Datasheet LT3014 (Analog Devices) - 10

FabricanteAnalog Devices
Descripción20mA, 3V to 80V Low Dropout Micropower Linear Regulator
Páginas / Página16 / 10 — APPLICATIONS INFORMATION. Calculating Junction Temperature
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APPLICATIONS INFORMATION. Calculating Junction Temperature

APPLICATIONS INFORMATION Calculating Junction Temperature

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LT3014
APPLICATIONS INFORMATION
Continuous operation at large input/output voltage dif- area. So the junction temperature rise above ambient will ferentials and maximum load current is not practical be approximately equal to: due to thermal limitations. Transient operation at high 0.52W input/output differentials is possible. The approximate • 50°C/W = 26°C thermal time constant for a 2500sq mm 3/32" FR-4 board The maximum junction temperature will then be equal to with maximum topside and backside area for one ounce the maximum junction temperature rise above ambient copper is 3 seconds. This time constant will increase as plus the maximum ambient temperature or: more thermal mass is added (i.e. vias, larger board, and TJMAX = 50°C + 26°C = 76°C other components). Example 2: Given an output voltage of 5V, an input voltage For an application with transient high power peaks, average of 48V that rises to 72V for 5ms(max) out of every 100ms, power dissipation can be used for junction temperature and a 5mA load that steps to 20mA for 50ms out of every calculations as long as the pulse period is signifi cantly less 250ms, what is the junction temperature rise above ambi- than the thermal time constant of the device and board. ent? Using a 500ms period (well under the time constant of the board), power dissipation is as follows:
Calculating Junction Temperature
P1(48V in, 5mA load) = 5mA • (48V – 5V) Example 1: Given an output voltage of 5V, an input volt- + (100μA • 48V) = 0.22W age range of 24V to 30V, an output current range of 0mA to 20mA, and a maximum ambient temperature of 50°C, P2(48V in, 20mA load) = 20mA • (48V – 5V) what will the maximum junction temperature be? + (0.55mA • 48V) = 0.89W The power dissipated by the device will be equal to: P3(72V in, 5mA load) = 5mA • (72V – 5V) + (100μA • 72V) = 0.34W IOUT(MAX) • (VIN(MAX) – VOUT) + (IGND • VIN(MAX)) P4(72V in, 20mA load) = 20mA • (72V – 5V) where: + (0.55mA • 72V) = 1.38W IOUT(MAX) = 20mA Operation at the different power levels is as follows: VIN(MAX) = 30V 76% operation at P1, 19% for P2, 4% for P3, and IGND at (IOUT = 20mA, VIN = 30V) = 0.55mA 1% for P4. So: P EFF = 76%(0.22W) + 19%(0.89W) + 4%(0.34W) + 1%(1.38W) = 0.36W P = 20mA • (30V – 5V) + (0.55mA • 30V) = 0.52W With a thermal resistance in the range of 40°C/W to The thermal resistance for the DFN package will be in the 62°C/W, this translates to a junction temperature rise range of 40°C/W to 62°C/W depending on the copper above ambient of 20°C. 3014fd 10