# High freq. response of op amp using Miller

I am trying to find the -3dB freq. denoted $f_H$ in the following circuit, where $R$ has been replaced with a capacitor $C$. I tried doing that in two ways which yielded different results. I'd appreciate some feedback on which of the two is correct:

Applying Miller's theorem:

$G_v=(A/sC)/(1/sC + R_{sig}(1-A))$

First method:

$f_H=1/(2\pi R_{sig}C(1-A)) = 1.58KHz$

Second method:

I tried to determine the frequency at which the voltage gain decreases by a factor of $\sqrt2$, hence:

$(A/sC)/(1/sC + R_{sig}(1-A)) = A/\sqrt 2$

Substituting $s=j\omega$, yielded $f_H=4.1KHz$

• Your text talks about an opamp yet your schematic shows an amplifier (not an opamp). Is is unclear what the sign of A is which is quite essential as it changes the functionality of the circuit. – Bimpelrekkie Jun 7 '17 at 14:04
• @Bimpelrekkie Sorry about that.. A is negative. – peripatein Jun 7 '17 at 17:51
• @peripatein 3dB frequency is equal to $f_H= \frac{1}{2\pi R_{sig}C(1+|A|)}$ – G36 Jun 7 '17 at 18:54
• @G36 Great, thanks! How may I now find the frequency of the unity gain? Must I first determine the BW? – peripatein Jun 7 '17 at 19:01
• For the ideal integrator with op-amp the unity gain frequency is equal to $F = \frac{1}{2 \pi R_{sig}C}$. And your gain $A$ is frequency dependent or not ? – G36 Jun 7 '17 at 20:15

To be able to find the unity-gain frequency you can use this trick that will work for this practical circuit.

simulate this circuit – Schematic created using CircuitLab

Hence $V_X = \frac{\frac{1}{s C1 (1+A)}}{R1+\frac{1}{s C1 (1+A)}}$

And because we are interested only in magnitude we can write:

$$V_X = \frac{1}{\sqrt{1+(\frac{F}{F_H})^2}}$$

Where the $F$ - is a signal frequency and $F_H$ - is -3dB frequency.

And finally we have:

$$\frac{1}{\sqrt{1+(\frac{F}{F_H})^2}} * |A| = 1$$

All you need is to solve for $F$

Hence the unity-gain frequency is

$$F_T = \frac{1}{2 \pi R_1 C_1} * \frac{\sqrt{A^2 -1}}{|A|+1}$$

Of-course the unity-gain frequency can also be find from transfer function directly.

• how may I arrive at the same result using the transfer function? – peripatein Jun 10 '17 at 14:15
• First you need do find the transfer function – G36 Jun 10 '17 at 14:36