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MCP1701 датащи(PDF) 11 Page - Microchip Technology |
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MCP1701 датащи(HTML) 11 Page - Microchip Technology |
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11 / 20 page ![]() 2004 Microchip Technology Inc. DS21874A-page 11 MCP1701 5.0 THERMAL CONSIDERATIONS 5.1 Power Dissipation The amount of power dissipated internal to the LDO linear regulator is the sum of the power dissipation within the linear pass device (P-channel MOSFET) and the quiescent current required to bias the internal refer- ence and error amplifier. The internal linear pass device power dissipation is calculated as shown in Equation 5-1. EQUATION 5-1: The internal power dissipation, which is due to the bias current for the LDO internal reference and error ampli- fier, is calculated as shown in Equation 5-2. EQUATION 5-2: The total internal power dissipation is the sum of PD (Pass Device) and PD (Bias). EQUATION 5-3: For the MCP1701, the internal quiescent bias current is so low (2 µA, typical) that the PD (Bias) term of the power dissipation equation can be ignored. The maximum power dissipation can be estimated by using the maximum input voltage and the minimum output voltage to obtain a maximum voltage differential between input and output. The next step would be to multiply the maximum voltage differential by the maximum output current. EQUATION 5-4: To determine the junction temperature of the device, the thermal resistance from junction-to-ambient must be known. The 3-pin SOT-23 thermal resistance from junction-to-air (RθJA) is estimated to be approximately 335°C/W. The SOT-89 RθJA is estimated to be approximately 52°C/W when mounted on 1 square inch of copper. For the TO-92, RθJA is estimated to be 131.9°C/W. The RθJA will vary with physical layout, airflow and other application-specific conditions. The device junction temperature is determined by calculating the junction temperature rise above ambient, then adding the rise to the ambient temperature. EQUATION 5-5: JUNCTION TEMPERATURE - SOT-23 EXAMPLE: EQUATION 5-6: JUNCTION TEMPERATURE - SOT-89 EXAMPLE: EQUATION 5-7: JUNCTION TEMPERATURE - TO-92 EXAMPLE: PD (Pass Device) = (VIN – VOUT) x IOUT PD (Bias) = VIN x IGND PTOTAL = PD (Pass Device) + PD (Bias) PD = (VINMAX – VOUTMIN) x IOUTMAX Given: VIN = 3.3V to 4.1V VOUT =3.0V ± 2% IOUT = 1 mA to 100 mA TAMAX = 55°C PMAX = (4.1V – (3.0V x 0.98)) x 100 mA PMAX = 116.0 milliwatts TJ PDMAX RθJA TA + × = TJ 116.0 milliwatts 335°C/W 55°C + × = TJ 93.9°C = TJ 116.0 milliwatts 52°C/W 55°C + × = TJ 61°C = TJ 116.0 milliwatts 131.9°C/W 55°C + × = TJ 70.3°C = |
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