0
\$\begingroup\$

I was reading this document talking about the use of superposition with dependant sources.

At example 14 they state the following : enter image description here

I don't understand how they pulled the equation for Vo by inspection only. They seem to be using the equations for the non-inverting and inverting amplifiers. However, for the non-inverting step, it's as if Vo is grounded (zero). For the inverting step, they seem to consider the V+ of the op-amp as grounded (zero), which should not be the case because of the RC network.

What's going on behind their reasonings? Could someone explain it to me step by step? Thanks!

\$\endgroup\$

2 Answers 2

2
\$\begingroup\$

I agree with the comments that most op amp-based all-pass filters show a pull-down resistor after the capacitor rather than having it connected to the output as in Leach's example. Nevertheless, we can try to determine the transfer function of this circuit applying superposition. The trick to split \$V_i\$ in two separate sources is good to solve these types of circuits. We thus have two circuits to which we can apply superposition also. The first circuit appears below:

enter image description here

What you need to do is determine the value of \$\epsilon\$ in all configurations: write the voltage at (-) and (+) nodes and write \$\epsilon=V(+)-V(-)\$. In the first configuration in which \$V_o\$ is considered as an individual source and set to 0 V while \$V_i\$ is on, you have \$\epsilon_1=-\frac{V_i}{2}\$ while \$\epsilon_2=V_o\frac{1}{1+sRC}-\frac{V_o}{2}\$. Sum up these two values and solve \$0=(\epsilon_1+\epsilon_2)\$ considering an infinite-gain op. If everything goes well, you have \$V_{01}(s)=V_i(s)\frac{1+sRC}{1-sRC}\$.

For the second part, ground the left side of \$R_i\$ and consider the below circuit:

enter image description here

If you follow the above steps, then you should obtain \$V_{02}(s)=V_i(s)\frac{2sRC}{sRC-1}\$. You can now write that \$V_o=V_{o1}+V_{o2}\$ and if you do the maths ok, then you find \$V_o=V_i\$ confirming the comments left by the two readers. Now, if you correctly connect the resistor after the cap. to ground, you have the following circuit:

enter image description here

You can split \$V_i\$ as recommended and you can inspect the circuit as the first configuration is an inverting configuration having a gain of -1 while the second one is a non-inverting configuration have a gain of \$1+\frac{R_1}{R_1}\$ and amplifying the differentiator output.

enter image description here

So the output is equal to \$V_o=V_i(-1+\frac{sRC}{1+sRC}(1+\frac{R_1}{R_1}))\$ which is what Mr. Leach wrote but from a wrong sketch unfortunately. If you solve this expression, you have \$H(s)=-\frac{1-\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}\$ in which \$\omega_z=\omega_p=\frac{1}{RC}\$.

This expression is made of a RHP zero and a LHP pole tuned at the same frequency canceling each other in magnitude. It gives a flat 0-dB response. However, as the RHP zero lags in phase, it sums up with that of the LHP pole and you create a delay. This is the practical realization of the 1st-order Padé approximant of \$H(s)=e^{-s\tau}\$ with the pole/zero pair tuned at \$\frac{2}{\tau}\$.

enter image description here

\$\endgroup\$
1
\$\begingroup\$

I do get the point they are making, but I don't think their derivation is correct for the following reason:

At very low frequencies, the capacitor will act as an open circuit, disconnecting the path between \$V_i\$ and \$v_+\$. No current will be able to flow through \$R\$ which makes \$v_+ = V_o\$. In other words, you get the following circuit at DC (\$s = 0\$):

schematic

simulate this circuit – Schematic created using CircuitLab

This system doesn't even have negative feedback...

I think what he meant to use was this circuit:

schematic

simulate this circuit

On first sight, the equations seem to fit that circuit AND they can be derived by inspection.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Yes - Sven is correct. The drawings a) and b) show a positive feedback path, which is NOT correct. For an allpass function the resistor R must be at ground as shown (last figure) by Sven. \$\endgroup\$
    – LvW
    Jan 21, 2018 at 13:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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