# Find resistors for an inverting op amp band pass filter

I'm trying to save a problem that shows the following band pass filter:

Note: the op amp doesn't need to be a TL081, it's an abstract, ideal op amp.

I need to find resistors $$\R1\$$ and $$\R2\$$ so that the cutoff frequencies of such filter are approx. 10 Hz and 20 kHz. Also, the peak gain should be achieved at around 1 kHz.

I have only 2 possible capacitor values: 1.5uF and 220pF, so if C1 is set to to 1.5uF then C2 must be 220pF and vice versa.

What I have done so far is write the gain of the circuit:

$$Z_{R_1C_1} = \frac{1}{\frac{1}{R1} + j\omega CR_1}$$

$$Z_{R_2C_2} = \frac{1}{jwC_2} + R_2$$

The inverting op. amp is:

$$v_{OUT} = v_{IN} \frac{-Z_{R_1C_1}}{Z_{R_2C_2}} = v_{IN} \frac{\frac{1}{\frac{1}{R1} + j\omega CR_1}}{\frac{1}{jwC_2} + R_2}$$

Simplifying the above:

$$\frac{v_{OUT}}{v_{IN}} = \frac{jR_1\omega C_2}{(1 + j\omega C_1R_1) (1 + j\omega R_2C_2)}$$

From here I get a bit lost.

I have tried to do $$\C_1=1.6\mu F\$$ and $$\C_2 = 220pF\$$, and find $$\R_1\$$ and $$\R_2\$$. However, in the process, I get a system of non-linear equations, so I strongly suspect I'm not doing it as expected. I also have spent quite a lot of time guessing different values and plotting the magnitude vs frequency chart, but haven't got anywhere.

Looking at the equation above, it seems to me it could be expressed as a product, so that would be combining 2 non-resonant, simpler filters. Is this correct? I'm not very sure how to do this.

What can I do?

I have tried the solution suggested below. It works as shown below:

• Grey: 0.707 * peak_gain line
• Green: 10Hz and 20kHz lines
• Yellow: 1kHz line (the max. gain should should be arround here
• Red: $$\\omega _0\$$ line: $$\\frac{1}{\sqrt{R_4 C_4 R_3 C_3}}\$$

Here is a link to Sage to experiment with this:

Sage

# EDIT 2 - Still have questions

I understand that the cutoff frequency of an RC filter (either low or high pass) is $$\\frac{1}{2\pi RC}\$$.

However, the transfer function of my problem is different from the classic RC filter. How can one be sure that the above is the actual cutoff frequency formula?

I mean, my transfer function can be divided in 2 filters, one high pass and one low pass. However, the numerator is very different from a normal RC filter. It also clearly affects the gain. How can it be known that the formula above still works?

• You have an extra R1 in the 1st equation, but the error didn't seem to propagate to the bottom. Your Vout/Vin equation seems to be correct; you need to learn how to interpret it. Or, learn the shortcuts as user350400 described. Commented Nov 30, 2023 at 0:25
• You said in the question that the low frequency 0.707 point is 1 Hz not 10 Hz.
– user350400
Commented Nov 30, 2023 at 10:53
• ..... If the lower cut-off frequency should in fact be 10 Hz then R2 becomes 10.6k and the mid-band gain becomes 3.4.
– user350400
Commented Nov 30, 2023 at 11:16
• Note that your passband gain is fixed by your capacitor value constraints. Commented Nov 30, 2023 at 13:26

The input pair of components create a high pass filter, the feedback pair of components create a low-pass filter.

Using f(-3dB)=1/(2 * pi * R * C) for each pair of components (input pair and feedback pair)

Transposing....

R2 = = 1/(2 * pi * 1Hz * 1.5uF) = 106.1k

EDIT IN RESPONSE TO OP'S "STILL HAVE QUESTIONS"

You could derive values for R1 & R2 via your transfer function technique if you wanted, to give you more confidence in the values already obtained, if the simulation is not enough evidence for you.

Your transfer function is correct (even though there is an error in the derivation which disappears).

Now you have the transfer function you need to derive an expression for the magnitude of the transfer function in which all the js have disappeared. This magnitude expression should then be equated to 0.707(R1/R2). The problem now comes that you have one equation with two unknowns in it, R1 & R2. So you need to generate another equation and solve it for R2.

The two input components form a high pass filter so, you can write down the transfer function for this high pass filter, jwRC/(1 + jwRC), find the magnitude expression from this transfer function, equate it to 0.707 and then solve for R2.

Now you have a value for R2 you can plug it into the equation which you obtained earlier (magnitude expression of the amplifier's transfer function equated to 0.707(R1/R2)) and solve for R1.

I haven't actually done this but the technique outlined above is the way I would attempt to go about it.

• I have tested this; It works to make the gain of 10Hz and 20kHz below 0.707, but it does the same with the other frequencies. I will edit my answer to ask more questions about this.
– Dan
Commented Nov 30, 2023 at 10:43
• @Dan The cut-off frequencies are 0.707 of the mid-band gain. The mid-band gain is set by the ratio of the two resistors over a frequency band where C2 can be treated as a short and C1 can be treated as an open.
– user350400
Commented Nov 30, 2023 at 11:05
• with R2 = 10.6k, the following gains are obtained: gain(10Hz) = 2.41, gain(1kHz) = 3.38, gain(20kHz) = 2.41. Indeed, the max. gain (3.38) times 0.707 gives 2.39. Both cutoff frequencies are below this value. But then, the cutoff frequency is not plain 0.707, but 0.707 times the maximum gain? (a.k.a. the gain at w0) This would explain a lot of the problems I'm having...
– Dan
Commented Nov 30, 2023 at 11:27
• @Dan Yes, a cut-off is usually defined as the frequency where the gain has dropped to 0.707 of its mid-band value. It is usually referred to as the "-3 dB point" or "-3 dB frequency" because the gain has dropped 3 dB from its mid-band value. 20log(0.707/1) = approximately -3 dB.
– user350400
Commented Nov 30, 2023 at 11:42

