PSpice and calculations don't add up for an active band pass filter

Question

I designed a basic active band pass filter and used these equations to come up with a center frequency of 29.9 kHz:

$$f_0 = \frac{1}{2\pi\sqrt{(R_1||R_2)*R_2*R_3*C_1*C_2}}$$

In PSpice, it looks like the center frequency is almost 36 kHz.

Did I use the wrong equations to calculate the center frequency?

Per @Sphero's recommendation, this is what the waveform looks like:

Answer 1

Here is the result if you follow my comment above:

I'm not going to check your calculations (at least right now) but maybe try simulating it with a better op-amp model and higher supply voltages. 30kHz is pretty high for a 741 (look at all that required gain!) and +/-5V is right at the minimum. Something like an AD8034 off +/-10V. And look at the output in a transient analysis at the center of the passband before trusting the AC analysis.

Pretty much bang-on.

Answer 2

However, in Pspice it looks like the center frequency is almost 36 KHz. Did I use the wrong equations to calculate the center frequency?

Your calculations assume an ideal op amp. In practice the center frequency should be lower due to the LM741's limited bandwidth.

However in your case it's higher because you have the inputs swapped around. In AC analysis this 'works' because Pspice calculates frequency response based on the initial operating point and doesn't consider transient effects. A transient analysis will show the operating point rapidly changing until the output hits a supply rail, due to the positive feedback through R3.

Answer 3

Try doing a transient analysis. You'll probably find your output slammed in to one of the rails since the opamp inputs are flipped (positive input goes to ground).

Answer 4

First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).

Well, we are trying to analyze the following opamp-circuit:

^{simulate this circuit – Schematic created using CircuitLab}

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\text{I}_2+\text{I}_3+\text{I}_4\\ \\ 0=\text{I}_3+\text{I}_4+\text{I}_5 \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_1-\text{V}_3}{\text{R}_4}\\ \\ \text{I}_3=\frac{\text{V}_2-\text{V}_3}{\text{R}_5} \end{cases}\tag2 $$

Now, we can put equations \$(1)\$ into \$(2)\$ to get:

$$ \begin{cases} \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_1-\text{V}_2}{\text{R}_3}+\frac{\text{V}_1-\text{V}_3}{\text{R}_4}\\ \\ 0=\frac{\text{V}_1-\text{V}_2}{\text{R}_3}+\frac{\text{V}_1-\text{V}_3}{\text{R}_4}+\text{I}_5\\ \\ \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_2-\text{V}_3}{\text{R}_5}+\frac{\text{V}_1-\text{V}_3}{\text{R}_4}\\ \\ 0=\frac{\text{V}_2-\text{V}_3}{\text{R}_5}+\frac{\text{V}_1-\text{V}_3}{\text{R}_4}+\text{I}_5 \end{cases}\tag3 $$

Now, using an ideal opamp, we know that:

$$\text{V}_+=\text{V}_-=\text{V}_2=0\space\text{V}\tag4$$

So, we can rewrite \$(3)\$ as follows:

$$ \begin{cases} \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_1}{\text{R}_3}+\frac{\text{V}_1-\text{V}_3}{\text{R}_4}\\ \\ 0=\frac{\text{V}_1}{\text{R}_3}+\frac{\text{V}_1-\text{V}_3}{\text{R}_4}+\text{I}_5\\ \\ \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}-\frac{\text{V}_3}{\text{R}_5}+\frac{\text{V}_1-\text{V}_3}{\text{R}_4}\\ \\ 0=\text{I}_5-\frac{\text{V}_3}{\text{R}_5}+\frac{\text{V}_1-\text{V}_3}{\text{R}_4} \end{cases}\tag5 $$

Hence, we can solve for the transfer function:

$$\mathcal{H}:=\frac{\text{V}_3}{\text{V}_\text{i}}=-\frac{\text{R}_2\text{R}_4\text{R}_5}{\text{R}_1\text{R}_2\left(\text{R}_3+\text{R}_4+\text{R}_5\right)+\text{R}_3\text{R}_4\left(\text{R}_1+\text{R}_2\right)}\tag6$$

Where I used the following Mathematica-code:

In[1]:=Clear["Global`*"];
V2 = 0;
FullSimplify[
 Solve[{I1 == I2 + I3 + I4, 0 == I3 + I4 + I5, I1 == (Vi - V1)/R1, 
   I2 == V1/R2, I3 == (V1 - V2)/R3, I4 == (V1 - V3)/R4, 
   I3 == (V2 - V3)/R5}, {I1, I2, I3, I4, I5, V1, V3}]]

Out[1]={{I1 -> ((R3 R4 + R2 (R3 + R4 + R5)) Vi)/(
   R1 R3 R4 + R2 R3 R4 + R1 R2 (R3 + R4 + R5)), 
  I2 -> (R3 R4 Vi)/(R1 R3 R4 + R2 R3 R4 + R1 R2 (R3 + R4 + R5)), 
  I3 -> (R2 R4 Vi)/(R1 R3 R4 + R2 R3 R4 + R1 R2 (R3 + R4 + R5)), 
  I4 -> (R2 (R3 + R5) Vi)/(
   R1 R3 R4 + R2 R3 R4 + R1 R2 (R3 + R4 + R5)), 
  I5 -> -((R2 (R3 + R4 + R5) Vi)/(
    R1 R3 R4 + R2 R3 R4 + R1 R2 (R3 + R4 + R5))), 
  V1 -> (R2 R3 R4 Vi)/(R1 R3 R4 + R2 R3 R4 + R1 R2 (R3 + R4 + R5)), 
  V3 -> -((R2 R4 R5 Vi)/(
    R1 R3 R4 + R2 R3 R4 + R1 R2 (R3 + R4 + R5)))}}

My equation was also confirmed using LTspice.

