Question - Finding Feedback loopgain while refernce voltage is applied

Thread Starter

yaron02

Joined Jan 29, 2022
4
Hello everyone!
I would like to know, Does this reference voltage (vcc/2) effects the value of beta ?
beta = (ve/vo while vin=0)
thank you !

1657611560118.png 1657611569364.png
 
Last edited by a moderator:

MrAl

Joined Jun 17, 2014
9,549
Your opamp is missing a very important resistor:
I am guessing that this is a highly theoretical question where practical things like that are often left out.
If you think about what that extra resistor actually does to the output response, it impairs the circuit from being an ideal differentiator. Although this is necessary in a practical circuit, in a theoretical one it just makes the analysis more difficult and also distracts from the main idea of the circuit.

To look at it theoretically, a finite step input voltage of even 1 volt applied to an ideal differentiator would produce an infinite spike output. With a practical circuit with no extra resistor the output would simply pin to one of the supply rails provided the max input voltage spec was not exceeded. With the extra resistor the output will ramp up fast but not as fast as with no resistor so it begins to look a little less like an ideal differentiator. That's usually necessary similar to how a resistor in parallel to the capacitor in an integrator circuit is often necessary in a practical circuit but just messes the theory up in a completely theoretical analysis.

With R1 the feedback resistor and R2 the extra resistor, with R2=0 we have:
Vout=Vin*s*R1*C
which is an idea differentiator.

With R2 included, it changes to:
Vout=Vin*(s*C*R1)/(s*C*R1+1)
and we can see now that we have the original differentiator along with a low pass filter convolved with the differentiator.

We can easily analyze both cases, but the first is more straightforward for a study in theory so we will see things like this a lot in academic works.
 
Last edited:

MrAl

Joined Jun 17, 2014
9,549
Maybe ve translates to "V input" and beta translates to "gain"?

Isn't the differentiator a highpass filter?
Hi,

Strictly speaking, the transfer function for an ideal differentiator is:
s*R*C

while for an ideal high pass filter it is:
(s*C*R)/(s*C*R+1)

So there is a difference, and that difference is in the shape of the amplitude vs frequency plot of the output. The second is a bounded curved, the first is a straight line with no bounds as frequency rises.
 

LvW

Joined Jun 13, 2013
1,561
Hi,

Strictly speaking, the transfer function for an ideal differentiator is:
s*R*C
Yes - that`s correct.
But it is to be noted that in reality this circuit (without the stabilizing resistor) will not work.
Due to the real opamp gain function the circuit will oscillate.
 

MrAl

Joined Jun 17, 2014
9,549
Yes - that`s correct.
But it is to be noted that in reality this circuit (without the stabilizing resistor) will not work.
Due to the real opamp gain function the circuit will oscillate.
Hi,

Yeah sure, that's why i always try to remember to specify 'ideal' or 'theoretical' when i speak of those kinds of functions.
Not sure about the oscillation, but there are a lot of other things to consider too but in the ideal case meant for study we dont have to include those extras that would come in a lab section of the course.
 
Top