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MIC5219-2.8BM5 Datasheet(PDF) 10 Page - Micrel Semiconductor |
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MIC5219-2.8BM5 Datasheet(HTML) 10 Page - Micrel Semiconductor |
10 / 13 page MIC5219 Micrel MIC5219 10 April 2001 Peak Current Applications The MIC5219 is designed for applications where high start- up currents are demanded from space constrained regula- tors. This device will deliver 500mA start-up current from a SOT-23-5 or MM8 package, allowing high power from a very low profile device. The MIC5219 can subsequently provide output current that is only limited by the thermal characteris- tics of the device. You can obtain higher continuous currents from the device with the proper design. This is easily proved with some thermal calculations. If we look at a specific example, it may be easier to follow. The MIC5219 can be used to provide up to 500mA continuous output current. First, calculate the maximum power dissipa- tion of the device, as was done in the thermal considerations section. Worst case thermal resistance ( θ JA = 220°C/W for the MIC5219-x.xBM5), will be used for this example. P = T – T D(max) J(max) A JA () θ Assuming a 25 °C room temperature, we have a maximum power dissipation number of P = 125 C – 25 C C/W D(max) °° () ° 220 P D(max) = 455mW Then we can determine the maximum input voltage for a five- volt regulator operating at 500mA, using worst case ground current. P D(max) = 455mW = (VIN – VOUT) IOUT + VIN IGND I OUT = 500mA V OUT = 5V I GND = 20mA 455mW = (V IN – 5V) 500mA + VIN × 20mA 2.995W = 520mA × V IN V 2.955W 520mA 5.683V IN(max) == Therefore, to be able to obtain a constant 500mA output current from the 5219-5.0BM5 at room temperature, you need extremely tight input-output voltage differential, barely above the maximum dropout voltage for that current rating. You can run the part from larger supply voltages if the proper precautions are taken. Varying the duty cycle using the enable pin can increase the power dissipation of the device by maintaining a lower average power figure. This is ideal for applications where high current is only needed in short bursts. Figure 1 shows the safe operating regions for the MIC5219-x.xBM5 at three different ambient temperatures and at different output currents. The data used to determine this figure assumed a minimum footprint PCB design for minimum heat sinking. Figure 2 incorporates the same factors as the first figure, but assumes a much better heat sink. A 1" square copper trace on the PC board reduces the thermal resistance of the device. This improved thermal resistance improves power dissipation and allows for a larger safe operating region. Figures 3 and 4 show safe operating regions for the MIC5219- x.xBMM, the power MSOP package part. These graphs show three typical operating regions at different tempera- tures. The lower the temperature, the larger the operating region. The graphs were obtained in a similar way to the graphs for the MIC5219-x.xBM5, taking all factors into con- sideration and using two different board layouts, minimum footprint and 1" square copper PC board heat sink. (For further discussion of PC board heat sink characteristics, refer to Application Hint 17, “Designing PC Board Heat Sinks”.) The information used to determine the safe operating regions can be obtained in a similar manner to that used in determin- ing typical power dissipation, already discussed. Determin- ing the maximum power dissipation based on the layout is the first step, this is done in the same manner as in the previous two sections. Then, a larger power dissipation number multiplied by a set maximum duty cycle would give that maximum power dissipation number for the layout. This is best shown through an example. If the application calls for 5V at 500mA for short pulses, but the only supply voltage available is 8V, then the duty cycle has to be adjusted to determine an average power that does not exceed the maximum power dissipation for the layout. Avg.P = % DC 100 V – V I V I DIN OUT OUT IN GND () + 455mW = % DC 100 8V – 5V 500mA 8V 20mA () +× 455mW = % Duty Cycle 100 1.66W 0.274 = % Duty Cycle 100 % Duty Cycle Max = 7.4% 2 With an output current of 500mA and a three-volt drop across the MIC5219-xxBMM, the maximum duty cycle is 27.4%. Applications also call for a set nominal current output with a greater amount of current needed for short durations. This is a tricky situation, but it is easily remedied. Calculate the average power dissipation for each current section, then add the two numbers giving the total power dissipation for the regulator. For example, if the regulator is operating normally at 50mA, but for 12.5% of the time it operates at 500mA output, the total power dissipation of the part can be easily determined. First, calculate the power dissipation of the device at 50mA. We will use the MIC5219-3.3BM5 with 5V input voltage as our example. P D × 50mA = (5V – 3.3V) × 50mA + 5V × 650µA P D × 50mA = 173mW However, this is continuous power dissipation, the actual on-time for the device at 50mA is (100%-12.5%) or 87.5% of the time, or 87.5% duty cycle. Therefore, P D must be multiplied by the duty cycle to obtain the actual average power dissipation at 50mA. |
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