Bode plot of Opamp integrator

[zzlong94]

If the Rf (feedback resistor) value small, the phase plot of the integrator will start at a higher degree.
Or vise versa.
Any equation or concept to prove that the smaller the value of Rf, the bigger the value of the starting phase?

Picture below : Rf =100k

[Jay_Diddy_B]

Hi,
If the Integrating capacitor is very small, then you are left with an inverting amplifier with a gain of R2/R1.

This is written gain = -R2/R1

since it is inverting the phase shift is 180 degrees.

A frequency of 1/(2 x pi x R2 x C) the phase shift will be -135 degrees

At higher frequencies the phase shift is 90 degrees.

It will remain at 90 degrees until the frequency response of the op-amp limits the performance.

Jay_Diddy_B

[zzlong94]

clear n good explaination! thx a lot bro!

[zzlong94]

one more thing bro.
if i wn to calculate the phase shift
the equation is 1/(2x pi x r2 x c)
the value i get is 1.519
how i get 135 degree?

[Jay_Diddy_B]

one more thing bro.
if i wn to calculate the phase shift
the equation is 1/(2x pi x r2 x c)
the value i get is 1.519
how i get 135 degree?

1.59 is the Frequency.

Jay_Diddy_B

[zzlong94]

based on the frequency tat i get, how cn i get the phase shift value of 135 degree?

[rstofer]

http://www.analog.com/en/analog-dialogue/articles/phase-response-in-active-filters-2.html

The first section talks about an op amp low pass filter.  The phase shift equation is right after Figure 1.  However, the inverting amplifier starts with 180 degrees of phase shift so when you get to the center frequency, you have 180 degrees minus 45 degrees or 135 degrees of phase shift  Later in the article they talk about the non-inverting case.

[Jay_Diddy_B]

Hi,

So the phase response is:


output phase = 180 -tan^-1(2 x pi x F x R2 x C1)

At F=0.159 Hz

Output phase = 174 ^o

At F=1.59 Hz

Output phase = 135^o

At F= 15.9Hz

Output phase = 95.7^o

This circuit is not an integrator, it is an inverting single pole low pass filter.

Jay_Diddy_B

[rstofer]

based on the frequency tat i get, how cn i get the phase shift value of 135 degree?

Well, as shown above, somewhere around 1.59 Hz should have the 135 degree phase shift.  I didn't do the arithmetic, that's up to you.

That is such a slow frequency you can watch paint dry between cycles.  I would change the component values to move the corner frequency up to 1 kHz or so.

[MrAl]

If the Rf (feedback resistor) value small, the phase plot of the integrator will start at a higher degree.
Or vise versa.
Any equation or concept to prove that the smaller the value of Rf, the bigger the value of the starting phase?

Picture below : Rf =100k

Hi,

The time response of this circuit is:
Vout/Vin=-(w*C1*R2^2*e^(-t/(C1*R2)))/(w^2*C1^2*R1*R2^2+R1)+(w*cos(t*w)*C1*R2^2)/(w^2*C1^2*R1*R2^2+R1)-(sin(t*w)*R2)/(w^2*C1^2*R1*R2^2+R1)

where w=2*pi*f.

If we set t=0 and do not set any other values, we get:
Vout/Vin=0

That means for an ideal op amp under ideal conditions we get zero out at t=0 always, no matter what the values of the components are.  This means that the output will always start at zero unless there are parasitics that cause something else to happen.

Once the exponential part has died down, the phase becomes:
Ph=atan2((w*C1*R1)/(w^2*C1^2*R1^2+1),-1/(w^2*C1^2*R1^2+1))

and that is when the phase really becomes relevant.

If there is some reason why you are looking at this that might help better explain what you need here.