Design Feature: October 26, 1995

When an IC is operating, its on-chip devices dissipate power as heat. In a regulator, most of the dissipated power comes from the power or pass transistor. As any electronic engineer knows, the current passing through the power transistor (the load current) and the voltage across it determine the amount of heat generated. The heat generated on the die flows through its package and into the surrounding air. The heat must be removed at a sufficient rate to keep the regulator's transistor junction temperatures below the temperature at which damage may occur. Most manufacturers of silicon-based ICs specify this temperature, T_J(max), as 150°C. The maximum power dissipation of an IC package is

where T_J(°C)=maximum recommended junction temperature, T_A(°C)=ambient temperature of the application, and RØ_JA(°C/W)=junction-to-ambient thermal resistance of the package, and any other heat dissipating materials.

Manufacturers list both T_J(max) and T_A(max) in the absolute-maximum-ratings section of the device's data sheet. T_A(max) is usually between 70°C and 125°C for regulators.

An IC's package determines its RØ_JA. This value quantifies how much the die's junction temperatures will rise for each watt the IC dissipates into still air. The RØ_JA value is listed in the package section of the data sheet for each available package. From Eq 1 above, a package with a lower RØ_JA has better power handling capability. (Note, however, that some IC data sheets specify a derating factor, GQ, that is the reciprocal of RQ.)

Selecting the right package for a regulator requires calculating the regulator's maximum power dissipation and checking that result against Eq 1 to ensure that the junction temperatures in the IC will remain below 150°C.

The maximum power dissipation for a single-output regulator is:

where V_IN(max) is the maximum input voltage, V_OUT (nom) is the nominal output voltage, I_OUT(max) is the maximum output current, and IQ is the quiescent current the regulator consumes at I_OUT(max). Once the value of P_D(max) is known, you can rewrite Eq 1 to determine the maximum permissible value of RØ_JA or:

You can then compare the value of RØ_JA with what is in the package section of the data sheet. Those packages with RØ_JAs less than the value you calculate from Eq 3 will keep the die temperature below 150°C. In some cases, none of the packages will dissipate all of the heat the IC generates and you'll need an external heat sink.

Choose a single-output regulator package

Assume that a linear regulator, such as the CS-8121, will operate under the following conditions: V_IN(max)=10V, V_OUT(nom)=5V, I_OUT(max)=200 mA, T_A(max)=85°C. Use the following steps to ensure that you choose the right package:

Step 1: Determine IQ from the data-sheet performance curve (Fig 1). Find the maximum output current on the graph's X axis. Locate the intersecting point on the curve whose temperature is the same or lower than the application'sT_A(max). In this case, I_Q=2.25 mA for I_OUT=200 mA at T_A(max)=25°C and I_Q=1.7 mAT_A(max)=125°C. For 85°C, the value of I_Q=2.25 mA is the worst-case possibility.

Step 2: Calculate the maximum power dissipation in the IC using Eq 2 and the values above.

Step 3: Calculate the maximum allowable RØ_JA using Eq 3.

The CS-8121 is available in three packages. Each package has the thermal characteristics specified in Table 1.

Step 4: Choose the correct package. Compare the value of RØ_JA from the calculation in step 3 with the ratings for the available package options.

According to Table 1, the only package that meets this power dissipation without a heat sink (the only package with a lower RØ_JA) is the TO-220 package with its RØ_JA=50°C/W. Although one of the packages in this example can handle the power-dissipation requirement, in some instances, none of the packages have a sufficiently low RØ_JA. In these cases, you'll have to attach a heat sink to the package. The heat sink increases the surface area of the package and thereby improves the flow of heat away from the IC.

All material in the heat-flow path between the IC and the outside environment has thermal resistance. Like electrical resistances, summing these series resistances determines the value of RØ_JA.

where: RØ_JC=the junction-to-case thermal resistance, RØ_CS=the case-to-heat-sink thermal resistance, andRØ_SA=the heat-sink-to-ambient thermal resistance.

RØ_JC appears in the package section of the data sheet. Like RØ_JA, it too is a function of package type.RØ_CS andRØ_SA are functions of the package type, heat sink, and the interface between them. These values appear in heat-sink data sheets. For these examples, assume that the package and the heat sink are in still air.

To select a package and a heat sink for a single-output regulator, assume that you are still using the CS-8121 but have increased the maximum input voltage to 12V. The new requirements are: V_IN(max)=12V, V_OUT(nom)=5V, I_OUT-(max)=200 mA, andT_A(max)=85°C. With these new requirements , I_Q(typ) remains at approximately 2.25 mA (Fig 1). The higher input voltage increases power dissipation. Using Eq 2: P_D(max)=(12V-5V)×250 mA+12V×(2.25 mA) =1.78W and the maximum allowable RØ_JA becomes RØ_JA= (150°C-85°C)/1.78W =36.52°C/W.

From Table 1, you can see that none of the package options have a low enough RØ_JA. You'll have to use a heat sink. To select the appropriate heat sink and interface, determine the value of the term RØ_CS+RØ_SA.

