Skip to Content

Power - Available

While I touched on it in my last post, I never really got into the specifics of available power. For probably 90% of the cases in circuits, you’re going to want to deliver the maximum possible power to the load. The answer is not necessarily the one you may feel is the most intuitive. Let’s take a basic case: voltage source and two resistors. How would we maximize the power to the second resistor?


Was your gut reaction “MAKE R2 BIGGER”? It makes sense doesn’t it? More power means we need more voltage, so we increase the resistance. The problem is, with two series resistors, by increasing their values, we decrease the amount of current flowing, and therefore have less power ($P=IV$, remember).

It makes sense - if we increase R2 to infinity, we’ll have an open circuit, which obviously draws no current and subsequently, no power. Curiously, no one ever tries to decrease R2 to get more current, although that’s just as [in]valid a guess. The real answer, unfortunately, lies with math. We want to maximize P, and to do that, let’s take a derivative (eww, calculus) and set it to 0 to find some maximums:

$$ P_{R_2} = V_{R_2} I_{R_2} = V_{in}\dfrac{R_2}{R_1+R_2} \cdot \dfrac{V_{in}}{R_1+R_2} = {V_{in}}^2 \dfrac{R_2}{(R_2+R_1)^2} $$

Using the product rule, with $ f(x) = R_2, g(x) = (\dfrac{V_{in}}{R_1+R_2)})^2 $

$$ \dfrac{dP}{dR_2} = f’(x)g(x) + f(x)g’(x) = 1 \cdot (\dfrac{V_{in}}{R_1+R_2)})^2 + R_2*-2\dfrac{{V_{in}}^2}{(R_1+R_2)^3} $$

We can set one side to zero, and move half the expression over:

$$ (\dfrac{V_{in}}{R_1+R_2)})^2 = 2*R_2\dfrac{{V_{in}}^2}{(R_1+R_2)^3} $$

$$ 1 = \dfrac{2R_2}{R_1+R_2} $$

$$ R_1+R_2 = 2R_2 $$

$$ R_1 = R_2 $$

To maximize power, we set them equal! Now, what does that mean? If we say R1 is a battery, and R2 is some device like a video game controller, that means we waste as much power just plugging the battery in as we do powering our toy - and that’s the most optimal case! This also explains why these devices get so hot - in this case, it’s due to they’ll dissipate just as much power to thermal waste as they will electrical purpose. This is a huge driving behind buffer circuits and all kinds of isolated sources.