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TDF8599BTH датащи(PDF) 35 Page - NXP Semiconductors |
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TDF8599BTH датащи(HTML) 35 Page - NXP Semiconductors |
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35 / 54 page ![]() NXP Semiconductors TDF8599B I2C-bus controlled dual channel 43 W/2 Ohm, single channel 85 W/1 Ohm class-D power amplifier with load diagnostics TDF8599B All information provided in this document is subject to legal disclaimers. © NXP Semiconductors N.V. 2016. All rights reserved. Product data sheet Rev. 2 — 23 August 2016 35 / 54 Table 22. Filter component values Load impedance (Ω) LLC (μH) CLC (μF) 1 2.5 4.4 2 5 2.2 4 10 1 Remark: When using a 1 Ω load impedance in Parallel mode, the outputs are shorted after the low-pass filter switches two 2 Ω filters in parallel. 14.5 Heat sink requirements In most applications, it is necessary to connect an external heat sink to the TDF8599B. Thermal foldback activates at Tj = 140 °C. The expression below shows the relationship between the maximum power dissipation before activation of thermal foldback and the total thermal resistance from junction to ambient: (7) Pmax is determined by the efficiency (η) of the TDF8599B. The efficiency measured as a function of output power is given in Figure 43. The power dissipation can be derived as a function of output power (see Figure 42). Example 1: • VP = 14.4 V • Po = 2 × 25 W into 4 Ω (THD = 10 % continuous) • Tj(max) = 140 °C • Tamb = 25 °C • Pmax = 5.8 W (from Figure 42) • The required Rth(j-a) = 115 °C / 5.8 W = 19 K/W The total thermal resistance Rth(j-a) consists of: Rth(j-c) + Rth(c-h) + Rth(h-a) Where: • Thermal resistance from junction to case (Rth(j-c)) = 1 K/W • Thermal resistance from case to heat sink (Rth(c-h)) = 0.5 K/W to 1 K/W (depending on mounting) • Thermal resistance from heat sink to ambient (Rth(h-a)) would then be 19 - (1 + 1) = 17 K/W. If an audio signal has a crest factor of 10 (the ratio between peak power and average power = 10 dB) then Tj will be much lower. Example 2: • VP = 14.4 V • Po = 2 × (25 W / 10) = 2 × 2.5 W into 4 Ω (audio with crest factor of 10) • Tamb = 25 °C • Pmax = 2.5 W • Rth(j-a) = 19 K/W • Tj(max) = 25 °C + (2.5 W × 19 K/W) = 72 °C |
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