Electronic – Increasing open loop gain of opamp by chaining

Closed loop gain is the gain that results when we apply negative feedback to "tame" the open loop gain. The closed loop gain can be calculated if we know the open loop gain and the amount of feedback (what fraction of the output voltage is negatively fed back to the input).

The formula is this:

$$ A_{closed} = \frac{A_{open}}{1 + A_{open} \cdot Feedback} $$

The open-loop gain affects the performance generally like this. Firstly, look at the above formula. If the open loop is huge, like 100,000, then the 1 + does not matter. \$A_{open} \cdot Feedback\$ is a large number, and it doesn't matter whether or not we add 1 to this large number: it is like a drop in a bucket. Thus the formula reduces to:

$$ \begin{align} A_{closed} &= \frac{A_{open}}{A_{open} \cdot Feedback} \\ &= \frac{1}{Feedback}\\ \end{align} $$ So, with a huge open-loop gain, we can easily get the closed loop gain if all we know is the negative feedback: if it just the reciprocal. If the feedback is 100% (i.e. 1) then the gain is 1, or unity gain. If the negative feedback is 10%, then the gain is 10. With a huge open-loop gain, we can precisely set up gains: as precisely as we care to design and build our feedback circuit. With open-loop gain which is not that large, we may not be able to ignore that 1 +. All the more so if \$Feedback\$ is small.

Okay, so far that's more of an issue of clean math and design convenience. Big open loop gain: closed loop gain is simple. But, practically speaking, small open-loop gains means that you must use less negative feedback to achieve a given gain. If the open loop-gain is a hundred thousand, then we can use 10% feedback to get a gain of 10. If the open loop gain is only 50, then we must use much less negative feedback to get a gain of 10. (You can work that out with the formula.)

We generally want to be able to use as much negative feedback as possible, because this stabilizes the amplifier: it makes the amplifier more linear, gives it a higher input impedance and lower output impedance and so on. From this perspective, amplifiers with huge open loop gains are good. It is usually better to achieve some necessary closed loop gain with an amplifier that has huge open loop gain, and lots of negative feedback, than to use a lower gain amplifier and less negative feedback (or even just an amplifier with no negative feedback which happens to have that gain open loop). The amp with the most negative feedback will be stable, more linear, and so on.

Also note that we don't even have to care how huge the open loop gain is. Is it 100,000 or is it 200,000? It doesn't matter: after a certain gain, the simplified approximate formula applies. Amplifiers based on high gain and negative feedback are therefore very gain-stable. The gain depends only on the feedback, not on the specific open-loop gain of the amplifier. The open loop gain can vary wildly (as long as it stays huge). For instance, suppose that the open loop gain is different at different temperatures. That does not matter. As long as the feedback circuit is not affected by temperature, the closed-loop gain will be the same.

You can determine the transfer function of this system using the fast analytical circuits techniques or FACTs. First, you start with \$s=0\$, shorting inductors and opening capacitors. The dc gain is simply

\$H_0=-\frac{R_2}{R_1}\$

Then, you look at the resistance offered by the energy-storing elements when temporarily removed from the circuit. You should find:

\$\tau_1=\frac{L_1}{R_1}\$ then \$\tau_2=C_1*0\$ and \$\tau_3=\frac{L_2}{R_{inf}}=0\$

Then, you determine the resistance seen from the energy-storing elements when one of them is set in its high-frequency state (inductors replaced by open circuit and capacitors replaced by short circuits). You should find:

\$\tau_{12}=C_1R_1\$ then \$\tau_{13}=\frac{L_2}{R_{inf}}=0\$ and \$\tau_{23}=\frac{L_2}{R_{inf}}=0\$

Finally, you determine the resistance seen from \$L_2\$ while \$L_1\$ and \$C_1\$ are set in their high-frequency state (inductors replaced by an open circuit and capacitors replaced by short circuits). You have:

\$\tau_{123}=\frac{L_3}{R_{inf}}=0\$

The denominator is thus equal to

\$D(s)=1+s(\tau_1+\tau_2+\tau_3)+s^2(\tau_1\tau_{12}+\tau_1\tau_{13}+\tau_2\tau_{23})+s^3(\tau_1\tau_{12}\tau_{123})\$

The zero exists when the impedance made of \$L_2\$ and \$R_2\$ becomes a transformed short circuit. This occurs when \$\omega_z=\frac{R_2}{L_2}\$. The complete transfer function is defined as

\$H(s)=H_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}\$ with \$H_0=-\frac{R_2}{R_1}\$, \$\omega_z=\frac{R_2}{L_2}\$, \$\omega_0=\frac{1}{\sqrt{L_1C_1}}\$ and \$Q=R_1\sqrt{\frac{C_1}{L_1}}\$

The complete Mathcad file appears below. I have purposely changed the labels so that time constant labels match that of the components but results are similar:

It looks a bit mysterious but FACTs are easy to learn and apply. Check out this APEC 2016 presentation

http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

and all these examples solved in the book

http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf