SNOSAL8D April   2006  – September 2021 LMH6321

PRODUCTION DATA  

  1. 1Features
  2. 2Applications
  3. 3Description
  4. 4Revision History
  5. 5Specifications
    1. 5.1 Absolute Maximum Ratings
    2. 5.2 Operating Ratings
    3. 5.3 Thermal Information
    4. 5.4 ±15 V Electrical Characteristics
    5. 5.5 ±5 V Electrical Characteristics
    6. 5.6 Typical Characteristics
  6. 6Application Hints
    1. 6.1  Buffers
    2. 6.2  Supply Bypassing
    3. 6.3  Load Impedence
    4. 6.4  Source Inductance
    5. 6.5  Overvoltage Protection
    6. 6.6  Bandwidth and Stability
    7. 6.7  Output Current and Short Circuit Protection
    8. 6.8  Thermal Management
      1. 6.8.1 Heatsinking
      2. 6.8.2 Determining Copper Area
      3. 6.8.3 Procedure
      4. 6.8.4 Example
    9. 6.9  Error Flag Operation
    10. 6.10 Single Supply Operation
    11. 6.11 Slew Rate
  7. 7Device and Documentation Support
    1. 7.1 Receiving Notification of Documentation Updates
    2. 7.2 Support Resources
    3. 7.3 Trademarks
    4. 7.4 Electrostatic Discharge Caution
    5. 7.5 Glossary
  8. 8Mechanical, Packaging, and Orderable Information

Package Options

Refer to the PDF data sheet for device specific package drawings

Mechanical Data (Package|Pins)
  • KTW|7
  • DDA|8
Thermal pad, mechanical data (Package|Pins)
Orderable Information

6.8.4 Example

Assume the following conditions:

V+ = V− = 15 V, RL = 50 Ω, IS = 15 mA TJ(MAX) = 125°C, TA(MAX) = 85°C.
  1. From Equation 7
    • PD(MAX) = IS (2 V+) + V+2/4RL = (15 mA)(30 V) + 15 V2/200 Ω = 1.58 W
  2. From Equation 8
    • TR(MAX) = 125°C - 85°C = 40°C
  3. From Equation 9
    • θJA = 40°C/1.58 W = 25.3°C/W

Examining Figure 6-4, we see that we cannot attain this low of a thermal resistance for one layer of 1 oz copper. It will be necessary to derate the part by decreasing either the ambient temperature or the power dissipation. Other solutions are to use two layers of 1 oz foil, or use 2 oz copper (see Table 6-1), or to provide forced air flow. One should allow about an extra 15% heat sinking capability for safety margin.

GUID-2B8DE760-86D7-4AAD-B3FC-0A03CA7D944D-low.gif Figure 6-4 Thermal Resistance (Typical) for 7-L DDPAK Package Mounted on 1 oz. (0.036 mm) PC Board Foil
GUID-14BE21AE-DF03-4F5F-8955-D359D48CCEB3-low.gif Figure 6-5 Derating Curve for DDPAK package. No Air Flow
Table 6-1 θJA vs. Copper Area and PD for DDPAK. 1.0 oz cu Board. No Air Flow. Ambient Temperature = 24°C
Copper Area θJA at 1.0W
(°C/W)		 θJA at 2.0W
(°C/W)
1 Layer = 1”x2” cu Bottom 62.4 54.7
2 Layer = 1”x2” cu Top and Bottom 36.4 32.1
2 Layer = 2”x2” cu Top and Bottom 23.5 22.0
2 Layer = 2”x4” cu Top and Bottom 19.8 17.2

As seen in the previous example, buffer dissipation in DC circuit applications is easily computed. However, in AC circuits, signal wave shapes and the nature of the load (reactive, non-reactive) determine dissipation. Peak dissipation can be several times the average with reactive loads. It is particularly important to determine dissipation when driving large load capacitance.

A selection of thermal data for the SO PowerPAD package is shown in Table 6-2. The table summarizes θJA for both 0.5 watts and 0.75 watts. Note that the thermal resistance, for both the DDPAK and the SO PowerPAD package is lower for the higher power dissipation levels. This phenomenon is a result of the principle of Newtons Law of Cooling. Restated in term of heatsink cooling, this principle says that the rate of cooling and hence the thermal conduction, is proportional to the temperature difference between the junction and the outside environment (ambient). This difference increases with increasing power levels, thereby producing higher die temperatures with more rapid cooling.

Table 6-2 θJA vs. Copper Area and PD for SO PowerPAD. 1.0 oz cu Board. No Airflow. Ambient Temperature = 22°C
Copper Area/Vias θJA at 0.5W
(°C/W)		 θJA at 0.75W
(°C/W)
1 Layer = 0.05 sq. in. (Bottom) + 3 Via Pads 141.4 138.2
1 Layer = 0.1 sq. in. (Bottom) + 3 Via Pads 134.4 131.2
1 Layer = 0.25 sq. in. (Bottom) + 3 Via Pads 115.4 113.9
1 Layer = 0.5 sq. in. (Bottom) + 3 Via Pads 105.4 104.7
1 Layer = 1.0 sq. in. (Bottom) + 3 Via Pads 100.5 100.2
2 Layer = 0.5 sq. in. (Top)/ 0.5 sq. in. (Bottom) + 33 Via Pads 93.7 92.5
2 Layer = 1.0 sq. in. (Top)/ 1.0 sq. in. (Bottom) + 53 Via Pads 82.7 82.2