# filter capacitor circuit for full wave rectifier

Discussion in 'Homework Help' started by kunalpat, Aug 7, 2014.

1. ### kunalpat Thread Starter New Member

Aug 7, 2014
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0
In a typical filter circuit involving a filter cap. we see ripples being generated..i.e capacitor starts discharging (as if there were no external power supply), in the decreasing half of each half cycle.This can happen only if the voltage across the diode drops to something below its barrier potential.But assuming a straightforward approach, why isnt the charge profile of the capacitor similar to that of the voltage output?In most books, its starts with the cap. discharging(its never highlighted why) and then because the rate of change of voltage across the cap is less than the external power supply, the diode is off.its never mentioned that only because the diode becomes reverse biased that the cap. starts discharging like in the absence of a power supply.WHY SHOULD THE DIODE TURN OFF IN THE FIRST PLACE?
this thing has been haunting me for over a week, havent found a satisfactory answer.

2. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,071
Think about what is upsteam from the rectifier(s).

Usually the power comes from a secondary winding on a tranformer, so is really a 50 or 60Hz Sine wave. The rectifier conducts only when the instantaneous transformer voltage is higher than the voltage left on the filter capacitor...

3. ### kunalpat Thread Starter New Member

Aug 7, 2014
3
0
exactly, and that is the case till the peak, but then WHY does the diode have to switch off then(ideally), why cant the charge just drop sinusoidally again?why do we assume the cap starts discharging non-sinusoidally and consequently infer that the diode is switched off??

4. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
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Well, if there is any load on the power supply (including a bleeder resistor), then the filter capacitor does discharge between the positive peaks of the input waveform...

Look at this simple simulation: The blue trace shows current through the diode. When the diode current is zero, it is because the diode is not forward biased, which is the case for most of the ac cycle. The diode is forward biased only when the green trace V(s) is greater (by 0.65V) than the voltage left in the filter capacitor V(f) red trace.

Note the large peak current that flows in the diode. Compare that to the average current that flows in the load resistor

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Last edited: Aug 7, 2014
5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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In an ideal resistive loaded capacitive filter rectifier with zero forward diode voltage drop, the point of deviation from the rectified sinusoid envelope doesn't actually occur at the rectified sinusoidal peak. Depending on the capacitor value the point of deviation may occur well beyond the peak - with the delay after peak increasing with decreasing capacitance.

6. ### MrAl Distinguished Member

Jun 17, 2014
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Hi,

You can get a (half) sinusoidal cap voltage no problem. All it takes is a capacitor value that is too low for use in a typical power supply and a little load resistance. But who wants that? A well designed or even moderately designed power supply will have a large enough capacitor such that the waveform across the cap is more like a ramp than sinusoidal, and that makes it work better.

It is true that the point where the cap voltage dv/dt becomes greater than the sinusoidal dv/dt is the point where the cap starts to discharge, but that's what we want to happen. If instead the cap voltage follows the sine too far down then we did something wrong because we'll have too much ripple.

So solve this equation:
dVs/dt=dVc/dt {90<Theta<180 degrees}

For example, at 50Hz and Rload=100 ohms and C=1000uf we get an approximate solution of 91.7334 degrees, which is just to the right of the peak.
For 100 ohms and C=100uf, we get approximate solution of 100.494 degrees, which is a bit farther from the peak so the cap voltage follows the sine for a longer time period.

This is assuming an ideal diode, and we usually dont get into the details of the diode too much for this kind of problem unless we want to do a more careful analysis in which case we have to use the diode curve. These are also first approximations which do not attempt to iterate or solve for the exact initial voltage of the cap.

Last edited: Aug 7, 2014
7. ### alfacliff Well-Known Member

Dec 13, 2013
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now do it again for a full wave rectifier like the op said at first. less ripple and twice the frequency.

8. ### MrAl Distinguished Member

Jun 17, 2014
3,715
788

Hi,

It is the same equation for half wave or full wave. The frequency is the line frequency not the pseudo pulse frequency. The only real difference is we could solve for the solution also between 180+90 degrees and 360 degrees instead of just 90 to 180 degrees.

Another interesting point to solve for is the point where the cap voltage again coincides with the sine, only this time the solution would lie somewhere between 0 and 90 degrees unless it was half wave and the cap was too small and then it would look almost like it reached zero volts and then it would start up again at 0 degrees.

Last edited: Aug 7, 2014

Mar 6, 2009
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10. ### alfacliff Well-Known Member

Dec 13, 2013
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sctuallly the difference between a half wave and full wave rectifier is that filtering is simpler and usus less capacitance because there are twice as any pulses, and almost no dead time between them. 120 hz hum in the load, vs 60hz hum.

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I use a different approach which I believe gives a more accurate estimate.

The ideal rectifier output current will drop to zero at an angle θ after the sinusoidal voltage peak where

$\theta=atan(\frac{1}{\omega R C})$

So at 50 Hz or ω=314 rads/sec, R=100 Ω and C=100uF

θ=atan(0.319)=17.7°

or at 107.7° rather than 100.494°.

The result may be confirmed by simulation of the ideal full wave rectifier.

Last edited: Aug 7, 2014
12. ### MrAl Distinguished Member

Jun 17, 2014
3,715
788
Hi there,

I'll try to check out your solution and compare results.

I also saw the other post with the interesting links, however a few years back i ran into this same problem, where we had no analytical solution for something that we all use: the cap filtered rectifier. This led me to find a solution for this simple but complex circuit myself because i could not find anything i felt was worthwhile. By worthwhile i mean if we are going to take the time to do a more complex calculation, then it had better return more accurate results. What this led to was a model that included all the strange stuff like cap ESR, inductance, and the Spice model of the diode. I cant stress enough that the Spice model of the diode is a necessary feature and there's no way to get accurate results without it. This is why in a cursory discussion i always present a rough estimation, because the real result is going to take a lot more of a calculation than most people want to get into and also some extra measurements that nobody wants to perform. The more 'ideal' case is usually acceptable in order to just get an idea what is going on, so this is what people often seem to find more acceptable.

Checking out your result, i have to agree that using the fact that the current would go through zero if it could when the cap starts to discharge (obviously since it stops charging and starts to discharge at that point) is a good way to do this idealized calculation. I came up with a slightly different result:
Angle=pi-atan(w*R*C)

but because you later modify your result this means they should both lead to the same result numerically (107.657 degrees, for R=100 ohms, C=100uf, f=50Hz).

I also have to add that this makes the slope comparison method look less accurate in general, and the current vector method look better overall. I'll have to check to see if the slope method is an analytically less accurate method or it is just caused by limited numerical precision.

Last edited: Aug 8, 2014