---

When we want to apply the derivation from above to your circuit we need to use Laplace transform (I will use lower case function names for the functions that are in the (complex) s-domain, so \$\text{y}\left(\text{s}\right)\$ is the Laplace transform of the function \$\text{Y}\left(t\right)\$):

• $$\text{R}_3=\frac{1}{\text{sC}_1}\tag7$$`
• $$\text{R}_4=\frac{1}{\text{sC}_2}\tag8$$

So, we can rewrite the transfer function as:

$$\mathscr{H}\left(\text{s}\right)=-\frac{\text{R}_2\cdot\frac{1}{\text{sC}_2}\cdot\text{R}_5}{\text{R}_1\text{R}_2\left(\frac{1}{\text{sC}_1}+\frac{1}{\text{sC}_2}+\text{R}_5\right)+\frac{1}{\text{sC}_1}\cdot\frac{1}{\text{sC}_2}\cdot\left(\text{R}_1+\text{R}_2\right)}=$$ $$-\frac{\text{C}_1\text{R}_2\text{R}_5\text{s}}{\text{C}_1\text{C}_2\text{R}_1\text{R}_2\text{R}_5\text{s}^2+\text{R}_1\text{R}_2\left(\text{C}_1+\text{C}_2\right)\text{s}+\text{R}_1+\text{R}_2}\tag9$$

Now, when working with sinusoidal signals we can use \$\text{s}:=\text{j}\omega\$ (where \$\text{j}^2=-1\$ and \$\omega=2\pi\text{f}\$ with \$\text{f}\$ is the frequency of the input signal in Hertz). So, we get:

$$\underline{\mathscr{H}}\left(\text{j}\omega\right)=-\frac{\text{C}_1\text{R}_2\text{R}_5\text{j}\omega}{\text{C}_1\text{C}_2\text{R}_1\text{R}_2\text{R}_5\left(\text{j}\omega\right)^2+\text{R}_1\text{R}_2\left(\text{C}_1+\text{C}_2\right)\text{j}\omega+\text{R}_1+\text{R}_2}=$$ $$-\frac{\text{C}_1\text{R}_2\text{R}_5\omega\text{j}}{\text{R}_1+\text{R}_2-\text{C}_1\text{C}_2\text{R}_1\text{R}_2\text{R}_5\omega^2+\text{R}_1\text{R}_2\left(\text{C}_1+\text{C}_2\right)\omega\text{j}}\tag{10}$$

So, the absolute value if given by:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\text{C}_1\text{R}_2\text{R}_5\omega}{\sqrt{\left(\text{R}_1+\text{R}_2-\text{C}_1\text{C}_2\text{R}_1\text{R}_2\text{R}_5\omega^2\right)^2+\left(\text{R}_1\text{R}_2\left(\text{C}_1+\text{C}_2\right)\omega\right)^2}}\tag{11}$$

We can solve for the maximum using:

$$\frac{\partial\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|}{\partial\omega}=0\space\Longrightarrow\space\hat{\omega}:=\omega=\sqrt{\frac{\text{R}_1+\text{R}_2}{\text{C}_1\text{C}_2\text{R}_1\text{R}_2\text{R}_5}}\tag{12}$$

Now, we can find the cut-off frequencies using:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\left|\underline{\mathscr{H}}\left(\text{j}\hat{\omega}\right)\right|}{\sqrt{2}}\space\Longrightarrow\space$$ $$\omega_\pm:=\omega=\frac{\sqrt{\left(\text{C}_1+\text{C}_2\right)^2+\frac{4\text{C}_1\text{C}_2\text{R}_5\left(\text{R}_1+\text{R}_2\right)}{\text{R}_1+\text{R}_2}}\pm\left(\text{C}_1+\text{C}_2\right)}{2\text{C}_1\text{C}_2\text{R}_5}\tag{13}$$

And the bandwidth is given by:

$$\mathcal{B}=\left|\omega_--\omega_+\right|=\frac{\text{C}_1+\text{C}_2}{\text{C}_1\text{C}_2\text{R}_5}\tag{14}$$

And the quality factor of the filter is given by:

$$\mathcal{Q}:=\frac{\hat{\omega}}{\mathcal{B}}=\frac{1}{\text{C}_1+\text{C}_2}\cdot\sqrt{\frac{\text{C}_1\text{C}_2\text{R}_5\left(\text{R}_1+\text{R}_2\right)}{\text{R}_1\text{R}_2}}\tag{15}$$