Electronic – How to verify the max power theorem in a buffer opamp

I didn't follow the link, but air conditioners can move more heat than the power it takes to do the moving.

This is a case where the Carnot efficiency works in your favor. Carnot says that the maximum possible theoretical efficiency of a heat engine is Tdiff / Thot, where the temperatures are expressed in a absolute linear scale, like Kelvin. For example, if you had a reservoir of boiling water (373 K) and icewater (273 K), even a ideal heat engine can't be more than 100 K / 373 K = 27% efficient. Put another way, for a heat flow of 100 W in this setup, you can't ever extract more than 27 W of work.

However, heat pumps do the reverse. A perfect heat pump would only require 27 W of work input to move 100 W of heat from 273 K to 373 K. This is not a violation of conservation of energy because work can't be extracted from this heat without hitting the 27% limit as above.

Air conditioners are heat pumps. They move heat from the room to the outside, presumably from cooler to warmer. The temperature difference is a lot less than the difference between boiling and frozen water, so the Carnot efficiency is much less too. That is actually good news for air conditioners, because they benefit according to the reciprocal of the Carnot efficiency. Of course real air conditioners are real systems with inevitable real inefficiencies, so you can't just take the reciprocal of the Carnot efficiency to determine their power input requirements. But, this still works in their favor, and the systems are good enough for reasonable temperature differences to still require less work in than the amount of heat they move around.

For more details, look up "Carnot efficiency". Surely there is much written about it out there.

If we let R1 = R2 = R and C1 = C2 = C which is usually the case for this type of oscillator (same values). We can see that C1, C2, R1 and R2 form a potential diver.

The impedance of the top part R2 in series with C2 is:

$$ Z_{top} = R + \frac{1}{j \omega C} = \frac{1+j\omega C R}{j\omega C} $$

The impedance of the bottom part R1 in parallel with C1 is:

$$ Z_{bot} = \frac{R \frac{1}{j \omega C}}{R + \frac{1}{j \omega C}} = \frac{R}{1 + j \omega C R} $$

So the gain of this potential divider is:

$$\begin{align} G & = \frac{Z_{bot}}{Z_{top}+Z_{bot}} \\ & = \frac{\frac{R}{1 + j \omega C R}}{\frac{1+j\omega C R}{j\omega C}+\frac{R}{1 + j \omega C R}} \\ & = \frac{\frac{R}{1 + j \omega C R}}{\frac{(1 + j \omega C R)^2 + j \omega C R}{j \omega C(1+j\omega C R)}} \\ & = \frac{j \omega C R}{1 + 3j \omega C R -\omega^2 C^2 R^2} \\ & = \frac{\omega C R}{-j(1 + 3j \omega C R -\omega^2 C^2 R^2)}\\ & = \frac{\omega C R}{(\omega^2 C^2 R^2 -1)j + 3 \omega C R} \end{align}$$

If we chose a frequency such that the imaginary (j) part is zero then the Gain G is 1/3. Now for an oscillator to work it needs a gain of exactly one so the gain of the amplifier needs to be 3 since 3 times 1/3 = 1. Typically there is a non linear element such as a PTC or bulb in the feedback path R6, R3 to provide a simple auto gain control to ensure this.

The frequency when this will oscillate is:

$$ \omega^2 C^2 R^2 - 1 = 0 \Rightarrow \omega = \frac{1}{C R} \Rightarrow f = \frac{1}{2 \pi C R} $$

Finally consider what would happen if the output were exactly zero volt. All inputs to the opamp would be zero and the circuit would never start. However there is always some noise in the system to start it going.

Edit:

If you are simulating this Remove R4 and change R3 to 20k. This should make the gain exactly 3 as required. You will also need to add an initial charge to either C1 or C2 to get it started otherwise the inputs will be at 0V and it wont start.

If you are testing with real hardware you will need to play with the setting of the pot to much or too little gain and it wont work.