DC active filter

Hi everyone..

i would like some suggestion for my project since this project little hard for me to find the circuit

my topology

power supply=>rectifier=>active filter=>buck converter=>load

input voltage=415v
output voltage should be=220v,7a

my problem at dc active filter where i have to eliminate ripple by not using super conductor or passive filter.how??

hope u alll can help me..thanks

If the voltages weren't so high I'd suggest this old thing.



The apparent capacitance is β χ C.

Doesn't handle surges well though.

If the question is hypothetical, use a bulky, heavy, and expensive inductor. If the question is practical, I will learn something new from the answer.

of course the conventional way is using bulky capacitor. but for my project, this new filter is a technique by not using the expensive capacitor. in my opinion:

f(t)=Vdc+ v1(sin nwt+cosnwt)


son by injecting harmonic:
v1(-sin nwt-cosnwt)=v1( (sin nwt+180)+(cos nwt+180)
we can can get pure Vdc

so how can i perform this technique

What you illustrate looks remarkably like the full wave rectification of the original 3 phases. Dos your technique propose to achieve this same effect by out-of-phase power injection, or will it act upon the rectified waveform so as to substitute for the filtering action of capacitors?

It is hard to imagine the harmonic injection you speak of - do you have some ideas to share with us?

yes,it will act upon rectified waveform so as to substitute for the filtering action of capacitors. my objective here to eliminate those harmonics came from rectified voltage..

anyone..please...

The big bulky capacitor you are trying to remove isn't only important for filtering... it also plays a large role in assisting large and quick current draws from the load.

If you are able to ignore this and focus on voltage rectification, I don't see why you couldn't ac couple and phase shift 180 degrees with an active driver with a gain of 1 to elliminate your ripple. This technique would cripple under load I imagine.

Any thoughts?

The general technique is to use a capactors to remove the ripple, it does this by providing power during the points the source doesn't. You can also use a battery for something similar, but without the ability to vary the voltage. You have to have something to store juice, be it a cap, coil, or battery. Capacitors aren't that expensive.

I'm going to say "very big inductor" again.

Actually, how about an AC motor driving a homopolar generator? This would be even more costly in practice, but might get extra points for style.

How would a single pole motor help smooth the ripple? i iz curious about that one.

of course my lecturer said that this project is not practical and not many done my project..that is why i'm stuck now to do the project =(..too bad..i hope anyone could help me for this project..i don't have any idea right now

So what kind of project are you looking for. What level are you at, be it college or high school?

Once you start it, assuming you continue posting here, I'd start it as a new thread (with a possible reference to this one).

The easiest solution mate is to measure your ripple and generate an equally negative voltage to it. This won't be practical for using it as a reliable high power source, but should smooth it out.

how do i generate negative voltage??i try to search on internet about creating negative voltage but in practical i dont find one

my level is college level

Here is a circuit I just drew up to demonstrate what I am thinking. Mind the notes.

R2 >> R1

The intent is that INVERTED_RIPPLE will be an inverted version of RIPPLE with a much larger gain, so once you feed INVERTED_RIPPLE back into Vac after the decoupling cap C2, it should overpower RIPPLE and pull RIPPLE back to 0.

Somehow I think there will be stability problems.

You must also note that INVERTED_RIPPLE must greatly overpower Vac so as to effectively remove RIPPLE.

In reality, just using a filter CAP on the rectified output is much much much easier and stabler. You'll need as big of reservoir caps on +- 24VDC as you would on Vac.

In summation, if you use this implementation, you'll end up with less stability, more complexity, more larger capacitors.

If you are just trying to show off to a professor, this might win you a few bonus points.

i want to share to you all from my reference book
("UPS and ACTIVE FILTER", Ali Emadi)

In DC/DC converters, there is a high-frequency ripple voltage at the converter
output. This ripple is larger in high-current, low-output voltage converters.
In modern DC/DC converters, which need smooth and ripple free
outputs, this problem should be solved properly. Traditionally, LC passive
filters have been used to compensate this ripple. For proper mitigation, this
passive filter must have a wide frequency band. This characteristic increases
the size and, consequently, the weight and cost of the passive filters. As mentioned
for the EMI filters, second-order passive filters cannot mitigate ripples
properly. With increasing size and attenuation of the passive filter, the
chance of an interaction between the control system and filter increases. On
the other hand, changes in the parameters of the passive filters with time and
environmental conditions decrease their attenuation [45].
Another passive method for decreasing current ripple in DC/DC converters
is the use of coupled inductors. Using coupled inductors, the voltage
over output inductor of the converter can be forced to zero and, consequently,
the ripple of the current will be zero. Major savings in size, weight,
and cost can be achieved using this method. Unfortunately, this method is
not applicable to all types of DC/DC converters. Additionally, changes in
converter parameters with time and temperature do not allow proper cancellation
of the ripple current.
In the same manner as in EMI cancellation, active ripple filters with robust
characteristics and better attenuation can provide an alternative for passive
filters.

are active ripple filters any different than just 'normal' active filters?

The downside to active is that the active filter must provide all of the original power of the dc-dc, which generally with passive components is easier to achieve.

You can not have an "active" filter without a power supply. This is putting the cart before the horse.

Your book is slightly off in one detail, that LC filter in a power supply isn't just a filter, it provides power to fill the gaps. You can treat it as a filter with math, but you have to keep in mind that the capacitor, or the inductor, or both, is providing stored energy. An "active" filter emulates a real filter, by emulating it can remove all those bulky components. It does this by using the power supply to source the power the inductor and capacitor used to.

I think you have a serious misconception about active filters. I can make a gyrator that emulates 1000 Henries of inductance, but if I try to use it as a flyback it will start to react, and hit the power supply rail. This is because it isn't really storing any energy. It only pretends to, and as long as the energy needed is within the power supply parameters you won't have a problem. The same arguement could be used for a gyrator that emulates 1000 Farads. I can use a 1KF cap for power, almost like a battery, but turn the power off an active filter and what have you got?

tq for your explanation..u really open my eye about this subject..hope u can share more knowledge in this thread