Calculating Fc (cutoff) of series to RC to ground on inverting input of non-inverting amplifier

Question

Regarding the chosen answer for: How does this OP-AMP non-inverting amplifier work?

The following circuit is given, and it says that $$F_c = \frac{1}{2\pi R_1C_1}$$

^{simulate this circuit – Schematic created using CircuitLab}

However, this doesn't seem right to me, even though I have found the same example and formula elsewhere on the web (e.g., "Non-inverting amplifier – alternative" at http://stompville.co.uk/?p=470).

To me it doesn't take into account the value of the op amp feedback resistor (\$R_2\$ in the sketch above).

Would the cut-off frequency not be based upon (\$R_1+R_2\$)? If not, why not?

My point is that the only way the capacitor can charge or discharge is through the output of the op-amp, so surely the feedback resistor is also critical?

Answer 1

The cutoff frequency is the frequency at which the gain has fallen by -3dB over the gain at high frequencies.

(for the purposes of the below discussion, we'll ignore the high pass network R and C since the effects of R1/C1 are being examined)

If the signal was applied to C1 (rather than C1 being grounded) R2 would not matter at all- gain would be -R2/R1 for high frequencies. When the reactance of the C1 equals the resistance of R1 in magnitude you have a reduction in voltage of 1/sqrt(2) which is -3dB. R2 will determine the gain, but not the cutoff frequency.

However it's a bit more complex in this case, because the op-amp is being fed from the non-inverting input, and the gain is actually 1 + R2/R1 at high frequencies, so the frequency at which the gain is reduced by 1/sqrt(2) will actually depend on R2.

In an extreme case, if you short R2, it becomes a voltage follower and then Fc is limited only by the amplifier!

Answer 2

This sort of depends on the Gain bandwidth of the op-amp, but if we're assuming an ideal amp, you can think about it this circuit as a simple voltage follower at the output of a passive RC filter. Thus, the cutoff of the circuit is simply the cutoff of the RC filter.

At much higher frequencies (MHz) in a real-world example, the op-amp will start to attenuate the output.

Answer 3

Based on the definition of the 3dB cut-off you can calculate the corresponding frequency:

wc1=K/R1C1 with K=SQRT[1-(2/Amax²)] and Amax=(1+R2/R1)

That means: wc1~1/R1C1 (with K~1) for Amax>5.

More than that, in case the C-R highpass at the non-inv. input has to be also taken into account, for the total cut-off you can use the approximation

wc=SQRT(wc1²+wc2²) with wc2=1/RC.

Answer 4

In the op-amp circuit you refer to, \$ C_1 \$ (there it's named \$ C_2 \$) is merely a de-coupling capacitor and is not intended for filtering.

I have never seen this circuit so I'd like to just calculate the transfer function and see where it brings us.

So, let's neglect \$ R \$ and \$ C \$ and assume the Op-Amp ideal, i.e. the voltage on the non-inverting pin equals that on the inverting pin and its input impedance is infinitely high.
Then in the frequency domain

$$ I_{R_2} = I_{R_1} = I_{C_1} = V_i/Z_1 = \frac{V_i}{R_1 + 1/j \omega C_1} = \frac{j \omega C_1}{1+ j \omega R_1 C_1} V_i = \frac{1}{R_1} \frac{j \omega R_1 C_1}{1 + j \omega R_1 C_1} V_i = \frac{1}{R_1} \frac{j \omega/\omega_1}{1 + j \omega/\omega_1} V_i $$

with \$ \omega_1 = \frac{1}{R_1 C_1} \$. Now we continue by writing the output voltage

$$ V_o = V_i + R_2 I_{R_2} = V_i + \frac{R_2}{R_1} \frac{j \omega/\omega_1}{1 + j \omega/\omega_1} V_i = \left[1 + k \frac{j \omega/\omega_1}{1 + j \omega/\omega_1} \right] V_i $$

with \$ k = \frac{R_2}{R_1} \$, and the transfer function is

$$ H(j \omega) = \frac{V_o}{V_i} = 1 + k \frac{j \omega/\omega_1}{1 + j \omega/\omega_1} = \frac{1 + j \omega/\omega_1 + k j \omega/\omega_1}{1 + j \omega/\omega_1} = $$

which for \$\omega = 0 \$ has a gain

$$ |H(0)| = 1$$

for \$ \omega = \infty \$

$$ |H(j \infty) = 1 + k $$

and for \$ \omega = \omega_1 \$.

$$ |H(j \omega_1)| = \left| 1 + k \frac{j}{1+j} \right| = \left| 1 + k \frac{j}{1+j} \frac{1-j}{1-j} \right| = \left| 1 + k \frac{1+j}{2} \right| = \left| 1+k/2 + jk/2 \right| = \frac{2}{k} \left| (1+2/k) + j \right| = \frac{2}{k} \sqrt ( 1 + (1 + 2/k)^2 ) = \frac{1}{\sqrt2 k} \sqrt(1 + 2/k + 2/k^2) $$ $$ = \sqrt(k^2/2 + k + 1) $$

I don't think the circuit does what it is assumed to do and the only way the cut-off frequency (where \$ \left| H(j \omega) \right| = 1/\sqrt2 \$) would be when \$ \frac{1}{k} \sqrt(1 + 2/k + 2/k^2) = 1 \$ or \$ k^2 + 2 k + 2 = (k+1)^2 + 1 = 1\$ and that would require \$ k = -1 \$, which is impossible with this circuit.

Edit:

After reading the link you gave, it appears there is a misunderstanding. It's about the non-ideal op-amp with bias current where \$ R \$ in your drawing should match \$ R_1||R_2 \$. If you adapt R to match this, you'll change the input decoupling circuit \$ R C \$ into a low-pass filter with the cut-off frequency \$ f = \frac{1}{2 \pi R C} \$, not \$ \frac{1}{2 \pi R_1 C_1} \$.