Electronic – Why exactly does maximum power transfer happen at 50 ohms (matched impedance)

circuit analysishomeworkoperational-amplifier

As the title mentioned – I am not sure why exactly the maximum power delivered to the load will be max when R_L is 50 ohms.

If I guess why, it's because if the resistance was greater than 50 ohm then the current will be lower, but if it was less than 50 ohm (e.g. 25 ohm), then the constant 50 ohm resistor would deliver the majority of the power instead of going to the load.

Why does the maximum power transfer happen at 50 ohms?

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Best Answer

The power delivered to the load is from the Joule heating effect:

\begin{equation} P=\dfrac{\Big(\dfrac{R_L}{R_L + 50}G\,V_{IN}\Big)^{2}}{R_L} \end{equation}

So from differential calculus we know that a function reaches its maximum or minimum value when we differentiate and equal it to zero. In this case we'll have a maximum value, thus:

\begin{equation} \dfrac{dP}{dR_L} = (G\,V_{IN})^2 \dfrac{((R_L+50)^2 - 2\cdot R_L\cdot(R_L + 50))}{(R_L+50)^4} \end{equation}

Finally by making \$ \dfrac{dP}{dR_L} = 0 \$ we only need to care about the numerator since the denominator does not make the equation be equal to zero from any real value. Thus we have:

\begin{equation} (R_L+50)^2 - 2\cdot R_L\cdot(R_L + 50) = 0 \implies \, (R_L)^{2} = 2500 \implies R_L = 50\,\Omega \end{equation}