Hello,

I'm modifying a device that no works very well. This device includes one amplifier TDA7384A, and in the old design I found several mistakes or parameters that haven't calculated. I saw the datasheet and i didn't know how can i calculate some parameters with the information in the datasheet:

- If the power supply is 12V, How can I calculate the power dissipated?

- Apart of power dissipated, for calculate the size of the sink, Is not necessary to know the Rcd of the Flexiwatt25?

- The device have a load resistance of 8 Ohm and the TDA7384A is designed for 4 Ohm. Can a 8 Ohm load modify the response of the amplifier or only decrease the power?

Thanks

Carlos

I'm modifying a device that no works very well. This device includes one amplifier TDA7384A, and in the old design I found several mistakes or parameters that haven't calculated. I saw the datasheet and i didn't know how can i calculate some parameters with the information in the datasheet:

- If the power supply is 12V, How can I calculate the power dissipated?

- Apart of power dissipated, for calculate the size of the sink, Is not necessary to know the Rcd of the Flexiwatt25?

- The device have a load resistance of 8 Ohm and the TDA7384A is designed for 4 Ohm. Can a 8 Ohm load modify the response of the amplifier or only decrease the power?

Thanks

Carlos

the power dissipation could be calculated with below formula , TDA7384 is single supply BTL output configuration power amplifier , then the power dissipation is :Pd= 2*(VCC^2)/((3.14^2)*RL).

to calculate the heatsink size , you need to check the thermal impedance between heatsink and ambient , then you can consider that the power dissipation is as current , thermal impedance is as resistor , then ambient temperature , IC case temperature and IC junction temperature as voltage potential level. when the target temperature had been set , you can choose suitable heatsink. the Rth of package is 1C/W , means that the temperature difference between junction and case will be 10C , when dissipation power is 10W .

the 8 ohm load will just decrease the output power .

Thanks ,

YT