3
\$\begingroup\$

I need to calculate the gain of the filter without explicitly calculating the transfer function, ie I have to replace the capacitors with an open circuit. The problem is that I've been trying to solve the problem but I do not know how to start. Thanks.

enter image description here

enter image description here

\$\endgroup\$
8

4 Answers 4

1
\$\begingroup\$

I realize that I derived the transfer function of this bandpass filter in Transfer function of a bandpass filter without documenting how I got \$H_0\$. Let's consider the circuit in which I replaced the resistive divider made of \$R_4\$ and \$R_5\$ by a ratio \$k\$ equal to \$\frac{R_5}{R_5+R_4}\$. The schematic is below:

enter image description here

First, considering an open-loop gain \$A_{OL}\$, I can determine the voltage at the low-side op amp output: \$V_{o2}=V_{out}(k-1)A_{OL}\$. Then, by using superposition, I can get the voltage at node (2), the junction of \$R_1\$ and \$R_2\$: \$V_{(2)}=V_{in}\frac{R_2}{R_2+R_1}+V_{o2}\frac{R_1}{R_2+R_1}\$. Finally, the output voltage is the voltage at node (2) minus \$V_{out}\$ times the open-loop gain because no current flows through \$R_3\$ considering infinite input resistances for both op amps. Substitute and rearrange to obtain the definition of \$H_0\$:

\$H_0=\frac{R_2}{\frac{R_2}{A_{OL}}+\frac{R_1}{A_{OL}}+R_1+R_2+R_1A_{OL}(1-k)}\$

With a 100-dB (100k) open-loop gain for the op amps and a 1-V bias, the output is 64 µV as confirmed by simulation and Mathcad:

enter image description here

The dc gain in this case, when \$s=0\$ is -83.9 dB. As \$A_{OL}\$ approaches infinity, the output voltage is 0 V. A tricky little circuit! : )

\$\endgroup\$
1
  • \$\begingroup\$ In the above derivation and generally speaking with op amps, I prefer to first consider a finite open-loop gain \$A_{OL}\$ in the calculations because I can easily relate the op amp output to its bias on the (-) and (+) pins and check that simulation matches what I found. It's only later, when the final expression is derived, that I push \$A_{OL}\$ to infinity and obtain the equation with perfect op amps. Here, you can see this network (in dc at least) as a null double injection circuitry, biasing the low-side of \$R_2\$ to null \$V_{out}\$. If \$k\$ approaches 1, it no longer works. \$\endgroup\$ Jul 4, 2017 at 15:06
0
\$\begingroup\$

To start, the ONE point you should start with is the noninverting input of the bottom op amp. Regardless of anything else going on, that voltage is $$\frac{V_{out} R_5}{R_4+R_5}$$

Assuming you have negative feedback in all the op amps, that same voltage will be at all the input terminals of both op amps, and thus appears at \$V_2\$.

\$\endgroup\$
3
  • \$\begingroup\$ And what is that voltage? \$\endgroup\$ Apr 13, 2017 at 21:07
  • \$\begingroup\$ But the gain of the circuit must be zero, right? \$\endgroup\$ Apr 14, 2017 at 12:57
  • \$\begingroup\$ Looks that way --- IF there is valid feedback, which I haven't spent the time to convince myself. . Eventually, you see that there can't be any current into R3, so V1= V2= Vout, and the only value that satisfies the equation above is Vout=0. Alternatively, there is no feedback, and the voltage at the input terminals is not the same, and the op amps are saturated. Another possibility is that when you include the caps, there is AC gain you need to think about. \$\endgroup\$ Apr 14, 2017 at 13:31
0
\$\begingroup\$

There's no current through \$R_3\$, so \$V_1=V_o\$; hence the output of the lower op-amp is at or near the negative supply voltage, say \$ -V_s\$.

The upper op-amp is a buffer, therefore \$V_o=V_2=V_1\frac{R_2}{R_1+R_2}-V_s\frac{R_1}{R_1+R_2}\$

\$\endgroup\$
2
  • \$\begingroup\$ But the gain of the circuit must be zero, right? \$\endgroup\$ Apr 14, 2017 at 12:57
  • \$\begingroup\$ If the overall gain is zero, Vo is zero, and if Vo is zero, V- of the upper op-amp is zero. Also, if Vo is zero, the output of the lower op-amp is zero, so V2 (= V+ of the upper op-amp) is not zero. If V+ is not zero, and V- is zero, then Vo cannot be zero. Hence there's a conflict. Therefore the proposition that the overall gain is zero cannot be true. \$\endgroup\$
    – Chu
    Apr 14, 2017 at 15:41
0
\$\begingroup\$

The shown circuit resembles an RLC-Bandpass because the active part is nothing else than an "active inductor". Hence the wanted gain of the circuit is the "midband gain". Therefore, we must NOT replace the capacitors with open circuits.

Realizing that the parallel combination of L and C is a tank circuit (infinite impedance at w=wo), the midband gain at the non-inv. opamp input is unity and the gain at the opamp output is (1+R4/R5).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.