# How filters work

Discussion in 'General Electronics Chat' started by stellarpower, Feb 20, 2009.

1. ### stellarpower Thread Starter Member

Feb 19, 2009
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Can anyone explain how filters, e.g. bandpass, work? Thanks.

2. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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Why certainly....one of my favorite topics.

Stand by, I need some coffee.

Apr 5, 2008
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4. ### beenthere Retired Moderator

Apr 20, 2004
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Last edited: Feb 20, 2009
5. ### speeed_777 Active Member

Oct 5, 2008
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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.

6. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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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)

7. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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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.

8. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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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?

9. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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P.S.

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

Eric

10. ### speeed_777 Active Member

Oct 5, 2008
54
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I have attached some simplest form of filter circuits,without use of inductor....

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11. ### stellarpower Thread Starter Member

Feb 19, 2009
28
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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.

12. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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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

13. ### stellarpower Thread Starter Member

Feb 19, 2009
28
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What does this bit mean?

Apr 5, 2008
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15. ### stellarpower Thread Starter Member

Feb 19, 2009
28
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Sorry, could you explain that, as it's a bit confusing. I'm not good with maths.

16. ### wilson479 New Member

Feb 23, 2009
5
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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.

17. ### italo New Member

Nov 20, 2005
205
2
IT work by signal leads and lags .