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

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?

• The math at stompville is simplified. For more information on this circuit topology, please see Mancini, Op Amps For Everyone, section A.3.19 (p439 of 464 in PDF) Dec 6, 2016 at 18:02

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!

• Ah, so is it correct to think of it like this?: The whole of the original signal will pass through (since gain cannot be less than 1 for any frequency going through a non-inverting amplifier (within the amp's internal bandwidth)), and the -3dB cut-off will only apply to the extra amplification that is being given. i.e., it can only affect the R2/R1 part of the gain equation (G = 1 + R2/R1). Is that correct? Jul 7, 2015 at 1:07
• Yes, that's correct. Jul 7, 2015 at 1:19

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.

• Sorry, not what I was asking. And it's not a voltage follower because it isn't unity gain. Jul 7, 2015 at 1:10
• Guess my point wasn't clear, my intent was to get you to think about why the gain of the amp after the filter would have any effect on the cutoff of the filter, regardless of gain.
– RYS
Jul 7, 2015 at 7:41

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.

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} \$$.