# Need Help With Cutoff Frequency

Discussion in 'Homework Help' started by Mawangs1, Apr 23, 2013.

1. ### Mawangs1 Thread Starter New Member

Mar 26, 2012
23
0
I'm getting confused. I'm trying to make a sallen key low pass filter where it's been established that R1 = R2 and C1 = C2.

ω = 1/sqrt(R*C)

Let's say I want to make an active lowpass filter that cuts out all frequencies above 400Hz.

With this in mind, I would do:

(1) 400 = 1/sqrt(R*C)

(2) plug in a random C value

(3) solve for the R.

I'm being told I'm off though. I hear that I actually need to do:

(4) ω = 2*∏*f

(5) put in 400 for the f and calculate ω

(6) do steps (2) and (3) as before

Steps (4) to (6) make no sense to me though. If I do this, wouldn't it would make my signal take longer to attenuate?

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,706
1,323
The correct answer is step 4 to 6.

3. ### Youssef.zein86 New Member

Apr 24, 2013
5
1
Indeed the correct answer is step 4; however you should be logical in choosing the values of R and C.
The salen key design assums that the imput resistance of the op-amp is infinity, so you have to choose R relatively small to the imput resistance of the op-amp.
Typical voltage controlled op-amps have imput impedances that range 1 - 2 Mohms, so choosing R to be 50 times smaller will give good results, so R should not be higher of about 40 K, and you will find C as the last unknown of the equation.