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TPA6204A1DRB データシート(PDF) 11 Page - Texas Instruments |
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TPA6204A1DRB データシート(HTML) 11 Page - Texas Instruments |
11 / 15 page TPA6204A1 SLOS429 − MAY 2004 www.ti.com 11 FULLY DIFFERENTIAL AMPLIFIER EFFICIENCY AND THERMAL INFORMATION Class-AB amplifiers are inefficient. The primary cause of these inefficiencies is voltage drop across the output stage transistors. There are two components of the internal voltage drop. One is the headroom or dc voltage drop that varies inversely to output power. The second component is due to the sinewave nature of the output. The total voltage drop can be calculated by subtracting the RMS value of the output voltage from VDD. The internal voltage drop multiplied by the average value of the supply current, IDD(avg), determines the internal power dissipation of the amplifier. An easy-to-use equation to calculate efficiency starts out as being equal to the ratio of power from the power supply to the power delivered to the load. To accurately calculate the RMS and average values of power in the load and in the amplifier, the current and voltage waveform shapes must first be understood (see Figure 18). V(LRMS) VO IDD IDD(avg) Figure 18. Voltage and Current Waveforms for BTL Amplifiers Although the voltages and currents for SE and BTL are sinusoidal in the load, currents from the supply are different between SE and BTL configurations. In an SE application the current waveform is a half-wave rectified shape, whereas in BTL it is a full-wave rectified waveform. This means RMS conversion factors are different. Keep in mind that for most of the waveform both the push and pull transistors are not on at the same time, which supports the fact that each amplifier in the BTL device only draws current from the supply for half the waveform. The following equations are the basis for calculating amplifier efficiency. Efficiency of a BTL amplifier + P L P SUP Where: P L + V L rms2 R L , and V LRMS + V P 2 , therefore, P L + V P 2 2R L PL = Power delivered to load PSUP = Power drawn from power supply VLRMS = RMS voltage on BTL load RL = Load resistance VP = Peak voltage on BTL load IDDavg = Average current drawn from the power supply VDD = Power supply voltage ηBTL = Efficiency of a BTL amplifier and PSUP + VDD IDDavg and IDDavg + 1 p p 0 V P R L sin(t) dt +* 1p V P R L [cos(t)] p 0 + 2V P p R L Therefore, P SUP + 2V DD V P p R L substituting PL and PSUP into equation 6, Efficiency of a BTL amplifier + V P 2 2R L 2V DD VP p R L + p V P 4V DD V P + 2P L R L Where: h BTL + p 2P L R L 4V DD Therefore, (4) (5) |
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