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AAT3201IGV-2.7-T1 Datasheet(PDF) 11 Page - Advanced Analogic Technologies |
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AAT3201IGV-2.7-T1 Datasheet(HTML) 11 Page - Advanced Analogic Technologies |
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11 / 14 page  AAT3201 150mA OmniPower™ LDO Linear Regulator 3201.2002.3.0.91 11 TA = 25°C. Given TA = 85°, the maximum package power dissipation is 267mW. At TA = 25°C°, the maximum package power dissipation is 667mW. The maximum continuous output current for the AAT3201 is a function of the package power dissi- pation and the input to output voltage drop across the LDO regulator. Refer to the following simple equation: IOUT(MAX) < PD(MAX) / (VIN - VOUT) For example, if VIN = 5V, VOUT = 2.5V and TA = 25°, IOUT(MAX) < 267mA. The output short circuit protec- tion threshold is set between 150mA and 300mA. If the output load current were to exceed 267mA or if the ambient temperature were to increase, the inter- nal die temperature will increase. If the condition remained constant and the short circuit protection did not activate, there would be a potential damage hazard to LDO regulator since the thermal protection circuit will only activate after a short circuit event occurs on the LDO regulator output. To figure what the maximum input voltage would be for a given load current refer to the following equa- tion. This calculation accounts for the total power dissipation of the LDO Regulator, including that caused by ground current. PD(MAX) = (VIN - VOUT)IOUT + (VIN x IGND) This formula can be solved for VIN to determine the maximum input voltage. VIN(MAX) = (PD(MAX) + (VOUT x IOUT)) / (IOUT + IGND) The following is an example for an AAT3201 set for a 2.5 volt output: From the discussion above, PD(MAX) was deter- mined to equal 667mW at TA = 25°C. VOUT = 2.5 volts IOUT = 150mA IGND = 20µA VIN(MAX)=(667mW+(2.5Vx150mA))/(150mA +20µA) VIN(MAX) = 6.95V Thus, the AAT3201 can sustain a constant 2.5V out- put at a 150mA load current as long as VIN is ≤ 6.95V at an ambient temperature of 25°C. 5.5V is the max- imum input operating voltage for the AAT3201, thus at 25°C, the device would not have any thermal con- cerns or operational VIN(MAX) limits. This situation can be different at 85°C. The follow- ing is an example for an AAT3201 set for a 2.5 volt output at 85°C: From the discussion above, PD(MAX) was deter- mined to equal 267mW at TA = 85°C. VOUT = 2.5 volts IOUT = 150mA IGND = 20µA VIN(MAX)=(267mW+(2.5Vx150mA))/(150mA+20µA) VIN(MAX) = 4.28V Higher input to output voltage differentials can be obtained with the AAT3201, while maintaining device functions in the thermal safe operating area. To accomplish this, the device thermal resistance must be reduced by increasing the heat sink area or by operating the LDO regulator in a duty cycled mode. For example, an application requires VIN = 5.0V while VOUT = 2.5V at a 150mA load and TA = 85°C. VIN is greater than 4.28V, which is the maximum safe continuous input level for VOUT = 2.5V at 150mA for TA = 85°C. To maintain this high input voltage and output current level, the LDO regulator must be operated in a duty cycled mode. Refer to the following calculation for duty cycle operation: PD(MAX) is assumed to be 267mW IGND = 20µA IOUT = 150mA VIN = 5.0 volts VOUT = 2.5 volts %DC = 100(PD(MAX) / ((VIN - VOUT)IOUT + (VIN x IGND)) %DC=100(267mW/((5.0V-2.5V)150mA+(5.0Vx20µA)) %DC = 71.2% For a 150mA output current and a 2.5 volt drop across the AAT3201 at an ambient temperature of 85°C, the maximum on time duty cycle for the device would be 71.2%. The following family of curves shows the safe oper- ating area for duty cycled operation from ambient room temperature to the maximum operating level. |
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