Calculating Heat Dissipation from Power and Efficiency Data

Hello! This course will build from a thermal description of a heat-generating PCB in a sealed enclosure toward a defensible CFD model. Before deciding how heat travels through the board, air, and enclosure, you need to know how many watts are generated, and where. This lesson turns electrical power and efficiency data into a heat-source inventory—the starting point for later calculations and Fluent simulations.

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Set the boundary before calculating heat

Imagine drawing a boundary around a voltage regulator. Electrical power enters; useful electrical power leaves for its load. The difference is the regulator’s power loss, which ordinarily becomes heat in or near the regulator:

$$P_{\mathrm{loss}}=P_{\mathrm{in}}-P_{\mathrm{out}}.$$

For steady DC operation, electrical power is voltage multiplied by current:

$$P_{\mathrm{in}}=V_{\mathrm{in}}{I}_{\mathrm{in}},P_{\mathrm{out}}=V_{\mathrm{out}}{I}_{\mathrm{out}}.$$

The unit is the watt: one volt multiplied by one ampere gives one watt. Convert milliamperes to amperes before calculating; $200\mathrm{mA}=0.200\mathrm{A}$.

Watch the short opening explanation in Power Electronics WK4 2a – Efficiency and Loss of a DC-DC Converter – Conduction Losses, from Power Electronics with Dr. K. It establishes the power balance and, importantly, shows that a converter’s loss can occur in several physical parts.

The boundary matters. If you expand it to surround both regulator and load, the power delivered to the load has not disappeared: it may subsequently become heat in the load. Thus, “regulator loss” and “total heat generated inside the enclosure” are different quantities. A sealed enclosure has no ventilation opening, but it can still pass heat through its walls; sealed does not mean thermally insulated.

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Convert efficiency into converter loss

Converter efficiency is the fraction of input power delivered as useful output power:

$$\eta =\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}.$$

Use efficiency as a fraction in calculations: $90\%$ means $\eta =0.90$, not $90$. If output voltage, output current, and efficiency are known, you can find the loss without knowing input current:

$$P_{\mathrm{out}}=V_{\mathrm{out}}{I}_{\mathrm{out}},P_{\mathrm{in}}=\frac{P_{\mathrm{out}}}{\eta },P_{\mathrm{loss}}=P_{\mathrm{out}}\left(\frac{1}{\eta }-1\right).$$

Notice the denominator. Ten percent below perfect efficiency does not mean the loss is ten percent of output power. At $90\%$ efficiency, the loss is $11.1\%$ of output power, or $10\%$ of input power.

The following short passage from Understanding Thermal Dissipation and Design of a Heatsink, a Texas Instruments application report, applies this balance to a buck converter and identifies an inductor as a separate heat source.

Let us calculate that example while retaining more digits than the report’s approximate arithmetic. A buck converter supplies $2.5\mathrm{V}$ at $1.0\mathrm{A}$, with assumed efficiency $0.90$:

$$P_{\mathrm{out}}=2.5(1.0)=2.50\mathrm{W},$$ $$P_{\mathrm{in}}=\frac{2.50}{0.90}\approx 2.778\mathrm{W},P_{\mathrm{loss}}\approx 0.278\mathrm{W}.$$

The report uses approximate values of $2.75\mathrm{W}$ input and $0.25\mathrm{W}$ loss. Those are useful order-of-magnitude estimates, but $2.50/0.90$ is approximately $2.778$, not $2.75$. Keeping the power balance consistent is worthwhile when allocating small losses among components.

Suppose the inductor’s winding resistance is $0.100\Omega$ and its current is approximately $1.0\mathrm{A}$. Neglecting current ripple for this estimate, its resistive loss is

$$P_{\mathrm{inductor}}\approx I^{2}R=(1.0{)}^2(0.100)=0.100\mathrm{W}.$$

If we assume the remaining converter losses occur in its IC, the estimated IC heat source is

$$P_{\mathrm{IC}}\approx 0.278-0.100=0.178\mathrm{W}.$$

That last allocation is an assumption, not a consequence of efficiency alone. An actual design may also have losses in capacitors, external switches, or other parts. Efficiency establishes the total converter loss; component-level information establishes where to place it.

