| Electronics > Beginners |
| Add a zero to this RC transfer function |
| << < (4/6) > >> |
| Wimberleytech:
--- Quote from: bonzer on January 07, 2019, 08:36:59 pm ---Meanwhile, @Wimberleytech I took a look at Symspice and it seems to be interesting for this stuff but it's not free. Anyway do you know other software like this that generate you the transfer function from the circuit? I know only how to do bode diagram. It would be really helpful to get the actual formula with components when dealing with complex rc networks to verify the result. I'll be back with updates about my circuit in the next days. --- End quote --- They have a free version that is limited to six nodes. That is what I use. Anything beyond six nodes seems too cumbersome to deal with by hand...even with the matrix solved. |
| MrAl:
--- Quote from: Wimberleytech on January 07, 2019, 09:15:11 pm --- --- Quote from: bonzer on January 07, 2019, 08:36:59 pm ---Meanwhile, @Wimberleytech I took a look at Symspice and it seems to be interesting for this stuff but it's not free. Anyway do you know other software like this that generate you the transfer function from the circuit? I know only how to do bode diagram. It would be really helpful to get the actual formula with components when dealing with complex rc networks to verify the result. I'll be back with updates about my circuit in the next days. --- End quote --- They have a free version that is limited to six nodes. That is what I use. Anything beyond six nodes seems too cumbersome to deal with by hand...even with the matrix solved. --- End quote --- Hi, Yes that does look interesting. Only thing i dont like is they want your whole life history before you can download :-) I use a hand written analysis program i wrote myself which is not limited except by sheer time factor to solve the equations. Regular simulators also help here. Lots of people use LT Spice. |
| bonzer:
I'm back. Thank you guys, adding a capacitor in parallel to the first resistor works! I verified it with pspice and it gives me that zero at 1/(RC*2*pi). Getting to the final formula by hand is bloody hard because requires many steps so it's easier to make a mistake but from the bode diagram it seems to give me exactly what I was looking for. I see no other poles introduced by the capacitor even though I can't be sure at 100% as I didn't solve the circuit in theory so there could be one at higher frequency that I didn't see in my limited observation range inside the simulation. Also it seems to work for my compensation. I'm working on a derivative circuit. Anyway if you are curious here's the whole circuit with the capacitor. I still have to analyse if stability is ok about the phase and stuff but it showed very little overshoot so far. But as I said adding a capacitor is a bit more risky so I think that adding a resistor like in the first case would still be better but whatever it's still an alternative. |
| MrAl:
--- Quote from: bonzer on January 07, 2019, 10:42:07 pm ---I'm back. Thank you guys, adding a capacitor in parallel to the first resistor works! I verified it with pspice and it gives me that zero at 1/(RC*2*pi). Getting to the final formula by hand is bloody hard because requires many steps so it's easier to make a mistake but from the bode diagram it seems to give me exactly what I was looking for. I see no other poles introduced by the capacitor even though I can't be sure at 100% as I didn't solve the circuit in theory so there could be one at higher frequency that I didn't see in my limited observation range inside the simulation. Also it seems to work for my compensation. I'm working on a derivative circuit. Anyway if you are curious here's the whole circuit with the capacitor. I still have to analyse if stability is ok about the phase and stuff but it showed very little overshoot so far. But as I said adding a capacitor is a bit more risky so I think that adding a resistor like in the first case would still be better but whatever it's still an alternative. --- End quote --- Hi again, Is that the final circuit there? That looks much different than the original network. We can analyze that too though if you like. The circuit now looks like a high pass filter with cutoff frequency f0=3183.2649 Hz The theoretical time response is (subject to op amp output limitations): f(t)=(4*e^(-(40000*t)/53))/51-(5512*e^(-20000*t))/51 which says it is stable. Since the output is clamped and passband gain is over 100 though that should be investigated also. |
| bonzer:
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? |
| Navigation |
| Message Index |
| Next page |
| Previous page |