question about conjugate complex poles in miller opamp

[sunbeam]

hi, I have a question about conjugate complex poles in opamp.
Thank you all in advance.

when i adjust the size of miller cap in two-stage opamp, in some regions i get conjugate complex poles,which means the dominant and second pole turn into
conjugate complex pairs.
from the view of transfer function,it is a two order polynomial and it is possible in
the situation.
is there any intuitive ways to realize it?
Directly from the circuit, could we realize that we get a conjugate complex pair in an intuitive way?

Thank you!

 

[LvW]

As your opamp has - as you have mentioned - a second order transfer function it can be viewed as a second order low pass filter with a second order denumerator N(s)

N(s)=1+a1*s+a2*s²

The zeros of N(s) are the poles of the transfer function.
By writing this solution you will see, that for rather low values of a1 the roots become conjugate complex. Thus there is a pole pair in the s-plane with a common negative real part and different signs for both imaginary parts. This is identical to a low pass filter with a rather good selectivity (pole Q>o.5)

If then a1 (damping factor) increases, the imag. part decreases and the poles move to the real axis, meet there (pole Q=0.5) , and for further a1 increase you have to real and different poles on the neg. real axis (pole Q not defined, poor low pass selectivity). The BODE diagram reflects this poor selectivity caused by two real poles with a slope of -20dB/dec in the lower and app. -40 dB/dec. in the higher frequency region.
Hope this could help a bit.

 

[sunbeam]

Hi, LvW, thank you for your help. I got it.

 

[eagleliujun]

got it!

 

[ddrr]

As your opamp has - as you have mentioned - a second order transfer function it can be viewed as a second order low pass filter with a second order denumerator N(s)

N(s)=1+a1*s+a2*s²

The zeros of N(s) are the poles of the transfer function.
By writing this solution you will see, that for rather low values of a1 the roots become conjugate complex. Thus there is a pole pair in the s-plane with a common negative real part and different signs for both imaginary parts. This is identical to a low pass filter with a rather good selectivity (pole Q>o.5)

If then a1 (damping factor) increases, the imag. part decreases and the poles move to the real axis, meet there (pole Q=0.5) , and for further a1 increase you have to real and different poles on the neg. real axis (pole Q not defined, poor low pass selectivity). The BODE diagram reflects this poor selectivity caused by two real poles with a slope of -20dB/dec in the lower and app. -40 dB/dec. in the higher frequency region.
Hope this could help a bit.

That's a clear picture!

 

[amriths04]

For any physically real system, the poles or zeros will always occur in conjugates. Only then will the transfer function have real coefficients.

 

[LvW]

For any physically real system, the poles or zeros will always occur in conjugates. ....

...if they are complex! Otherwise they may be, of course, real.