Hi,
I need a good source to understand the derivation for cut-off frequency, relation of input to output by Kirchoff's laws and what is meant exactly by the impulse response of the amplifier (providing it is not made of a simple unity gain buffer).

What theories are behind it so I can go through them?


thanks

 

I found this app note attached but I can't figure out the part highlighted in red.

also in the first part Vi and Vo are considered as grounds in for the ac signal, but considering an ideal op amp, why isn't Z2 and Z3 considered in series for deriving eqn 1.(only z2) (providing op amp have infinite input impedance)

 

the analysis assumes that opamp is ideal. that means:
- no current flows in or out of inputs and
- voltage difference between inputs is zero.

using first assumption:

current through Z2 is

(Vf-Vp)/Z2

current through Z3 is

Vp/Z3

because there is no current in or out of opamp input, those two currents must be same:

(Vf-Vp)/Z2=Vp/Z3

if you separate variables Vp and Vf, you get form as KCL at Vp in the example:
Vf/Z2=Vp(1/Z2+1/Z3)


the other way is to use voltage divider:

Vp=Vf*Z3/(Z2+Z3)

rewrite in terms of Vf to get
Vf=Vp(Z2+Z3)/Z3

which is the same thing...

 

To make a sharp cutoff (instead of a droopy cutoff), a Sallen-Key filter uses some positive feedback to boost the -6dB level at the cutoff frequency of the two RC filters to -3dB. The positive feedback makes the filter ring a little at the cutoff frequency.

 

I'm assuming this is to get rid of your current source oscillation mentioned in this thread. As I said in post #24, your current source will probably not put out the correct average current when it is oscillating (it didn't in simulation). Did you try putting the series combination of 100nF and 100Ω in parallel with the inductor to kill the oscillation?

 

Hi Ron thanks for you interest but this is not for the same application I will try it out in practice after the Easter recess at school since I do not have access to a scope.

This problem was in my control systems and automation assignment and I literally have to go deep into the impulse response of this filter.

I understood all the kirchoff's part and now I getting familliar with the impulse response part.

Any one can guide me on what I should look for, will be of great help.

thanks

 

Hi Ron thanks for you interest but this is not for the same application I will try it out in practice after the Easter recess at school since I do not have access to a scope.

This problem was in my control systems and automation assignment and I literally have to go deep into the impulse response of this filter.

I understood all the kirchoff's part and now I getting familliar with the impulse response part.

Any one can guide me on what I should look for, will be of great help.

thanks

OK - never mind.

 

When you mention the ringing effect... are you refering to the cut off part in the bode plot or is it like feeding the particular cut off frequency and the attenuated output is added to some ringing??

 

The positive feedback at the cutoff frequency forms a resonant circuit that rings like a bell when the signal suddenly starts or stops. The amplitude bounces.

 

thanks for the diagrams.

So to clarify the situation: We are seeing this (ringing) at the output and for it to occur, the input signal has to contain a harmonic equal to the cutt-off frequency. When we are saying 'start' and 'stop', do we mean when this particular frequency is present at the input and when it is not? .. or is it that the first few cycles pass through until they are stopped by the filter?

 

Ringing in a Sallen-Key filter does not occur if the filter has a droppy Bessel cutoff. It occurs when the filter has a sharp Butterworth or "Shabbychev" cutoff.
It occurs with any frequency near the cutoff frequency.
It occurs if a tone near the cutoff frequency is suddenly started or stopped.

Audio crossover filters use Sallen-Key filters and the ringing is not heard.
Audiophools are purists who use a Linkwitz-Riley filter that is sort of droopy with no ringing.

 

okay intersting . how exactly is the Barkhausen satisfied?

 

The Barkhausen is satisfied because a Sallen-key filter has a low amount of positive feedback. Far lower than is needed to cause continuous oscillation. It is not unstable, it simply has a +3dB peak in its frequency response to boost the -6dB drop at the cutoff frequency caused by the two RC filters.

 

I found this app note attached but I can't figure out the part highlighted in red.

also in the first part Vi and Vo are considered as grounds in for the ac signal, but considering an ideal op amp, why isn't Z2 and Z3 considered in series for deriving eqn 1.(only z2) (providing op amp have infinite input impedance)

If you are familiar with Millman's Theorem, you will see that eq. 1 and the portion of eq. 2 that you have outlined in red are both examples of Millman's.

hgmjr

 

Hi,
after a some hours .. i managed to get something.
Can someone check my derivation. I tried to use the Conductance G to simplify things a little in the early steps.

Now I need to find the range of values of A for which the impulse response of the amplifier is not sinusoidal.