The new LDO linear regulator G series, H series, I series, product lineup total has 91 types. The output voltage has a variable type 0.8V to 13V, and a total of 14 voltages from 1V to 12V, and the output accuracy is up to 1%. 0.3A ~ 1.5A output current has a current total of 4, which form a strong matrix.
These series product lineup and specifications are as follows.
The table to obtain the maximum output current of 1.5A is H series products, e.g., products available 7V output of the output power of 10.5W.
However, in fact, affected by many other conditions may not reach the target, which we should have awareness. Representative power IC process substantially higher power IC, dependent on the limited power available to heat. Accurate to say that the chip temperature Tj should not exceed the maximum rated Tj max, so need to use less than Tj max. Linear regulator Tj is calculated as follows.
Tj = + Ta self-heating
= (Thermal power θja × W) + Ta
= {Θja × (difference between input and output voltage × output electric current) W} + Ta
In this formula, set forth for convenience, not included in the power consumption itself. Strictly speaking, it must be added (input voltage × current consumption itself) amount of power, but the self-consumption current is small, the output current is large when the output power is dominant.
Actual use, the output voltage must reach the specified voltage, the problem needs to be considered is whether the available output current that is required by the load current. Before making its calculation, we must take into account the presence of other items and items not adjustable / difficult to adjust in advance. Of course, Tj max can not be changed.
Input-output difference voltage is greater impact factor. However, the available voltage (input voltage) generally has been determined. To use a linear regulator, the condition becomes better prepared to make the input voltage source is not normal. In other words, the difference between the input and output voltage under normal circumstances is not possible to change the project.
Since the temperature Ta has equipment design specifications, therefore, the temperature limit is determined if the device is 0 ℃ ~ 50 ℃, the maximum value of Ta is 50 ℃, or taking into account within the housing, due to heat causes the temperature rise inside the housing, therefore this part plus Ta is calculated. By allowing cooling fan or the like when, under the conditions of Ta, Ta but basically that too many items can not be adjusted.
As a result, when the fever is usually subject to restrictions not get the expected output current, reduce the thermal resistance is the best policy. The method of reducing thermal resistance: the use of low thermal resistance of the IC package, PCB mounting plate using a thermally good product characteristics of the multilayer structure, and the radiator is mounted. Given the recent including dimensions, including space-saving requirements, in many cases difficult to install the heat sink, it is necessary to explore the package to the PCB heat sink.
The new series of linear regulator, has been used on the back surface of the heat radiating plate is exposed HTSOP-J8 package (4.9 × 6.0 × 1.0mm). By welding the radiator plate package in consideration of the PCB heat sink, the thermal resistance can be greatly improved. The following is HTSOP-J8 graph the power dissipation.
Examples of the foregoing it is assumed that the input voltage is output 7V /1.5A 12V, calculated as follows. Tj max of 150 ℃.
Output Power = (12V-7V) × 1.5A = 7.5W
It is seen from the graph, the maximum allowable loss of 3.76W (the condition ⑤, θja = 33.3W / ℃), so that the output can not be used under current. At this time, the self-heating of 7.5W × 33.3 ℃ ≒ 250 ℃, Ta prior to investigate as to be completely eliminated.
Suppose Ta is 50 ℃, the allowable power of the lowest thermal resistance under condition ⑤ is 3W. Inverse calculation shows 0.6A is the limit under this condition. If the thermal resistance can be further reduced, the same inverse calculation, (Tj max 150 ℃ -Ta 50 ℃) ÷ 7.5W = 13.3 ℃ / ℃, so if this can be achieved, can be obtained 1.5A.
Roughly speaking, the elements determine the actual output current is Tj, i.e. the heating and the ambient temperature. Incidentally, under this condition, and use the results of the use of the product 1A 1.5A product is the same, no change.
Edit: hfy, read full text
Our other product:
