The selection of sound or amplifier operating current and inductance is written too well
Time:2023-12-24
Views:447
Figure 1
Calculate the currents at points A, B, and C separately?
According to Figure 1, it is known that the power supply voltage of the amplifier is PVDD=12V; The battery is powered by a single lithium battery.
PVDD=Ub=12V
So Uc=6V
So AC_ Uc=Uc * 1.414=6 * 1.414=8.484V
So Ic=10W/8.484V=1.179A
So Ib=2 * Ic=2 * 1.179A=2.36A
Due to energy conservation, Pa=Pb (approximately), Pb=12V * 2.36A=28.32W;
So, Max Ia=28.32W/3.3V=8.58A;
Mix Ia=28.32W/4.2V=6.74A;
Summary of Key Component Selection:
1) The battery capacity should be selected with large capacity, strong discharge capacity, low internal resistance, and a lithium battery protection board with an overcurrent of 13A or above;
2) DC-DC, under the condition that the chip meets the requirements, the inductor should be selected with a rated current of 12A or above;
3) The voltage withstand value of the electrolytic capacitor at the output end of DC-DC should be selected as 25V, including the capacitance of the power amplifier supply pin;
4) The selection of LC requires the rated current of the inductor to be 2.3A or above, and the capacitor withstand voltage to be 16V or 25V;
Drawing inferences from one case to another
If we follow the circuit shown in Figure 1 and change the single section lithium battery to a double section lithium battery (please note whether the DC-DC and capacitor withstand voltage meet the requirements in practical applications), as shown in the following figure.
Figure 2
Just like above, the battery has been changed to a dual lithium battery for power supply.
Calculate the currents at points A, B, and C separately?
PVDD=Ub=12V
So Uc=6V (when the power amplifier is working)
So AC_ Uc=Uc * 1.414=6 * 1.414=8.484V
So Ic=10W/8.484V=1.179A
So Ib=2 * Ic=2 * 1.179A=2.36A
Due to energy conservation, Pa=Pb (approximately), Pb=12V * 2.36A=28.32W;
So, Max Ia=28.32W/6.8V=4.16A;
Mix Ia=28.32W/4.2V=3.37A;
Summary of Key Component Selection:
1) The battery capacity should be selected with large capacity, strong discharge capacity, low internal resistance, and a lithium battery protection board with an overcurrent of 7A or above;
2) DC-DC, under the condition that the chip meets the requirements, the inductor should be selected with a rated current of 7A or above;
3) The voltage withstand value of the electrolytic capacitor at the output end of DC-DC should be selected as 25V, including the capacitance of the power amplifier supply pin;
4) The selection of LC requires the rated current of the inductor to be 2.3A or above, and the capacitor withstand voltage to be 16V or 25V;
The voltage of the battery is different above, but below, we still choose a single lithium battery. We change the power supply of the amplifier, which means we need to change the output voltage of the DC-DC, as shown in the following figure.
Figure 3
Calculate the currents at points A, B, and C separately?
According to Figure 1, it is known that the power supply voltage of the amplifier is PVDD=15V; The battery is powered by a single lithium battery.
PVDD=Ub=15V
So Uc=7.5V
So AC_ Uc=Uc * 1.414=7.5 * 1.414=10.605V
So Ic=10W/8.484V=0.94A
So Ib=2 * Ic=2 * 0.94A=1.888A
Due to energy conservation, Pa=Pb (approximately), Pb=15V * 1.888A=28.32W;
So, Max Ia=28.32W/3.3V=8.58A;
Mix Ia=28.32W/4.2V=6.74A;
Summary of Key Component Selection:
1) The battery capacity should be selected with large capacity, strong discharge capacity, low internal resistance, and a lithium battery protection board with an overcurrent of 13A or above;
2) DC-DC, under the condition that the chip meets the requirements, the inductor should be selected with a rated current of 12A or above;
3) The voltage withstand value of the electrolytic capacitor at the output end of DC-DC should be selected as 25V, including the capacitance of the power amplifier supply pin;
4) The selection of LC requires the rated current of the inductor to be 2.3A or above, and the withstand voltage of the capacitor to be 25V;
The above text was written by Xin Ao Electronics engineers. If readers have any questions, they can leave a message.
After reading this article, we should know the following points:
The above parameters are only a rough estimate, and there may be slight deviations from the actual test results, which is considered normal;
The capacitance withstand voltage value is generally selected to be at least twice the circuit voltage. If the structural space is limited, the withstand voltage should be at least 1.5 times;
The rated current of the inductor is generally at least 1.5 times the circuit current, and the larger the inductor current, the more stable it is;
If readers have any questions about LC parameters, they can refer to previous articles;
The above parameters do not take into account the operating current, efficiency, and heat generation of the chip itself, so they can only be used as reference values.
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