Finding the -3db Frequency?

Thread Starter

ryanc

Joined Oct 22, 2011
3
Could somebody explain to me how calculate the -3db frequency of this circuit? I would appreciate being shown where I went wrong in my calculations (picture).

What I don't want: somebody to just say something to the effect of fo=1/2pi*C*R, like what every single search result on the history of the internet comes up with in every forum where the question is asked..

I'm using the point where the magnitude of the transfer function drops to 1/sqrt(2), and solving for wo manually, not just by 'inspection'. I have a feeling I made a simple error in my equation, and am not totally off...(answer i got is on the same order of magnitude..) :/

Rs=20k, Ri=80k, Ci=5pF

Thanks you
 

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Ron H

Joined Apr 14, 2005
7,012
Did you consider that the output is down 1.94dB at DC? It only has to drop another 1.07dB to be down 3.01dB (1/sqrt(2)). I didn't inspect your math, because it is sloppy and blurry, but I suspect that is where you made your mistake.
As your correct answer indicates, the actual 3dB frequency is relative to the gain at DC.
 

Thread Starter

ryanc

Joined Oct 22, 2011
3
Would it be correct to say when doing this sort of calculation, I should be removing any gains found at w=0 when I do the gain at DC then? (meaning I just take out 1+Rs/Ri factor) Sorry its blurry, it looks perfectly clear to me :p

My instructor(s) essentially keep telling me to get the 3db frequency I 'just make the magnitude 1/sqrt2', so I don't know any conditions that have to exist for me to actually do that..

I'm just trying to pin down a method for doing these kind of problems.. so what would I be solving for on the left hand side of the equation? The magnitude drop of gain at DC? Any other conditions?

Thanks for your help!
 
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Thread Starter

ryanc

Joined Oct 22, 2011
3
Does this help?
Yup, it does a little, but all you really did was reduce the circuit so it was easier to solve, I'm trying to get to the root of a method that makes solving all my other 3db frequencies I encounter (2nd order, SK, oscillators, etc) easier. I did find something in my textbook that used something along the lines of |dc gain|/sqrt(2)=|H| for an active filter, so I guess that would make sense..?
 

Ron H

Joined Apr 14, 2005
7,012
Yup, it does a little, but all you really did was reduce the circuit so it was easier to solve, I'm trying to get to the root of a method that makes solving all my other 3db frequencies I encounter (2nd order, SK, oscillators, etc) easier. I did find something in my textbook that used something along the lines of |dc gain|/sqrt(2)=|H| for an active filter, so I guess that would make sense..?
That sounds right. You have to normalize the response to the DC gain.
 

Audioguru

Joined Dec 20, 2007
11,251
The stray capacitance of the wiring and load will be at least 5pF so the actual -3dB frequency will be much lower than calculated.
 

Ron H

Joined Apr 14, 2005
7,012
The stray capacitance of the wiring and load will be at least 5pF so the actual -3dB frequency will be much lower than calculated.
Maybe 5pF IS the stray capacitance.
This is homework. Homework doesn't have stray capacitance.:p
 
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