Needing PSU!!! help

[KettermanJ]

I have a problem. I am building the TDA8920BJ amplifier and it requires a +/- 25v for operation. I dont have the time to build my own SMPS. Does anyone know where i can get a PSU that will power this bad boy. Its 2x100w rms. Im doing this for my senior project in high school.

-Josh

[AndrewT]

Hi,
buy a transformer, rectifier and smoothing capacitors.

for 200W total output, 200 to 300VA will do.
for +-25Vdc look for 18-0, 18-0Vac. If you can't find that, then 20-0, 20-0Vac will go a little higher.
25A 200V bridge rectifier.
4off 10mF 50V capacitors. 63V capacitors will do just as well but may be bulkier and more expensive. 40V caps are difficult to find cheaply and 35V are too low in case your voltage rises during light power demands (at night?).

BTW,
you can't believe that 100W spec from +-25Vdc, somebody is pulling your leg.

[juro]

it is D-class amplifier.. som thne 100w Rms its possible

[AndrewT]

Hi,
100W requires 40Vpk and 5Apk into an 8r0 load.
You can't get that out of 25Vdc.

Similarly 100W into 4r0 requires 28.3Vpk and 7.1Apk and that can't come from 25Vdc either.

[TheMG]

Simple formulas really.

Vrms=Vpeak/sqrt(2)

P=Vrms^2/R


So: Vrms=25V/sqrt(2)=17.68V
P=17.68V^2/4=78.15W

The highest power an amplifier with a +/- 25V supply can achieve into a 4 ohm resistive load theoretically is 78W RMS.

[AndrewT]

Originally posted by TheMG
Simple formulas really.

Vrms=Vpeak/sqrt(2)

P=Vrms^2/R


So: Vrms=25V/sqrt(2)=17.68V
P=17.68V^2/4=78.15W

The highest power an amplifier with a +/- 25V supply can achieve into a 4 ohm resistive load theoretically is 78W RMS.

any builder will be very "lucky" to get better than 50W into 4r0 (25W into 8r0) from +-25Vdc.
There are many very significant losses between the mains outlet socket and the driver terminals.

Juro,
you have gone very quiet.
Are you developing a new theory of the universe to show that digital amps can perform miracles.

[robertj88]

I have looked up the datasheet:




So is does 100Watt @ 4ohm @ 10% THD

But it can drive 3ohm loads, witch give:
P=(25/sqrt(2))^2/3=104Watt

But I go with AndrewT, it's quite impossible....

[AndrewT]

the clue is in the graphs.
using the 0.5% distortion limits you'll that double the power is delivered into double the impedance fir the BTL arrangement.
They are quoting bridged power.

So it looks like they are claiming about 55W into 4ohms for the 25Vdc supply.

Learn to read. Then learn to interpret. That why we are civilised and the animals aren't.

On interpretation.
Is the 55W claim based on an ideal underload supply voltage at the power input pins?

How high will the voltage be on the unloaded PSU when the amp is quiescent or disconnected?

[valleyman]

Looking at your calculations I notice what I think is an oversight though I am new to this so please correct me if I'm wrong. The OP says he wants a +/-25V supply and I think we can safely assume he means +/-25V DC.

With a DC voltage (assuming it's smooth) Vrms=Vpeak as the graph isn't a sine wave, its just a straight line.

This would imply that:

P= Vrms^2/R = 25^2/4 = 156W max

Please don't rip me to shreds if I'm wrong, this just seems to make sense from my (admittedly fairly meagre) knowledge about electronics and I'm hoping either to be right or to learn something


Edit: I understand better now. My calculations refer to the top case in the pic below, the previous calculations refer to the bottom case:
http://en.wikipedia.org/wiki/Image:T...entDiagram.png

Assuming a steady +/-25V supply as in the top case in that picture, is there any reason you couldn't get 100W from a +/-25V supply?

[AndrewT]

Hi,
100W into 8r0 requires 40Vpk and 5Apk.
You cannot get that with a +-25Vdc supply.
Most 100W into 8r0 amps runs on near +-50Vdc supplies.