CR · DC analysis

Maximum Power Transfer

How large should the load resistance be to send as much power as possible from a source to a load? This unit shows how to find that compromise, where both too small and too large lose out, by comparing the load with the source's internal resistance.

Too small loses, too large loses

The gold curve in the graph is the power delivered to the load as a function of load resistance, and the dot on the curve is the present load. With a load resistance of 0 there is no voltage across the load, and with an endlessly large one no current flows, so either way the power is 0. Find the single point between them where the power is largest, R_L = R_th.

Load resistance R_LR_L = 18.0 Ω

Load power and efficiency

P = 4.50 W  ·  η = 75%

Far from match

You cannot have both voltage and current

Power is the product of the voltage across the load and the current through it. A small load resistance draws a large current but has almost no voltage across it, while a large one has a high voltage but almost no current. The two trade off against each other, so leaning toward either side shrinks the product, and the power peaks when the two are in balance.

The balance is at R_L = R_th

Differentiate the power P = (V_th/(R_th+R_L))²·R_L with respect to R_L and set it to 0, and the maximum comes out exactly at R_L = R_th. This condition, making the load resistance equal to the source's internal resistance, is called matching. The voltage then splits evenly in two, V_th/2 appears across the load, and the maximum power is P_max = V_th²/(4R_th). The source in the plot has 12 V and 6 Ω, so the power peaks at 6 W when R_L is 6 Ω.

Maximum power is not maximum efficiency

At matching the load resistance and the internal resistance are equal, so the total power splits evenly between them. As much power is wasted as heat in the internal resistance as the load receives, so the efficiency is exactly 50%. When the goal is to draw out the most power, matching is the answer, but where efficiency matters, as in power transmission, the load resistance is deliberately made large to cut losses. Maximum power and maximum efficiency are different goals.

ObserveP = I² R_L

Power is voltage × current.

ChooseP_max → R_L = ?

Make the load equal to the internal resistance.

Fill inP_max = V_th² / ?

At matching, the voltage splits in half.

On your ownR_L = R_th → η = ?

The load and the internal resistance take equal shares of the power.

Back to the first screen

As you changed the load resistance, the load power started at 0, climbed to a peak, and came back down, and that peak sat exactly at R_L = R_th, the dashed line where the load equals the internal resistance. There the voltage split evenly in two and the efficiency was 50%. Once a circuit has been folded into one source and one resistor by its Thevenin equivalent, what to connect to receive the most power can be read at a glance like this.

Maximum power transfer is the theorem that a load connected to a Thevenin source (V_th, R_th) receives the most power when R_L = R_th (matching). The maximum power is then P_max = V_th²/(4R_th), and since the load and the internal resistance share the power equally, the efficiency is 50%. A load that is too small or too large upsets the balance of voltage and current and the power falls.