What Size Heatsink Do I Need? - Right now there are a big array of high power modules within the vary of 300 to 1,000-watts, each DC-DC converters in addition to AC-DC power modules, that are generally known as “bricks.” Although these gadgets feature high conversion efficiencies within the space of 85 to 90% (or larger), some power is misplaced within the type of warmth that should be handled with a view to maximize the lifespan of the top product. For instance, a 500-watt power module with 90% conversion effectivity would generate over 55-watts of wasted warmth inside the module that should be eliminated to maximise its reliability.
The concentrated power density (watts per cubic inch) inside these power modules make them a problem to chill in actual world functions. Most high power bricks are packaged in thermally conductive plastic or epoxy instances with integral metallic baseplates. The high power parts inside the bricks (i.e., semiconductors, inductors, transformers, and many others.) are thermally coupled to those baseplates, which in flip might be hooked up to external heatsinks or liquid-cooled chilly plates with a view to preserve the baseplate at or beneath its most working temperature (sometimes 85 to 100°C).
The utmost baseplate temperature is primarily decided by the utmost internal junction temperature of the semiconductors inside the power bricks. The time period “thermal administration” refers back to the designer’s problem of cooling these power bricks by contemplating the various ranges for warmth transfers by way of conduction (direct contact between solids), convection (contact with air or a fluid) and thermal radiation (electromagnetic infrared power), each internal and external to the power module.
The diagram above exhibits the series-connected thermal resistances that impede the movement of warmth from one stage to the following. These impedances have to be thought of, starting with the internal semiconductor’s junction temperature relative to its case, the thermoplastic module case and its metallic baseplate, and ending with a mechanically hooked up heatsink that conducts away the warmth from the baseplate to the encompassing ambient air by way of pure or pressured air convection cooling. Heatsinks are designed to cross thermal obstacles primarily by considerably rising the floor space that is available in contact with the ambient air, thereby offering enhanced convection cooling. As a result of the mating surfaces of the power module’s baseplates and heatsinks are usually not completely flat, some kind of thermally conductive interface materials is required to fill the tiny voids. This interface materials can vary from a skinny layer of thermal grease to a custom designed silicon pad.
Choosing the proper measurement and form of a heatsink and figuring out if pressured air cooling is required are among the many tradeoffs the designer wants to contemplate. This course of begins with an in depth review of the power module’s specifications and information of the top product’s warmth hundreds, internal and external working temperatures, house constraints, and accessible air movement sources, paths and restrictions.
The following step on this course of is to find out the quantity of power that will likely be misplaced (wasted) inside the power module, based mostly on its effectivity. This data for computing that is often listed on the power module’s datasheet or set up guide, but it surely may also be decided by precise measurements of the enter and output powers. For this instance, we are going to use a typical AC-DC power module with a 48V/10.5A, 504W output ranking, and a typical effectivity of 85% with a 120VAC enter. By the best way, the 85% effectivity ranking is excellent contemplating the truth that this module accommodates full-bridge rectification and lively power issue correction AC front-end circuits in addition to an integral DC to DC converter. As well as, this module has a most working baseplate temperature, as measured at its heart level, of 100°C.
Primarily based on the above data, to compute the internal power dissipated (wasted warmth); we will use the next components:
Pd = (Pout / η) – Pout
Definitions & Calculation Instance:
Pd : Inside Energy Dissipated (W)
Pout : Output Energy (504W)
η : Effectivity (85%) Pd = (504W / 0.85) - 504W = 88.9W
To calculate the required baseplate to ambient air thermal resistance that will be wanted for this application, the next components would apply:
θba = Tb - Ta / Pd
Definitions & Calculation Instance:
θba : Baseplate to Ambient Air Thermal Resistance (°C/W)
Tb : Baseplate Temperature (100°C)
Ta : Ambient Air Temperature (40°C)
Pd : Inside Energy Dissipated (88.9W)
θba = 100°C - 40°C / 88.9W = 0.67°C/W
On this instance, we would want a heatsink (with or with out air movement) that supplied a thermal resistance of 0.67°C/W. Nevertheless, until the heatsink features a thermal interface materials like thermal grease or a pad in its ranking, we have to account for this extra thermal contact resistance (θbs), which might be on the order of 0.1°C/W. Subsequently, the required thermal resistance of the heatsink itself, with the thermal interface materials included, might be calculated per this components and instance:
θba-bs = θba – θbs θba – θbs = 0.67°C/W - 0.1°C/W = 0.57°C/W
The following step on this course of is to review specifications for potential heatsinks which have a thermal resistance of 0.57°C/W. On this case, the power module has three non-obligatory heatsinks to select from as proven within the chart beneath.
The Y axis of this chart exhibits the thermal resistance between the heatsink and the air (°C/W) and the X axis exhibits the required airflow velocity for the three heatsinks. On this instance, we have to discover 0.57°C/W alongside the Y axis after which transfer to proper alongside the X axis to the place it intersects a heatsink curve. On this instance, 0.57°C/W intersects with the HAF-15T heatsink curve at in regards to the 1 m/s airflow velocity level. Subsequently, for this application we would choose the mannequin HAF-15T heatsink and must present pressured air cooling with an air velocity of 1 m/s. To translate m/s (meters/second) into LFM (linear ft/second), use this common conversion issue: 1 m/s = 200 LFM. On this instance, since 1 m/s = 200 LFM of forced-air velocity, the fan required for this application should present 200 LFM.
Primarily based on the above, we've now decided the necessities for cooling this power module with a heatsink, thermal compound and compelled air movement. If we needed to be extra conservative and enhance the MTBF of the module, we might recalculate the required heatsink with the belief that we needed to maintain the baseplate temperature at 85°C.
A phrase or warning ought to to be injected right here. After going by the thermal calculations and deciding on a heatsink, air movement, and many others., the following step is to substantiate the “paper-design” by operating precise assessments on a pattern unit. The difficult half is to get entry to the middle level of the power module’s baseplate so you'll be able to measure the temperature at that time whereas the module is working underneath load. A method to do that is to drill a gap within the heart of the heatsink so the leads from a thermocouple might be mounted on the module’s baseplate and routed to your temperature measurement device.
In abstract, we've proven methods to decide the proper heatsink for power module functions. Because the efficiencies of those gadgets enhance the necessity for cooling will scale back, however the designer ought to all the time concentrate on the heating results from not solely from the power module, but in addition from close by gadgets. Subsequently, it’s all the time best to run precise thermal assessments with thermocouples hooked up to the power module and inside the top product to insure the design will likely be as dependable as potential.