Filters, Low Pass and Others

[Winterstone]

Shouldn't it be:

[latex]H_f= \frac {R2||\frac {1}{sC2}}{R1+R2||\frac {1}{sC2}}[/latex]

and:

[latex] H_r = \frac {R1}{R1+R2||\frac {1}{sC2}}[/latex]

Of course, you are right. Thank you for correction.
W.

 

[MrAl]

Hello,

I think you meant this:
Vout/Vin=Hf*Ao/(1+Hr*Ao )

I like to show a general method that comes up from the roots of analysis so that the reader can handle just about anything that is thrown at them.
If you care to show an alternate method that's fine. Some circuits are much more difficult though like the twin T notch or bandpass and these circuits require basic analysis anyway at least to start. So we really need that basic analysis more than we need the shortcuts.

If i wanted to show a simpler method i would have shown state space. That's a more general approach that is also simpler for the more complex circuits too. But it's always good to refresh. In another thread someone brought up the age old method of using a certain Laplace operation more directly to solve for the time response solution of a circuit. I hadnt seen that approach in years but it was good to see it again.

BTW maybe my short cut method was missed: the response is -Zf/Zin. That was one of the points.
I also believe that if anyone is afraid of solving multiple dimension equations they should get to finding methods to handle these too. What would they do if they were faced with a 100x100 matrix equation.

 

[The Electrician]

I think when as much math as you gave in post #1 is in text form, it's really hard to follow.

With that much math, using Latex is appropriate, and it's not difficult.

For example, instead of: Vout/Vin=Hf*Ao/(1+Hr*Ao )

how about: [latex] \frac{V_{out}}{V_{in}} = \frac {H_f A_o}{1+ H_r A_o}[/latex]

 

[MrAl]

Hi,

I keep forgetting that i am one of the few people who are fluent in standard text equations. I use them all the time because they are the most basic.

But anyway here is a Latex rendition of the point i was trying to make in my original post:
[LATEX]\[\frac{Vout}{Vin}=-\frac{Z2}{Z1}\][/LATEX]

but one of the other points in that post was to avoid using ready made formulas unless we also memorize when they are applicable, and these should probably wait till a future time in the learning process after reaching some level of competence in standard analysis. In other words, standard basic analysis (which works in every case) first, shortcuts second. This is not to put shortcuts down, but just to hold off in the learning process.

I think when as much math as you gave in post #1 is in text form, it's really hard to follow.

With that much math, using Latex is appropriate, and it's not difficult.

For example, instead of: Vout/Vin=Hf*Ao/(1+Hr*Ao )

how about: [latex] \frac{V_{out}}{V_{in}} = \frac {H_f A_o}{1+ H_r A_o}[/latex]

Is that even correct? This is what i mean about applying shortcuts...we have to be very very careful.

I think this is what we were looking for:
[latex] \frac{V_{out}}{V_{in}} = -\frac {H_f A_o}{1+ H_r A_o}[/latex]

where [LATEX] A_o[/LATEX] is the internal gain of the op amp itself expressed as a positive number.

 

[Winterstone]

Hi MrAl,

thank you for your comments in post #22.

I like to repeat that I fully understand your intention - however, I don`t think that my approach in principle differs from yours.
The only difference is that for my approach the calculation is split into several parts that are combined at the end. There is no "shortcut" or a "trick" or something else.

Let me give a very simple example:
To find the voltage delivered by a simple resistive voltage divider everybody uses one single formula
V2=V1*R2/(R1+R2).
Is this a "trick" or a "shortcut"? As you know, it is a formula that results from a combination of two other basic formulas.

The same applies to Black´s feedback formula. It is also the result of basic current-voltage relations, which you are using.

Quote: Some circuits are much more difficult though like the twin T notch or bandpass and these circuits require basic analysis anyway at least to start. So we really need that basic analysis more than we need the shortcuts.

I think, also in case of Twin-T or other active filter circuits you can avoid the time-consuming analysis (starting with basic current-voltage relation). It is much more easy to split the calculation into several steps (like calculation of Hf, Hr and gain) and to combine the various results.

By the way: The general formula applicable to all kinds of feedback is

Vout/Vin=Hf*Ao/(1-Hr*Ao)

with:
Ao: Gain without feedback (always positive) ,
Hf: Factor, which defines the part of the input voltage Vin appearing at the amplifier input port for Vout=0 (Hf positive/negative for non-inverting/inverting operation),
Hr: Factor, which defines the part of the output voltage Vout appearing at the amplifier input port for Vin=0 (Hr positive/negative for positive/negative feedback operation).

For my opinion, using the above feedback formula in addition has the advantage that the user can gain a better understanding of terms "feedback factor" and "loop gain", which are essential for designing circuits with feedback.

 

[MrAl]

Hi MrAl,

thank you for your comments in post #22.

I like to repeat that I fully understand your intention - however, I don`t think that my approach in principle differs from yours.
The only difference is that for my approach the calculation is split into several parts that are combined at the end. There is no "shortcut" or a "trick" or something else.

Let me give a very simple example:
To find the voltage delivered by a simple resistive voltage divider everybody uses one single formula
V2=V1*R2/(R1+R2).
Is this a "trick" or a "shortcut"? As you know, it is a formula that results from a combination of two other basic formulas.

The same applies to Black´s feedback formula. It is also the result of basic current-voltage relations, which you are using.