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A linear regulator follows a different shortcut

A low-dropout linear regulator (LDO) reduces voltage by dissipating much of the voltage difference internally. When its input current is approximately equal to its output current, its heat dissipation is

$$P_{\mathrm{LDO}}\approx (V_{\mathrm{in}}-V_{\mathrm{out}})I_{\mathrm{out}}.$$

For comparison with the buck example, suppose an LDO takes $5.0\mathrm{V}$ and delivers $2.5\mathrm{V}$ at $1.0\mathrm{A}$:

$$P_{\mathrm{LDO}}\approx (5.0-2.5)(1.0)=2.5\mathrm{W}.$$

The load receives $2.5\mathrm{W}$, while the LDO dissipates another $2.5\mathrm{W}$. That differs sharply from the approximately $0.278\mathrm{W}$ total converter loss in the $90\%$-efficient buck example. Do not apply the LDO voltage-drop formula to a switching converter: its input and output currents generally differ.

If an LDO’s own supply, or ground, current is significant, return to the general power balance. For an arrangement where $I_{\mathrm{in}}=I_{\mathrm{out}}+I_{\mathrm{q}}$,

$$P_{\mathrm{LDO}}=V_{\mathrm{in}}(I_{\mathrm{out}}+I_{\mathrm{q}})-V_{\mathrm{out}}{I}_{\mathrm{out}}.$$

Here $I_{\mathrm{q}}$ is the regulator current not delivered to the load. The simpler formula is appropriate when its contribution can reasonably be neglected.

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Turn component data into an enclosure heat-source inventory

Not every component has a converter efficiency specification. For a resistive part supplied at DC, the electrical power it dissipates can be calculated as $VI$ or $I^{2}R$. For an electronic load such as a processing IC, supply power $VI$ is a useful heat estimate if nearly all its incoming electrical power is ultimately dissipated in that IC. If it delivers appreciable power to another component or outside the enclosure, account for that transfer instead of assigning all the input power to its own body.

For a sealed-PCB model, a compact inventory should record the operating condition, calculated heat, and assumed location. Consider the buck converter above feeding two loads that together consume its $2.50\mathrm{W}$ output. Assume each load ultimately dissipates all the electrical power it receives inside the enclosure:

Modeled heat source	Basis	Heat rate
Converter IC	Total converter loss less estimated inductor loss	$0.178\mathrm{W}$
Inductor	Estimated resistive loss	$0.100\mathrm{W}$
Load A	Its share of delivered output power	$0.800\mathrm{W}$
Load B	Its share of delivered output power	$1.700\mathrm{W}$
Total inside enclosure	Sum of distinct sources	$2.778\mathrm{W}$

The total matches the electrical input power, as it should under the stated assumptions. A common double count would be to assign $2.778\mathrm{W}$ as a heat source at the converter and then add the loads. The converter receives $2.778\mathrm{W}$ electrically, but most of that power leaves it electrically and becomes heat elsewhere.

This distinction will matter when constructing a thermal or CFD model. The table’s watts and locations are the physical inputs to preserve. Later, the geometry and source representation will determine exactly how those watts are applied in Fluent; a solver cannot repair an incorrect electrical power budget.

Use operating data deliberately. Converter efficiency depends on conditions such as load and input voltage, so use a curve or measurement near the intended operating point rather than assuming a headline efficiency applies everywhere. For equipment that alternates between operating modes, calculate each mode’s heat first. If an active mode occupies fraction $d$ of the time and idle occupies the remainder, its time-averaged heat is

$${\overline{P}}_{\mathrm{heat}}=d{P}_{\mathrm{heat},\mathrm{active}}+(1-d)P_{\mathrm{heat},\mathrm{idle}}.$$

That average may suit a long-term thermal estimate; the active-mode value remains relevant if the thermal objective concerns a sustained high-power interval. Record which condition each inventory represents rather than blending incompatible measurements or datasheet values.

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Closing perspective

Heat-source calculations begin with a power balance across a clearly chosen boundary. Use $VI$ to establish electrical power, efficiency to find a converter’s total loss, and component information to distribute that loss among physical parts. Count power delivered to a load as heat at the load only when it is actually dissipated there. Finally, check that the sum of distinct enclosure heat sources agrees with the power entering the enclosure after accounting for any useful power exported.

In the next lesson, those heat-source locations become the starting points for mapping how heat travels from components through the PCB, internal air, and enclosure toward ambient.