How to evaluate the dynamic efficiency of Class
The biggest feature of Class D audio power amplifier is high efficiency. The advantages of high efficiency are power saving and lower heat generation. If the efficiency of the power amplifier is 90% and the chip package can dissipate 1W, then the power amplifier can output about 10W of power, which provides great convenience in system design. The efficiency of Class D power amplifiers can be viewed from different perspectives. The main reason for the high efficiency is the low on-resistance Rds (on) of the output power transistor. If the on-resistance is 0.4 ohm and the speaker impedance is 4 ohm, the efficiency of the output transistor is equal to 91%. But there are other consumptions of the power amplifier, including the consumption of analog circuits, the consumption of analog-digital mixing circuits and the consumption of digital circuits. These consumptions are reflected in the no-load quiescent current Iq. If the power supply voltage is 5V and Iq is 5mA, the static power consumption is 25mW. Therefore, the calculation of the efficiency of the power amplifier needs to consider both the static consumption and the consumption of the output transistor. If a load is added to output 25mW of power and the efficiency of the output power transistor is 90%, the power transistor consumes about 2.5mW. Therefore, the total power consumption at the output of 25mW includes static consumption, power transistor consumption and output power, that is, 25mW + 25mW + 2.5mW = 52.5mW, the efficiency at this moment is 25mW / 52.5mW = 48%. In the same way, if the output is 250mW, the total power consumption is 25mW + 250mW + 25mW = 300mW, and the efficiency at this moment is 250mW / 300mW = 83%. Similarly, if the output is 2.5W, the total power consumption is 25mW + 2.5W + 250mW = 2.775W, and the current efficiency is 2.5W / 2.775W = 90%, which is close to the efficiency of the output power transistor. So when the output power is small, the consumption of the output transistor can be ignored and when the output power is large, the static consumption can be ignored. If the load is equal to 4 ohms and the output transistor with an efficiency higher than 90% has an on-resistance of only 0.4 ohms or less, if it is a BTL double-ended output, Rds (on) is caused by a PMOS and an NMOS, and each MOS has only about The on-resistance of 0.2 ohms can easily cause measurement errors. When measuring, use a thick wire for the large current path and solder it to reduce the wiring resistance. Maintain the stability of the power supply voltage when drawing large currents. Since the measurement of the output power transistor's on-resistance is prone to errors and its resistance is related to the measurement conditions such as current or voltage, the best way to observe the efficiency of the output power transistor is to view the efficiency at the large output power end of the efficiency curve. The efficiency at one point is very close to the efficiency of the output power transistor, but it should be noted that this curve must be obtained under the condition of using a resistive load. If Pi represents the power input power, Pq represents the static power consumption, Po represents the output power, Emos represents the efficiency of the output power transistor and Eff represents the total efficiency, then the relationship is Po = (Pi-Pq) x Emos and Eff = Po / Pi Combining the above formulas, the total efficiency Eff = (Po x Emos) / (Po + Emos x Pq) If the ratio of the output power Po to the static power consumption Pq is Poq, then Eff = (Poq x Emos) / (Poq + Emos) This simple formula makes it easy to observe the relationship between output power and total efficiency. So if the efficiency of the power transistor Emos = 90% and the output power Po is 10 times the static power consumption Pq, that is, Poq = 10, then the total efficiency Eff = (10 x 0.9) / (10 + 0.9) = 83%. If the efficiency of the power transistor Emos is also 90% and the output power Po is 8 times the static power consumption Pq, then the total efficiency Eff = (8 x 0.9) / (8 + 0.9) = 81%. The above example shows that the ratio of the output power to the static power consumption of the same class D amplifier determines the overall efficiency, that is, the static current consumption and total efficiency have a certain degree of influence. In addition, if the efficiency of the power transistor Emos = 90%, a static power consumption of 25mW or 5V voltage 5mA static current consumption power amplifier as long as the output power is greater than 25mA x 8 = 200 mW can have 81% efficiency. But if the static power consumption is 50mW or 5V voltage 10mA static current consumption power amplifier, the output power must be greater than 50mA x 8 = 400 mW to have the same efficiency, so the actual efficiency needs to consider the static current consumption. The above discussion does not include the consumption of the output filter. If the circuit diagram contains the output filter, the consumption of the output filter should be counted as the static power consumption.