these plots is the real and which is the approximated rectifier and filter. You for one can probably spot the real, but im sure you can notice the similarities too. (Also keep in mind we are looking for the steady state solution more than anything else).
 

Attachments

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MikeML

Joined Oct 2, 2009
5,444
If ever there was a case for simulation instead of a analytical solution, this is it:

To do this seemingly simple circuit justice, it is necessary to account for some parasitics that are inevitable. I model a typical transformer-powered full-wave bridge rectifier, a filter capacitor, and a constant current load. The reason I choose a constant-current load is because that is effectively what happens in a practical power supply, where there is an IC voltage regulator between the filter and the ultimate load.

I account for typical source resistance R1, and effective series filter capacitor resistance R2. The values for R1 and R2 are typical, based on my own experience. The rectifiers are typical 3A Silicon diodes, with a internal series resistance intrinsic in the model.

The simulation shows four different regions of interest, that would have to accounted for in a purely analytic solution.

226b.gif

Region B-C is where the load current (1A) is entirely supplied from the filter capacitor. All other currents are zero except for leakage currents. The current through R2 is -1A. The capacitor is discharging linearly at the familiar I/C rate, or 1/1000u = 1000V/s = 1V/ms.

Region C-D is where the transformer current (flows through R1, the forward-biased diodes, R2) just begins charging C1. This is indicated by the current increase in I(R2) at arrow C. At arrow D, the current I(R2) in/out the filter capacitor goes to zero, which is the end of where it is discharging, and it begins charging.

Region D-A on the subsequent half-cycle is where the transformer is supplying all of the load current (1A), and is supplying charging current for the filter capacitor.

Region A-B is where the transformer current gradually decreases from 1A to zero, and the current out of C1 increases from zero to |-1A| as the capacitor again picks up the total load.

An analytical solution would require you to write equations that account for the behavior in all four regions.

Selecting a filter capacitor value is all about making it large enough so that at its minimum, the voltage is high enough so as not to drop below the dropout voltage of the following regulator. Since I already have used the simulator, it is simple to ask it for the average and RMS value of V(out) see Inset. Note that this the voltage across the filter capacitor after all of the drops are accounted for. The open-circuit voltage of the transformer is 30V, but the max of V(out) is 26.5V, and min is ~20.6V

It is also easy to ask it to show us the FFT of V(out), which shows that there is energy at all even harmonics of the line frequency, but none at the fundamental line frequency or odd harmonics.

226f.gif
 
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MrChips

Joined Oct 2, 2009
30,714
I recall in my first analog electronics course we were given the exact same problem to solve. Every course in electronics has the same design and analysis problem:

Find the average output voltage, ripple voltage and peak diode current in a full-wave rectifier power supply circuit.

This problem is too complex to solve using straight linear algebra. One has to make some simplifications and approximations.

The first simplification I made at the time was to determine the reactance of the filter capacitor at twice line frequency, i.e. at 120Hz.
 

MrAl

Joined Jun 17, 2014
11,389
Hi,

Yes i have to agree that this should only be an exercise that is not intended to be super accurate nor widely applicable, unless we know much more about the parts involved.

There is one case however which is more close to the theoretical, and that is when there is an inductor too. The inductor simplifies all the problems with the impedance so we can calculate lots of stuff knowing the inductance. We actually did this right here on this forum in another thread. The idea is to take advantage of the theoretical fact that the volt seconds to charge the inductor is equal to the volt seconds to discharge the inductor to the same original state. We'd have to find this, and it is pretty interesting really mostly because it shows how the inclusion of an inductor actually simplifies the analysis.

If we are allowed to do it another way, we can solve for everything but we do have to know all the specs of all he parts as well as things like line inductance. For the diodes we would use the spice model for those diode parts and have to be satisfied with that.
 
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