Linear Regulator Heat Calculator (Power Dissipation, Efficiency, Junction Temperature & Heatsink Needed)

How it works

P_d = (V_in − V_out) × I_out + V_in × I_q · efficiency = V_out I_out ÷ (V_in (I_out + I_q)) · T_j = T_a + P_d × R_th(JA); with heatsink R_th(JA) = R_th(JC) + R_th(CS) + R_th(SA) · heatsink needed: R_th(SA) ≤ (T_j,max − T_a) ÷ P_d − R_th(JC) − R_th(CS) · dropout margin = V_in − V_out − V_dropout

A linear regulator makes a lower voltage by burning the difference as heat: the dissipation is simply the voltage dropped times the current, plus a little for its own consumption. That heat raises the chip's junction temperature by the thermal resistance of the package and whatever it is bolted to — 62 °C per watt for a bare TO-220, which means a single watt in a warm box already approaches the limit. The calculator finds the temperature, the heatsink that would fix it, checks the regulator has enough input headroom above its dropout, and shows how much less heat a switching converter or a series pre-drop would make.

Worked example

A 7805 dropping 12 V to 5 V at 500 mA dissipates 3.56 W (41% efficient). Bare in a 40 °C enclosure a TO-220 would reach about 260 °C — impossible, it will shut down. To hold 110 °C the total thermal path must be under 19.7 °C/W, so a heatsink of about 15 °C/W (a 25–35 mm finned type) is needed; a buck module would waste only 0.28 W instead, and a 9 Ω, 5 W series resistor ahead of the regulator would take 2.25 W (63%) of the heat off the chip while keeping 0.5 V above dropout.

Frequently asked questions