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I am trying to do some calculation for my post lab and i dont understand why the lab manual only talks about open loop phase response. are these the same?

I know the gain for open and closed loop differs quit a bit.

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  • \$\begingroup\$ This is a very general question, so I can only answer generally. The closed-loop response is what you get once you've attached a feedback network. The open-loop response is what you get without any feedback at all. For a more specific answer, give more details about the feedback network. \$\endgroup\$ May 25, 2015 at 7:13
  • \$\begingroup\$ Depends,how you use your OPAMP IC in the feedback.If you create a differentiators circuit then is just like having the s,If you create a integrator circuit with capacitor in the feedback then its just like having a 1/s,where s is Laplace parameter a complex number frequency,the exact place where the phase response will cut the axis can only be determined by the transfer function which will depend on the value of the resistances.I hope I was of help \$\endgroup\$
    – MaMba
    May 25, 2015 at 8:28

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The closed loop response of your op-amp completely depends on the circuit you're putting your op-amp in. So, there is not 'one' closed-loop response of your op-amp.

The open-loop response describes the response from the op-amp inputs to its outputs. This is a useful measure, because this is completely independent of the other components in your circuit - you're only measuring the op-amp itself.

I come from a mechanical engineering background, so op-amps aren't my area of expertise, but consider a ideal op-amp in buffer configuration (gain from input to output: 1) or a simple inverting amplifier (gain from input to output: -R2/R1). So here, the closed loop responses are completely different even though we have an ideal op-amp! So, it all depends on how exactly we 'close the loop'.

If you're unsure how to calculate the closed-loop response yourself, I think that's best asked in another question. I'll give one hint: in typical system block diagrams, the 'transfer function' of a block is exactly the open loop response of that block.

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Don't get me wrong,because it ain't usually easy to calculate the phase response of a closed -loop system,also there is a correction need when doing a second order or higher frequency plot of a closed loop system.Hence to make life simple we just use the Open -loop systems.


As you can see tranfer function :
H(s)=Y(s)/X(s)
is completely a related term in bode plot you calculate the decibel fall wrt to frequency in frequency response. Hence for a closed loop equation the number coefficients increase the the bode plot gets more crazier specially the phase response.

Depends,how you use your OPAMP IC in the feedback.If you create a differentiators circuit then is just like having the s,If you create a integrator circuit with capacitor in the feedback then its just like having a 1/s,where s is Laplace parameter a complex number frequency.

The exact place where the phase response will cross the axis can only be determined by the transfer function which will depend on the value of the resistances.I hope I was of help

But to visualize a simple integrator ckt. is given by a 1/s in the Zeroes and poles and the s is given as a differentiators ckt using the opamp.

While drawing the Open loop circuit you wont have any poles.

You can visualize the whole circuit using the Direct Form I and Direct Form II structures,and replace the 1/s by the integrators and s by the differentiators using the Opamp IC.

As for the frequency the complex number (s= σ + į ω) the complex frequency will depend upon your reactive components like capacitor and inductor,yes inductor can be used but they are always avoided.

Where σ is basically a attenuation constant and can be determined by solving the transfer function of the given circuit or by just solving the differential equation obtained using Ordinary Differential Equations.

Well the method using the Kirchhoff is not preferred as they are in time domain where all the calculations are hard,so we usually convert it into the frequency domain.

Using the bode plot you have to first calculate the transient response of the circuit,by using the circuit's transfer function.

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