Transfer Function of differentiator

[sudarshann]

Hi everyone,
I have a small problem in deriving the transfer function of the below attached circuit which is opamp differentiator.
I have been trying this for past 3 days still no good results. can anyone guide how can i derive this transfer function

Thank you very much in advance.

 

[LvW]

Hi sudarshann,

start with the basic equation

H(s)=Vout/Vin=Hf*A/(1-Hr*A)=Hf/(1/A-Hr) with A=open-loop gain, Hf=forward function and Hr=return function (feedback factor).

Then calculate
Hf=R1/(R1+R2)-(R1||X1)/(R2 + R1||X1) with X1=1/sC1

Hr=- (R2||X1)/(R1 + R2||X1).

After inserting for Hf and Hr and some (involved) manipulations you will arrive at the final formula.
Recommendation: From the beginning you should set A>>infinity and use instead: H(s)=-Hf/Hr.

 

[sudarshann]

Hi LvW,
Thank you very much for your answer i will trying solving it and let you know the result ASAP.

 

[LvW]

Hi LvW,
Thank you very much for your answer i will trying solving it and let you know the result ASAP.

Here comes an alternative solution:

Calculate separately the opamp input voltages Vp and Vn.
Because Vn is generated by two voltage sources (Vin and Vout) you have to use the superposition theorem.
Then:
Vp=Vn for opamp ideal (A infinite) or Vout=A*(Vp-Vn) for finite A.

However, I don't know which alternative is more convenient to you.

 

[sudarshann]

Thanks for the above Solution 1 it helped a lot.

I have this one more problem.
i need to find the Gain Bandwidth Product from the magnitude and/or frequency plots . i really dont understand how.
this plot i got from the simulation of the above circuit with B2spice .

please help me how can i do find Gain Bandwidth Product (GB)





Thank you

 

[lhlbluesky]

hi, sudarshann. how do you get the above simulation result using the above circuit? the circuit is a differentiator, but the magnitude and/or frequency plots seems like a BPF, am i right? and in the transfer function, what is the value of 'A'?

 

[sudarshann]

Hi lhlbluesky,
I got the simulation done in b2spice. i constructed the circuit with TL081 (which is 100% similar in specs to TL082)
yes the circuit is a Non Inverting Differentiator. the B2Spice had test called 'AC Sweep' in which i set the frequency in range of 1 kHz to 1 mHz and i got this plot.
i do have the sample waveform for the plot and the output i got resembles it most of the cases. i have attached it below


In the Transfer Function A is the Open loop Gain and its equal to GB/s where GB is Gain Bandwidth Product

Hope i have given you all the details

 

[lhlbluesky]

ok, sudarshann. in your opinion, A=GB/s, if so, i understand it well.

 

[sudarshann]

exactly A=GB/s

 

[lhlbluesky]

but i have another doubt, why A=GB/s? in low frequency, A is almost constant, above p1(dominant pole), it is about GB/s. right?

 

[LvW]

In the Transfer Function A is the Open loop Gain and its equal to GB/s where GB is Gain Bandwidth Product
Hope i have given you all the details

No, you didn't.
In your posting#5 you have shown us a simulation result - but you didn`t give us the most important information: Which circuit configuration?
But - in this case, I could imagine (also from the magnitude response) that you have simulated the complete circuit as shown in your first posting.
But for future questions: Present simulation results in relation to the corresponding circuit only.

Now to your question: The magnitude and phase responses are OK. Up to approx. 10 kHz the circuit resembles the typical differentiator behaviour.
For larger frequencies the open loop opamp gain cannot be considered as ideal and causes deviations from the desired response.
Starting with approx. 15 kHz the loop gain approaches unity and the magnitude response is nearly identical to the open-loop response A(s) of the opamp.
Thus, the wanted GBW (transis frequency) is identical to the frequency where the gain is 0 dB. Either you perform simulations up to 10 MHz (by the way: m is milli and M is mega)
or you try to continue the gain slope up to the 0 dB crossing point (which will be approx. at 10 MHz).

- - - Updated - - -

but i have another doubt, why A=GB/s? in low frequency, A is almost constant, above p1(dominant pole), it is about GB/s. right?

Yes, of course. The approximation A=GBW/s can be used only for frequencies far above the first pole frequency (and is valid only for unity-gain compensated opamps with one single pole in the active region).

 

[sudarshann]

Thanks for the helping me out

just tell me whether i have got it clear or not

GB is the point where the magnitude plot cuts o dB line to get the cutting point i must simulate till the 10kHz to 10 mHz

also i found that GB for TL082 under typical conditions is 3 mHz from its datasheet


when i simulate the circuit i get around 14.800 mHz is that right or i have done something wrong

Thank you

 

[LvW]

also i found that GB for TL082 under typical conditions is 3 mHz from its datasheet
when i simulate the circuit i get around 14.800 mHz is that right or i have done something wrong

Correct - the transit frequency (GBW) of the TL082 is between 2 and 4MHz. (I repeat: M=mega, not m !!).
Thus, the gain magnitude response cannot cross the 0 db line at 15 MHz. Something is wrong.
Question: Did you display the gain=output/input? It is recommended to use 1 volt input signal, in this case the output is identical to the gain value.

 

[sudarshann]

Hi
i have used TINA now i got the answer correct.
i got 2 MHZ



Thank you very much LvW