Author Topic: RC LPF Filter  (Read 1055 times)

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Offline timjTopic starter

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RC LPF Filter
« on: March 02, 2019, 03:14:45 pm »
Hello, I'm a beginner in electronics, sorry if what I'm asking is basic. I have a task for one of my assignments which I'm unsure about. It is about RC LPF filters. I am asked to modify a circuit so the cut-off frequency is doubled. How would I go about doing so? Would I simply reduce the resistor or capacitor value for the formula: F(cutoff)= 1/2piRC. Also how would I show this practically? My lecturer says to keep the amplitude of the input constant but I'm not sure what he means by that. The circuit consists of just a signal generator, supplying a resistor of 1k and capacitor of 100n and a Vout measurement between the resistor and capacitor. Any help would be great. I have also attached an image.
« Last Edit: March 02, 2019, 03:25:13 pm by timj »
 

Offline bson

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Re: RC LPF Filter
« Reply #1 on: March 04, 2019, 02:35:25 am »
It doesn't matter which you change; the cutoff frequency follows their product, R*C.  Reducing either will increase the cutoff frequency since it's reciprocal.  The choice of values is mainly practical; smaller values for R increase current consumption and power dissipation, large values for R increase noise, and larger capacitors are physically bigger and cost more.  Since the current values are presumably well-chosen, you could reduce both enough to double the cutoff frequency.  Of you could reduce the capacitor alone; usually smaller capacitors are a win, at least down to 100pF or so.
« Last Edit: March 04, 2019, 02:37:57 am by bson »
 
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Offline todorp

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Re: RC LPF Filter
« Reply #2 on: March 04, 2019, 08:35:10 am »
To keep the loading of the source similar only the cap C should be reduced (without touching the R).
 
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Offline timjTopic starter

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Re: RC LPF Filter
« Reply #3 on: March 04, 2019, 09:11:06 am »
So if the audio generator has an output impedance of 600ohm, does that mean I'll need to include that in my calculation for R1?
 

Offline timjTopic starter

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Re: RC LPF Filter
« Reply #4 on: March 04, 2019, 09:14:03 am »
Also I kept V_in the same amplitude (1v), but when I change frequencies from 100Hz to 1kHz the amplitude drops and is not constant. Shouldn't it be the same, as those frequencies are before the roll off?
 

Offline MrAl

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Re: RC LPF Filter
« Reply #5 on: March 04, 2019, 01:52:43 pm »
Hi,

The amplitude of the output of the RC low pass filter is:
Vout/Vin=1/sqrt(w^2*C^2*R^2+1)

and the -3db cutoff point is reached when:
1/sqrt(w^2*C^2*R^2+1)=1/sqrt(2)

Solving this for w^2 we get:
w^2=1/(R^2*C^2)

and this leads to one non negative solution:
w=1/(R*C)

and of course w=2*pi*f with 'f' the cut off frequency in Hertz.

So we have:
2*pi*f=1/(R*C)

or:
f=1/(2*pi*R*C)

Now since R and C are both in the denominator, that means making R or C smaller by a factor 'a' makes the frequency 'f' go up by the same factor 'a'.
This means to raise the cut off frequency you have to make either R or C (or both) smaller.
So if we make C smaller by 2 times, the cut off frequency will go up by 2 times.
 


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