TDA7384A: Sink size, power dissipation and load

[cperuj23]

Posted on July 31, 2015 at 18:58

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

[Nickname1623_O]

Posted on August 03, 2015 at 05:05

Hi 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

[cperuj23]

Posted on August 03, 2015 at 12:16

Hello YT!

Thanks for your quick response! It has been very useful

[cperuj23]

Posted on August 07, 2015 at 11:33

Hello,

I have another doubt of TDA7384A. For calculate the power supply:

First, I have looked the Figure 8 and for a power supply of 12V I have 18W of power output (THD 10%). Secondly, I have looked the Figure 13 and the total power dissipation (for the four channels in 13.2V) is 35W for 18W of power output. Is that means that I need a power supply of (PS = (18*4) + 35) 107W (approx.)? or Are there another method to calculate the power supply? (PS = Io^2*Rload)

Sorry for all of questions, the University teachs a lot of theoretical aspects but very little practical experience.

Thanks,

Carlos

[Nickname1623_O]

Posted on August 25, 2015 at 09:34

Hi Carlos ,

Yes, in theroy  you need 107W power supply to  provide enough power to amplifier for 4 channels 18W continous output power .

due to music signal is dynamic signal and the average power is around 1/3 continous output power ,or you can consider to delivery full power continously in stereo channel , so you can use a 70W (12V @ 5A ) supply to reach similar speaker output loudness, of cource , you need to put big VCC bypass capacitor close to power amplifier IC to provide high transient current.

Thanks,

YT

[vinod]

Posted on September 28, 2015 at 09:51

Hi,

 If we use this formula Pd= 2*(VCC^2)/((3.14^2)*RL) 

for VCC=13.2V and RL=4 ohm than Pd=8.82W and for 4 channel it will be 4X8.82W=35.30W.

But from datasheet fig 8 at 13.2V Pout will be 22Watt@10% THD and from Fig 13 power dissipation at 22W will 30Watt.

There is a ~5Watt difference between formula and datasheet. Please tell which Pd should be used?

regards

Vinod

[Nickname1623_O]

Posted on October 10, 2015 at 03:22

hi Sharma,

in fact ,there are no conflict between those data, since the max dissipation power is at 1/3 output power , and with max output power , the efficiency of ampilifer is best, then power dissipation had been reduced .

thanks

YT

[scott23]

Posted on May 09, 2016 at 09:02

[scott23]

Posted on May 14, 2016 at 10:11