Can anyone explain how filters, e.g. bandpass, work? Thanks.

 

Can anyone explain how filters, e.g. bandpass, work? Thanks.

Why certainly....one of my favorite topics.

Stand by, I need some coffee.

 

Hello,

A lot of information on filters can be found in the links of this page of the EDUCYPEDIA.
http://www.educypedia.be/electronics/analogfil.htm

Greetings,
Bertus

 

Can't be cheese, so it must be 42.

Still curious? Look into our Ebook chatper on filters - http://www.allaboutcircuits.com/vol_2/chpt_8/1.html.

 

Can anyone explain how filters, e.g. bandpass, work? Thanks.

fillter has diffrent construction according to it's band.suppose we are using high pass filtter ,then we can fillter all high frequency by connecting capacitor in series.

 

Okay, I'm back.

All righty then. A filter works because reactive components (inductors and capacitors) have an effective resistance that changes with frequency. In fact this is the very definition of REACTANCE.

Let's take a very very simple filter, a capacitor and a resistor in series.
Now, think of the capacitor and resistor as a voltage divider.

If we apply an alternating current to the input of the capacitor, and measure the voltage at the junction of the capacitor and the resistor. We will find that as we INCREASE the frequency, the voltage across the resistor increases. This is because the voltage drop across the capacitor DECREASES, according to the formula for capacitive reactance: Xc= 1/2pi*FC. This simple circuit creates a HIGH PASS filter, because at higher frequencies, we have more voltage across our load (the resistor)

 

Now, what if we put an inductor and a resistor in series? An inductor looks like just a piece of wire at low frequencies...the effective "resistance" (Inductive reactance) is very small. An inductor acts the opposite of a capacitor. The formula for Inductive Reactance is: XL=2 pi *FL. The reactance of an inductor goes up proportionally to the frequency.

At a frequency of 0 Hz (DC) the full voltage of the power supply appears across the resistor. However, as the frequency INCREASES, less voltage appears across the resistor, because the inductor now has more effective resistance (reactance). So as we increase the frequency, the "output" voltage of the filter goes down. This is a basic LOW pass filter.

 

Now, what happens if we take our primitive HIGH PASS filter, and follow it with a LOW PASS filter? Well, assuming the CUTOFF frequency is higher for the low pass filter than is is for the high pass filter, we now have a BANDPASS filter.

Pretty simple, eh?

 

P.S.

This is grossly oversimplified...we haven't even talked about the PHASE yet. But that will come.

Eric

 

I have attached some simplest form of filter circuits,without use of inductor....

 

Hang on a minute, my head is slowly working this out...011101011010110. beep beep beep.., and I'm back. So, to re-cap(acitor), a capacitor can work like a frequency-dependent-potentiometer. The higher the AC frequency, the more the capacitor resists (effectively). And with an inductor, the higher the frequency, the more current it lets through.

 

Hang on a minute, my head is slowly working this out...011101011010110. beep beep beep.., and I'm back. So, to re-cap(acitor), a capacitor can work like a frequency-dependent-potentiometer. The higher the AC frequency, the more the capacitor resists (effectively). And with an inductor, the higher the frequency, the more current it lets through.

You've sort of got the picture, but you've got it exactly backwards.

A capacitor has LESS resistance at higher frequency, and the inductor has MORE resistance at higher frequency.

eric

 

Xc= 1/2pi*FC.

What does this bit mean?

 

Hello,

Take a look at this page of the AAC eBook.
http://www.allaboutcircuits.com/vol_2/chpt_4/2.html

Greetings,
Bertus

 

Sorry, could you explain that, as it's a bit confusing. I'm not good with maths.

 

The equivalent AC resistance (how much the capacitor fights against the current) is calculated by:

Xc = 1 / (2*pi*f*c)

Xc is the reactance (equivalent AC resistance)
pi is the constant value 3.14159
f is the input signal frequency
c is the capacitor value in farads.

So if we have a 1uF capacitor and apply a signal of some volltage at 1kHz, the reactance (equivalent resistance) would be:

Xc = 1 / (2*3.14159*1000*0.001) = 0.16 ohms (almost short circuit)

If we reduce the applied frequency to 10Hz;

Xc = 1 / (2*3.14159*10*0.001) = 16 ohms

So you see the capacitor allows high frequency and fights against low frequency.

 

IT work by signal leads and lags .