Identifying pole/zero frequencies in an op amp stabilization In-the-loop compensation circuit - Analog Devices article

Question

I have been reading the following article by Analog, which discusses methods used to stabilize op amps when driving capacitive loads.

Analog: Practical Techniques to Avoid Instability Due to Capacitive Loading

One popular method discussed is the 'in the loop' compensation method, using a small resistor placed between the output of the op amp and the capacitive load and a feedback capacitor from the op amp's output to the inverting input. The circuit is shown below:

The feedback part of the circuit is shown below:

Two cases are then discussed in the article, where the pole and zero frequencies are deduced by finding the time constants for \$C_L\$ and \$C_F\$. The pole and zero frequencies for \$C_L\$ are found as:

$$\text{Pole Frequency} = \frac{1}{2\pi (R_O + R_X) C_L}$$ $$\text{Zero Frequency} = \frac{1}{2\pi R_X C_L}$$

I am having trouble following how the zero frequency has been obtained. \$R_F\$ and \$R_\text{IN}\$ disappear due to \$R_X << R_F\$, and \$R_O << R_\text{IN}\$.

I recently bought Christophe Basso's book, 'The Fast track to determining transfer functions of linear circuits'. I have been trying to use the FACTS method to determine the time constants in order to perhaps gain an understanding of the reasoning used in the Analog article. I began by setting the op amp output to zero and finding the resistances looking into each of the capacitor terminals. My working is shown below:

I then shorted \$C_F\$ and repeated the method for \$C_L\$, shown in my working below:

My problem is, I still have no idea how the zero frequency is derived in the article.

Answer 1

Here's your image where I've highlighted the two components that make the zero: -

If you worked out the transfer function you'd get this: -

$$TF_Z = \dfrac{\frac{1}{sC_L}}{R_X + \frac{1}{sC_L}} = \dfrac{1}{1+sC_LR_X}$$

What makes that equation zero is when s is very large i.e. the equation reduces to \$\dfrac{1}{sC_LR_X}\$. We can then say that the zero frequency is \$\dfrac{1}{2\pi C_LR_X}\$

Answer 2

I think, there is an error in the refrenced article. The pole frequency was confused with the zero (as far as the closed-loop gain Acl is concerned).

Using the simplifications as mentioned in the article (case 1), the feedback function is

VB/VA=Hf=[Rx+(1/sCL)]/[Ro+Rx+(1/sCL)]

From this, the closed loop gain of the opamp is

Acl=1/Hf=[1+(Ro+Rx)sCL]/[1+sRxCL]

Hence, the zero is at wz=1/(R0+Rx)CL and the pole wp=1/RxCL

EDIT/Explanation: The feedback scheme as given in the figure is based on the principle of "frequency compensated voltage division" with phase enhancement below the pole frequency. Here is a "rule of thumb" for dimensioning the circuit:

1. Assumptions: RF>>Rx, CL>>CF and Rx=Rout

2. Requirement: k(Rout+Rx)CL=(Rout+RF)CF with k=(1...1.3).

In this case, the zero fz will be somewhat smaller than the pole frequency (fz app. equal to fp/k).

Answer 3

As indicated in the comments section, I gave a detailed answer about the transfer function linking nodes A and B in SE a few years ago. I used the FACTs to deliver the exact expression.

For the purpose of this post, I have updated my answer with the determination of poles and zeroes in the final application around the op-amp. The first TF linking nodes A and B is derived below:

I then compare the exact simulation results from SIMetrix with the full-blown expression and its simplified version, in which I involve the low-\$Q\$ approximation:

Then I check the response at the op-amp output (before \$R_{out}\$) now that I have the TF linking nodes A and B:

However, in the end, we want the output at the load node, across \$C_L\$ and \$R_L\$. Without considering \$R_L\$, we can reuse the exact transfer function we have derived and add an output pole involving \$C_L\$ alone:

I can further simplify, drop one pole and one zero (\$C_L\$ no longer contributes a zero for instance), to end up with a simplified expression featuring two poles (we have two energy-storing elements) and one single zero:

As a preliminary conclusion, the determination of the TF linking nodes A to B, is a good exercise and it leads to the exact voltage delivered by the op-amp output at node A in the SIMetrix template. However, more work is needed to derive the complete transfer function of the closed-loop system for a response sensed across \$C_L\$ and \$R_L\$. I did use an approximation in the end - it's getting late : ) but, the best nerdy approach is to rewrite the time constants across the entire network, including the op-amp and its open-loop gain, then identify the poles and zeroes. If time permit, I will do it later.