how to read a datasheet of power amplifier

dayanand

Member
2012-07-30 1:30 pm
hi,
can anybody help me how to read a datasheet of power amplifier as
i already had a datasheet of TDA8954.I want to use this power amplifier IC as output of 330W in mono BTL mode.I am little confused with details given in datasheet.
1)In datasheet it is mentioned that in mono BTL mode,at THD of 0.5% a 330W output in dynamic characteristcs and at THD of 10% its output is 420W.

Can i configure this power amplifier ic for 330W?
if its possible to configure 330W then how can i do that?
 
Funny you should ask.

I am working with this IC and have made a board to experiment with.

The power output rating of an IC depends on how much distortion you want to allow for the rating.

Manufacturers seem to want to rate the power of IC's now days at 10% distortion levels.

I tend to take the view of power rating at 1% because that is close to the level seen on an oscilloscope where the peaks are just starting to clip...an approximation I know could start its own discussion but that is for another thread.

So when I look at an amplifier output power curve vs power supply, I look at the 1% point.

With this device operating in single BTL with an 8R load and +/- 41V regulated supply you will get lots of power. If you use a power supply with conventional cap filtering then the minimum voltage from the supply during loading will determine the power level output where it starts to clip...depends on power supply design...other threads for that elsewhere.

The difference between 330W and 420W is only 1dB. This is generally not a significant increase in sound level...many discussions on this elsewhere.

Hope that helps.
 

dayanand

Member
2012-07-30 1:30 pm
thanks for your reply:),

i want only 300watts of output from TDA8954TH ic in mono btl mode,whats in my mind is i want to power this ic with +/[email protected] of current such that it can give me 328 watts of output.
After excluding loss (93% efficiency) from this i can get 300watts output.

Is that correct what am i doing?Correct me if i am wrong?
 
That is pretty much what I will be doing.

What does your supply do at greater than 8A of load current?

Look at Fig. 25 and Fig. 16 and Fig. 23 (use line 2 (2 x 4R))

Your efficiency will be more like 90%.

So at +/-41V if you only drive it to 300W your power supply should be fine.

If you drive it more then it will start to clip and if the power supply is current limited then it will clip even more due to power supply voltage droop.

You will have to heat sink it for >30W of heat.

Sounds like a good match to me.

I am planning on stereo with one IC for each side...I will need two 8-9A or one 16-18A

:)
 

dayanand

Member
2012-07-30 1:30 pm
Thanks Dug,This gives me a nice idea
As per the fig 25 keeping low (THD+N =0.5%) noise into consideration,i thought for supplying a +/[email protected] to get 300W of output which avoids clipping the signal and i can have a noiseless 300w of amplifier.

and thanks for suggesting heatsink,anyway we must use a heatsink in order to have an High wattage(300W)effective amplifier.

If you find any more issues in this let me know......
 
we have very simple way to calculate a good conditions amp output power @0.5%THD+N, dc supply voltage times 0.6, for tda8954 application. works at SE mode,dc voltage +/- 40v , output rms voltage should be 24v, output power 2x144W/[email protected]%THD+N. of course some bad design can not get this value.

becuase single chip class D have small heatsink area, if run it with RMS rated power output, very good fan cooling system necessary. but no problems for audio playing, it only need 1/8 rated power.
 
Last edited:
thanks for your reply:),

i want only 300watts of output from TDA8954TH ic in mono btl mode,whats in my mind is i want to power this ic with +/[email protected] of current such that it can give me 328 watts of output.
After excluding loss (93% efficiency) from this i can get 300watts output.

Is that correct what am i doing?Correct me if i am wrong?

If you want 330watts into 8 ohms, you need a signal of 51V across that 8 ohms. #
That means that your power supply must be able to provide at least that, normally a bit more because you lose always a few volts in the actual output devices.

With a bridge tied load you have double the voltage across the load because not only does one side of the load get driven, the other side also gets driven in opposite way. So you would think each side only needs to provide half the level, but because the level figures as the square in the power equation # you actually need about 0.7 times the level for non-bridge, 0.7 * 51 = about 36V. With the losses, you get about the 41V mentioned before.

Note that this is only to determine the supply for that power output; it does not tell you whether the chip actually can handle it.

The other thing is the gain. For 50V (or 36) output, you need a signal input which is that, divided by the gain. For a usual gain of say 28, your input needs 51/28=about 1.8V. Again, this in itself doesn't determine the power you will be able to get from this chip, but it tells you that with a specific gain and output power you need this at the input.

Jan

#The equation is: P (330W) = V^2/R, so V^2 =P * R so V = SqrRt (P * R).