The biggest feature of Class D audio power amplifier is high efficiency, and the advantage of high efficiency is to save electricity and reduce heat. If the efficiency of the power amplifier is 90% and the chip package can dissipate 1W, the power amplifier can output about 10W of power, which provides great convenience for system design.

    The efficiency of Class D power amplifiers can be looked at from different angles. The main reason for the high efficiency comes from the low on-resistance Rds (on) of the output power transistor, if the on-resistance is 0.4 ohms and the horn impedance is 4 ohms, the efficiency of the output transistor is equal to 91%, but there are other power amplifier consumption, including the consumption of analog circuits, analog and digital hybrid circuit consumption and digital circuit consumption. These costs are manifested in the no-load static current Iq. If the supply voltage is 5V and the 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 output transistor consumption.

    If the load is added and the output power is 25mW and the efficiency of the output power transistor is 90%, the power transistor consumption is about 2.5mW. Therefore, the total power consumption under the output of 25mW includes static consumption, power transistor consumption and output power, that is, 25mW + 25mW + 2.5mW = 52.5mW, and the efficiency at this moment is 25mW/52.5mW = 48%. In the same way, if the output is 250mW, the total power consumed 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 efficiency at this moment is 2.5W /2.775W= 90%, which is close to the efficiency of the output power transistor. Therefore, the output transistor consumption can be ignored when the output power is small and the static consumption can be ignored when the output power is large.

    If the load is equal to 4 ohms, the output transistor with an efficiency higher than 90% has an on-resistance of 0.4 ohms or less, and if it is a double-ended output of BTL, the Rds (on) is caused by a PMOS and an NMOS, each of which has an on-resistance of about 0.2 ohms, so it is easy to cause measurement errors. When measuring the large current path, the thick wire should be selected and welded at the same time to reduce the connection resistance. Maintain the stability of the supply voltage when drawing large current. Since the measurement of the on-resistance of the output power transistor is prone to error and its resistance is related to the measurement conditions such as current flow or voltage, the best way to look at the efficiency of the output power transistor is to look at the efficiency of the large output power end of the efficiency curve, which is very close to the output power transistor efficiency. However, it should be noted that this curve must be obtained under the condition of using a resistive load.

    If Pi stands for power input, Pq stands for static power consumed, Po stands for output power, Emos stands for the efficiency of the output power transistor, and Eff stands for the total efficiency, then the relationship is

    Po = (pi-pq) x Emos and Eff = Po/Pi

    Combined with the above formulas, the total efficiency

    Eff = (Po x Emos)/(Po + Emos x Pq)

    If the ratio of output power Po to static power consumed Pq is Poq, then

    Eff = (Poq x Emos)/(Poq + Emos)

    This simple formula makes it easy to observe the relationship between output power and overall efficiency.

    So if the efficiency of the power transistor Emos = 90% and the output power Po is 10 times the static power consumed 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 consumed Pq, then the total efficiency Eff = (8 x 0.9)/(8 + 0.9) = 81%. The above example shows that the ratio of output power and static power consumption of the same class D amplifier determines the total efficiency, that is, the static consumption current pair has a certain degree of influence on the total efficiency. In addition, if the efficiency of the power transistor Emos = 90%, a power amplifier with a static consumption power of 25mW or a 5V voltage of 5mA static consumption current can have 81% efficiency as long as the output power is greater than 25mA x 8 = 200 mW. However, 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 real efficiency needs to consider the static current consumption. The above discussion does not include the output filter consumption. If the circuit diagram contains the output filter, the output filter consumption should be counted as static power consumption.