# Analysis of an RC low-pass filter

Discussion in 'Homework Help' started by ParaTrooperJoe, Apr 21, 2013.

1. ### ParaTrooperJoe Thread Starter New Member

Apr 21, 2013
6
0
Hello everyone! I am new to this forum but am glad I found it. I have a pretty good understanding of troubleshooting and analyzing circuits, but the one I am currently working on is proving a bit difficult, especially sense my book does a terrible job at explaining how to do this.

Anyhow, here is my assignment:

An RC low-pass filter has the following exact values: RF = 100 Ω, RS = 50 Ω, and CF = 10 μF. The filter is connected to a load that can vary over a range of 10 to 500 Ω. Calculate:

The range of Av for the circuit.
The range of fC for the circuit.

So, I know the formula for Av(max) = Req/(Req)+(Rs)
and I also know that Req = Rp || RL

However, I am lost as how to get started because, for one thing, I do not see a value for RL listed. Furthermore to find Rp, I would need to find Q, but to find Q I would need to calculate XL and again, no value is presented for an inductor.

Perhaps I am just looking at this assignment the wrong way?

- ParaTrooperJoe -

2. ### ParaTrooperJoe Thread Starter New Member

Apr 21, 2013
6
0
I could be wrong here, but I am assuming that Rs is the resistance offered by the Voltage source, and is (in effect) in parallel with the capacitor, therefore it can be represented as RL.

With that in mind I have done the following calculations:

Av(max) = Rs/[(Rs)+(RF)] = 50/(50+100) = 0.333

Then,
Fc = 1/

AV(max) = Rs/(Rs+Rf ) = 50Ω/(50Ω+100Ω) = 0.333

fc = 1/2@C99,RC = 1/(2π(33.33Ω)(10µF)) = 477.51 Hz

Am I even close to getting this one right?

3. ### ParaTrooperJoe Thread Starter New Member

Apr 21, 2013
6
0
Scratch my solution above, I must have been smoking something....

Here is what I think is the correct way to approach this.

Once I reread the question, I see that the load (RL) ranges from 10 to 500 ohms (DOH!)

So:

A:
When RL = 10Ω -

AV(max) = R_L/(R_L+R_F ) = 10Ω/(10Ω+100Ω) = 0.0909

When RL = 500Ω -

AV(max) = R_L/(R_L+R_F ) = 500Ω/(500Ω+100Ω) = 0.833

B:

When RL = 10Ω -
First, I calculated the value of R while RL = 10Ω:
R1 = 1/(1/Rf +1/RL1 +1/Rs ) = 1/(1/100+1/10+1/50) = 7.69
fc = 1/(2π(R1)(C)) = 1/(2π(7.69)(10µF)) = 2.07 kHz

When RL = 500Ω -
First, I calculated the value of R while RL = 500Ω:
R2 = 1/(1/Rf +1/RL2 +1/Rs ) = 1/(1/100+1/500+1/50) = 31.25
fc = 1/(2π(R2)(C)) = 1/(2π(31.25)(10µF)) = 509.3 Hz

I think that makes more sense, and hopefully I have done these calculations correct. If anyone has any advice to offer, I would greatly appreciate it!