89.2.4. Common Mode Gain Calculation

Here we do have this resistor it is practically , and if we apply same voltage here and same voltage here namely v . So, that makes this voltage and this voltage to be equal in_c so, even though we do have r connected here. But we do have voltage dependent current source which depends on this voltage, making this node very insensitive or a smarter way to calculate their common mode voltage we can say that common mode gain.

Common mode gain refers to the ability of a circuit to respond to signals that are common to both inputs. In this case, if the same voltage (v_in) is applied to both inputs of the amplifier, the common mode signals will mostly not be amplified due to the design of this amplifier configuration. The presence of a resistor (r) affects this gain, but the circuit's response is largely defined by its inherent characteristics, such as the voltage-dependent current source present, rendering the node high impedance.

Think of it like a two-person conversation where both speakers say the same thing at the same time; the listeners (the amplifier) only respond to the differences in their speech patterns and effectively ignore common statements. This mimics how a differential amplifier handles signals.

Essentially, it is and in this circuit since we are applying same voltage here and here v with same polarity. In that case this node of course, it is remaining high impedance or rather I should say degenerated node. So, we do have this resistored remaining there and it is degenerating the circuit and this gain as you have discussed before it will be the gain of this circuit which is getting degenerated by r or if I split the circuit it will be rather 2 r.

The common mode gain can be calculated assuming the applied voltage at both inputs remains the same. When the resistors are present, they provide degenerative feedback that reduces gain. Therefore, we can quantify the common mode gain by considering this degeneration, typically leading to a scenario where the effective impedance doubles, resulting in a lower gain overall. Essentially, it can be represented as the output gain of the circuit modified by the presence of the load resistors.

Imagine a playground swing where pushing with both hands (the inputs) creates oscillating motion, but only caring about the swings' sideways movements instead of pushing evenly. This uneven push leads to the swing amplifying side-to-side rather than up and down, somewhat similar to how the amplifier behaves with common mode signals.

So, the corresponding gain here it is g × so that is the impedance here divided by 1 + g (2r ). In fact, after removing this 1 you can remove this g also, so that gives us a common mode gain = . In fact, g it is given to us and also yeah so it is becoming this is . So, that gives us how much 26 × 10‒5.

The formula derived illustrates a relationship where the common mode gain reduces relative to the total impedance present thanks to the feedback offered by the resistors. Simplifying the calculations allows us to find that the common mode gain approximates to a low value (in this case, 26 × 10^-5), indicating that the circuit is well-designed to favor differential signals over common mode signals.

Picture a wind turbine designed to generate energy from wind against the resistance of the tower (representing common signals). The turbine is optimized to turn only by capturing wind from a particular direction (the desired signal), minimizing energy loss from any other directions (common signals), thus efficiently focusing on what truly matters.

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