Quote: Some circuits are much more difficult though like the twin T notch or bandpass and these circuits require basic analysis anyway at least to start. So we really need that basic analysis more than we need the shortcuts.

I think, also in case of Twin-T or other active filter circuits you can avoid the time-consuming analysis (starting with basic current-voltage relation). It is much more easy to split the calculation into several steps (like calculation of Hf, Hr and gain) and to combine the various results.

By the way: The general formula applicable to all kinds of feedback is

Vout/Vin=Hf*Ao/(1-Hr*Ao)

with:
Ao: Gain without feedback (always positive) ,
Hf: Factor, which defines the part of the input voltage Vin appearing at the amplifier input port for Vout=0 (Hf positive/negative for non-inverting/inverting operation),
Hr: Factor, which defines the part of the output voltage Vout appearing at the amplifier input port for Vin=0 (Hr positive/negative for positive/negative feedback operation).

For my opinion, using the above feedback formula in addition has the advantage that the user can gain a better understanding of terms "feedback factor" and "loop gain", which are essential for designing circuits with feedback.

Hello there Winterstone,

Actually it's nice you brought this additional method up here, as i like to refresh too. I find myself forgetting too much as time goes by

But one thing i think you need to do is to rectify your equation and get it right:
Vout/Vin=Hf*Ao/(1-Hr*Ao)

Did you try some real life values in that formula? Try a few and see if they work and i'll check too.

Also trying to remember some of the Latex command syntax

 

[Winterstone]

Hello MrAl,

what is wrong with my equation? Why do you ask for rectification? You can be sure that I tried "some real life values in that formula".
For Ao approaching infinite and negative feedback (Hr=-|Hr|) it simplifies to the classical formula used in practice

Vout/Vin=Hf/|Hr|

and the sign of Hf determines inverting/noninv. operation.
What is the reason for your objections?
W.

 

[The Electrician]

I think, also in case of Twin-T or other active filter circuits you can avoid the time-consuming analysis (starting with basic current-voltage relation). It is much more easy to split the calculation into several steps (like calculation of Hf, Hr and gain) and to combine the various results.

Add a capacitor C in series with the Vin terminal of the circuit shown in the attachment so that the new Vin is the left hand terminal of C.

What are Hf and Hr for the resulting circuit? Is determining them and then applying Black's general formula to get Vout/Vin easier than analyzing the circuit from general principles?

 

[MrAl]

Hello MrAl,

what is wrong with my equation? Why do you ask for rectification? You can be sure that I tried "some real life values in that formula".
For Ao approaching infinite and negative feedback (Hr=-|Hr|) it simplifies to the classical formula used in practice

Vout/Vin=Hf/|Hr|

and the sign of Hf determines inverting/noninv. operation.
What is the reason for your objections?
W.

Hello again Winterstone,

You posted this formula:
Vout/Vin=Hf*Ao/(1-Hr*Ao)

and noted that Ao is positive.

But Hf and Hr are also positive, so we wont see the proper output. That's if i understand your application technique properly, which maybe i dont. But it would seem more correct to write:
Vout/Vin=-Hf*Ao/(1+Hr*Ao)

Is this correct or can you clarify a little more?

 

[Winterstone]

Hello again Winterstone,

You posted this formula:
Vout/Vin=Hf*Ao/(1-Hr*Ao)

and noted that Ao is positive.

But Hf and Hr are also positive, so we wont see the proper output. That's if i understand your application technique properly, which maybe i dont. But it would seem more correct to write:
Vout/Vin=-Hf*Ao/(1+Hr*Ao)

Is this correct or can you clarify a little more?

As mentioned in post#25 the functions (factors) Hf and Hr are NOT always positive.
I repeat:

Ao: Gain without feedback (always positive) ,
Hf: Factor, which defines the part of the input voltage Vin appearing at the amplifier input port for Vout=0 (Hf positive/negative for non-inverting/inverting operation),
Hr: Factor, which defines the part of the output voltage Vout appearing at the amplifier input port for Vin=0 (Hr positive/negative for positive/negative feedback operation).

Thus, the formula works for positive and negative feedback (and a combination of both) as well as for non-inv and inverting operation (and a combination of both like differential amplifier).

 

[Winterstone]

Add a capacitor C in series with the Vin terminal of the circuit shown in the attachment so that the new Vin is the left hand terminal of C.

What are Hf and Hr for the resulting circuit? Is determining them and then applying Black's general formula to get Vout/Vin easier than analyzing the circuit from general principles?

Black`s formula can be applied only for circuits that can be reduced to an amplifier with a single feedback loop.
I never have claimed that all circuits involving feedback (in particular, circuits with nested loops) can be analyzed using Black´s formula.
Of course, I agree with you that your example circuit can and should be analyzed based on "general principles" - however, on a block-by-block basis (using gain expressions for inverting/non-inv. blocks) rather than starting with current-voltage relations.
Regards
W.

 

[MrAl]

Hello again,

Ok so you are saying that we would evaluate the forward path by shorting the output, then evaluate the feedback path by shorting the input, then decide what sign we need to use on the feedback and/or forward path. Sounds reasonable i guess

 

[Winterstone]

Hello again,

Ok so you are saying that we would evaluate the forward path by shorting the output, then evaluate the feedback path by shorting the input, then decide what sign we need to use on the feedback and/or forward path. Sounds reasonable i guess

Hello MrAl,

Yes- exactly as you said. It is nothing else than a specific application of the rules of superposition.