Step 1: Look up the package'sRØ_JC. The values appear in the package section of the data sheet (Table 1). RØ_JC for the TO-220 is 3.5°C/W.

Step 2: Determine the value ofRØ_CS+RØ_SA by substituting for theRØ_JC of a TO-220 package and the calculated value of RØ_JA from Eq 4.

The value of the heat sink plus the interface thermal resistance (RØ_CS+RØ_SA) must be less than or equal to this value.

Step 3: Determine the interface material, find itsRØ_SA, and subtract that fromRØ_CS to find the maximum acceptableRØ_CS of the heat sink. In this case, we selected an Aavid Engineering Kon-Dux pad (Ref 1). It has anRØ_SA of 4.19¡C/W.

Step 4: Select the right heat sink. Manufacturers organize their heat-sink products by the package type they fit. Each heat sink has a performance curve, which shows its heat dissipation capability in still air (Fig 2).

Continuing our example, you have already chosen a TO-220 package. To select an appropriate heat sink, consult the heat-sink performance curve, which shows the temperature rise on the package surface as a function of power dissipation. Therefore, if the packages dissipates 2W, the surface temperature will rise about 20°C above ambient (Fig 2).

Locate the value of the power dissipation, P_D (1.78W), on the X axis of the graph. Then locate the value ofRØ_SA (28.83°C) on the Y axis of the graph. Locate the intersection point on the graph. If the curve falls below this point, this heat sink will work in the application.

Heat-sinking surface-mount packages

Suppose that your application requires a surface-mount package rather than a through-hole package? Surface-mount packages offer several heat-sink alternatives. Beyond the traditional heat sink described above, you can draw out modest amounts of heat through the package ground pins into a large copper area on the pc board.

Consider an example. The micropower CS-8101 regulator in a surface-mount package must meet the following performance specifications: V_IN(max)=10V, V_OUT(nom)=5V, I_Q(typ)=70µA, I_OUT(max)=80 mA,T_A(max)=125°C.

Solving Eq 2 and 3 for this application, P_D=0.4W and RØ_JA=62.5°C/W. RØ_JA for the 16-lead batwing surface-mount package is 70°C/W. Therefore, you'll need a heat sink if you use the surface-mount package.

With the board's copper foil as a heat sink, copper draws the heat through the package's ground pins into the copper-foil sink on the surface of the pc board. (The foil is usually connected to electrical ground.) Figs 3a and 3b show a typical copper-foil geometry and the thermal resistance of a 35-mm-thick foil of this material as a function of its area. Here's how to determine the size of the copper-foil sink.

Step 1: Solve Eq 4 for the term RØ_CS+RØ_SA, substituting RØ_JC=20°C/W for a 16-lead batwing SOIC.

Step 2: Find the value of the thermal resistance you need (RØ_CS+RØ_SA) on the Y axis in Fig 3b. Find the intersection of the Y value with the curve and then read down to the X axis. The value on the X axis is the total area of the heat sink that is required. In this case, the area is 18 cm2. Each wing of the heat sink is 9 cm² or 3 cm on a side.

The previous examples present package and heat-sink selections for a single-output regulator. A dual-output regulator requires slightly different calculations. Again, assume the IC and heat sink operate in still air. Forcing air over the heat sink naturally improves heat dissipation. (Consult any heat-sink manufacturers' data book for details on heat-flow effects.) The calculation for a dual-output regulator is similar to that for a single but must include the power-dissipation term for the second output.

In the following example, notice the power-dissipation component contributed by the second output and how to determine whether the application requires a heat sink. Suppose the dual-output CS-8167 in a TO-220 power package must operate under the following system conditions: V_IN(max)=10V, V_OUT1(nom)=8V, I_OUT1(max)=250 mA, V_OUT2(nom)=5V, I_OUT2(max)=35 mA, I_Q(typ)=60 mA, and T_A(max)= 55°C.

Step 1: Calculate the power dissipation for the IC under worst-case conditions using Eq 5 and the system specifications.

Step 2: Calculate the maximum permissible thermal resistance for the IC package using Eq 3.

Step 3: Compare the calculated value of RØ_JA with the RØ_JA of the TO-220 package in Table 1. Because the TO-220 package RØ_JA is 50°C/W, the regulator will not require a heat sink in this application.

Determining an appropriate package for a regulator in a particular application depends on key system specifications. Among these specifications are maximum input voltage, maximum load current, highest ambient operating temperature and nominal output voltage. The maximum quiescent current, maximum power dissipation and maximum junction-to-ambient-thermal-resistance requirements follow from these specifications.

If the package you choose can't meet the worst-case thermal-resistance requirements, you'll need a heat sink to keep IC junction temperatures below 150°C. Heat sinking in the form of bolting to a heat rail, attaching a metal heat sink, or adding copper foil underneath surface-mount and through-hole packages are all options you'll need to consider. Failure to determine the proper operating specifications, miscalculation of the power-dissipation requirements or improper heat-sink selection can have catastrophic effects.

References

1. Aavid Engineering, Box 400, Laconia, NH 03247. Phone (603) 528-3400; fax (603), 528-1478.