Electronics > Beginners
Add a zero to this RC transfer function
The Electrician:
--- Quote from: bonzer on January 09, 2019, 02:48:00 pm ---I would say adding the resistor (Rc) because in reality they are more precise in value and the overall calculation of the formula is simpler.
Maybe there're other reasons I don't know, what do you think?
--- End quote ---
What formula are you referring to? Frequency response, time response?
bonzer:
frequency response
MrAl:
--- Quote from: bonzer on January 09, 2019, 02:48:00 pm ---Yes it's stable. It's the main reason why I introduced that zero (which in closed loop became a pole). I had to deal with the phase margin and an overshoot. Actually the whole circuit apart from being high pass - its main purpose is being a derivative. In fact I'm working with a real op amp (simulated) so high frequencies are getting attenuated with a -20dB slope after like 4.5 kHz because of the crossing of my theoretical gain with the op amp's open loop gain (which behaves as a low pass) so it actually doesn't let those high frequencies pass (Bode diagram becomes like a triangle).
Anyway BOTH ways work fine (adding the resistor or adding that capacitor like you told me as an alternative) for the compensation.
Both ways give me similar phase margins at the critical point (like 50°) and a low overshoot (like half of dB when compared with an ideal derivative circuit).
So my question is:
Which solution is better in your opinion?
I would say adding the resistor (Rc) because in reality they are more precise in value and the overall calculation of the formula is simpler.
Maybe there're other reasons I don't know, what do you think?
--- End quote ---
Hi again,
Which circuit are you talking about now.
Is it the op amp circuit?
If so, can you show where you intend to insert the resistor and what value it is?
This would give us more to work with to answer that question.
I would prefer to see the op amp circuit with the added resistor because the network without the op amp is not as practical when looking for actual differences.
The Electrician:
--- Quote from: bonzer on January 09, 2019, 02:48:00 pm ---So my question is:
Which solution is better in your opinion?
I would say adding the resistor (Rc) because in reality they are more precise in value and the overall calculation of the formula is simpler.
Maybe there're other reasons I don't know, what do you think?
--- End quote ---
Starting from the circuit you show in reply #17, here are the expressions for the frequency response for 3 cases. The expression for the case where Rc is added is more complicated. I don't know whether the overall calculation of this formula is simpler because I don't do these calculations by hand; I use a modern mathematical software.
MrAl:
--- Quote from: The Electrician on January 11, 2019, 10:20:01 pm ---
--- Quote from: bonzer on January 09, 2019, 02:48:00 pm ---So my question is:
Which solution is better in your opinion?
I would say adding the resistor (Rc) because in reality they are more precise in value and the overall calculation of the formula is simpler.
Maybe there're other reasons I don't know, what do you think?
--- End quote ---
Starting from the circuit you show in reply #17, here are the expressions for the frequency response for 3 cases. The expression for the case where Rc is added is more complicated. I don't know whether the overall calculation of this formula is simpler because I don't do these calculations by hand; I use a modern mathematical software.
--- End quote ---
Hello,
I am not sure what he is doing now. I guess we assume he is adding Rc in series with Cd2?
And i guess when that happens, we take out Cd1?
I am trying to relate the new circuit to the questions, but there seems to be no direct correlation.
Cd3 is in parallel to R3 which doesnt make sense either.
We should have a circuit diagram that explains explicitly which caps get added and what resistors get added, or what gets removed at the same time. We'd have to look at too many variations i think without this information.
Maybe you have a better idea?
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