Well, the transfer function of your circuit is given by:

$$\mathscr{H}\left(\text{s}\right):=\frac{\displaystyle\text{V}_\text{o}\left(\text{s}\right)}{\displaystyle\text{V}_\text{i}\left(\text{s}\right)}=-\frac{\displaystyle\text{R}_1\space\text{||}\space\frac{\displaystyle1}{\displaystyle\text{sC}_1}}{\displaystyle\text{R}_2+\frac{\displaystyle1}{\displaystyle\text{sC}_2}}=-\frac{\displaystyle\text{C}_2\text{R}_1\text{s}}{\displaystyle\left(1+\text{C}_1\text{R}_1\text{s}\right)\left(1+\text{C}_2\text{R}_2\text{s}\right)}\tag1$$

Where $$\\displaystyle\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$$.

So, when $$\\displaystyle\text{s}:=\text{j}\omega\$$ we get:

$$\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\displaystyle\text{C}_2\text{R}_1\omega}{\displaystyle\sqrt{1+\left(\text{C}_1\text{R}_1\omega\right)^2}\cdot\sqrt{1+\left(\text{C}_2\text{R}_2\omega\right)^2}}\tag2$$

Now, we can solve for the maximum:

$$\frac{\displaystyle\partial\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|}{\displaystyle\partial\omega}=0\space\Longleftrightarrow\space\omega:=\hat{\omega}=\frac{\displaystyle1}{\displaystyle\sqrt{\text{C}_1\text{C}_2\text{R}_1\text{R}_2}}\tag3$$

So, we get:

$$\left|\space\underline{\mathscr{H}}\left(\text{j}\hat{\omega}\right)\right|=\frac{\displaystyle\text{C}_2\text{R}_1}{\displaystyle\text{C}_1\text{R}_1+\text{C}_2\text{R}_2}\tag4$$

Now, we can solve for the cut-off frequencies:

$$\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\displaystyle\left|\space\underline{\mathscr{H}}\left(\text{j}\hat{\omega}\right)\right|}{\displaystyle\sqrt{2}}\space\Longleftrightarrow\space\omega:=\omega_\pm=\dots\tag5$$

Which when solved, gives:

$$\omega_\pm=\frac{\displaystyle2}{\displaystyle\sqrt{\left(\text{C}_1\text{R}_1\right)^2+\left(\text{C}_2\text{R}_2\right)^2+6\text{C}_1\text{C}_2\text{R}_1\text{R}_2}\pm\left(\text{C}_1\text{R}_1+\text{C}_2\text{R}_2\right)}\tag6$$

EDIT, so when we set $$\\displaystyle\hat{\omega}\$$ we can see that:

$$\hat{\omega}=\frac{\displaystyle1}{\displaystyle\sqrt{\text{C}_1\text{C}_2\text{R}_1\text{R}_2}}\space\Longleftrightarrow\space\text{C}_1\text{C}_2\text{R}_1\text{R}_2=\frac{\displaystyle1}{\displaystyle\hat{\omega}^2}\space\Longleftrightarrow\space\text{C}_1\text{R}_1=\frac{\displaystyle1}{\displaystyle\text{C}_2\text{R}_2\hat{\omega}^2}\tag7$$

So, we can rewrite $$\\displaystyle(6)\$$:

$$\omega_\pm=\frac{\displaystyle2}{\displaystyle\sqrt{\left(\frac{\displaystyle1}{\displaystyle\text{C}_2\text{R}_2\hat{\omega}^2}\right)^2+\left(\text{C}_2\text{R}_2\right)^2+\frac{\displaystyle6}{\displaystyle\hat{\omega}^2}}\pm\left(\frac{\displaystyle1}{\displaystyle\text{C}_2\text{R}_2\hat{\omega}^2}+\text{C}_2\text{R}_2\right)}\tag8$$

Now, when using your values, $$\\displaystyle\hat{\omega}=2000\pi\space\text{rad/sec}\$$ and $$\\displaystyle\omega_+=20\pi\space\text{rad/sec}\$$, we can solve for $$\\displaystyle\text{C}_2\text{R}_2\$$:

$$\text{C}_2\text{R}_2=\frac{\displaystyle9999\pm\sqrt{99940001}}{\displaystyle400000\pi}\tag9$$

$$\omega_-=200000\pi\space\text{rad/sec}=100\space\text{kHz}\tag{10}$$

Instead of the wanted $$\\displaystyle20\space\text{kHz}\$$.

And, when we turn it around. When using your values, $$\\displaystyle\hat{\omega}=2000\pi\space\text{rad/sec}\$$ and $$\\displaystyle\omega_-=40000\pi\space\text{rad/sec}\$$, we can solve for $$\\displaystyle\text{C}_2\text{R}_2\$$:

$$\text{C}_2\text{R}_2=\frac{\displaystyle399\pm\sqrt{157601}}{\displaystyle80000\pi}\tag{11}$$

$$\omega_+=100\pi\space\text{rad/sec}=50\space\text{Hz}\tag{12}$$

Instead of the wanted $$\\displaystyle10\space\text{Hz}\$$.