Needing PSU!!! help

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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
 
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.
 
TheMG said:
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.
 
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?
 
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:TypesOfDirectCurrentDiagram.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 said:

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.


Which is why I said theoretically. ;)

Although now that I think about it, 100W would be plenty possible from a full-bridge configuration (which the amp being discussed is not).

From the datasheet it specifies +/- 27V rails, with 110W output power with a 3r0 load, at 10% THD, which I guess is more reasonable.
 
TheMG said:
From the datasheet it specifies +/- 27V rails, with 110W output power with a 3r0 load, at 10% THD, which I guess is more reasonable.
not reasonable.
110W into 3r0 requires 25.7Vpk and 8.6Apk.
Using a regulated +-27Vdc, one could get that power, but I suspect that neither the regulator nor the amplifier can pass 8.6Apk.
With an unregulated +-25Vdc the supply is likely to sag to around 20 to 22Vdc average when delivering full power.
From this one has to subtract an allowance for supply ripple when delivering high current from the smoothing caps. Let's assume that the ripple is 2Vpp. Subtract half ripple from average to obtain the supply voltage minimum at the PSU. I'll guess around 21Vmin.
Now take off the losses through all the cabling and amplifier. This could be around 4 to 6V.
This leaves about 16Vpk at the load.
maximum output power = Vpk*Vpk/R/2. In this example that is 16^2/16=16W into 8r0.
To get 25W into 8r0 one would need 20Vpk at the load and when all the losses are taken into account it will take a fair bit of build/design expertise to keep the supply rails @ +-25Vdc and still deliver that target power.

BTW,
the more usual distortion limit for maximum power is 0.1% not 10%.
only disreputable manufacturers who have no conscience about advertising inflated power claims would use 10% distortion.
 
valleyman said:
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:TypesOfDirectCurrentDiagram.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?

Hi valleyman,

You could probably get something like those output-power figures with **DC** output signals, for a load between V+ and ground for example (if you also take into account the maximum output voltage swing versus the power rails' voltages, for the chipamp you're using).

But for typical audio amplifier output-power specs, an **AC** output signal is assumed.

For example, if your amplifer were powered by +/-25 VDC, and you somehow managed to get it to swing from rail to rail (normally it won't even get close), then the RMS voltage of the AC output would be 25/sqrt(2) = 17.68 Vrms, and power would be Vrms^2/R, i.e. 78 Watts into 4 Ohms or 39 Watts into 8 Ohms.

Working the other way, to get 100 Watts, i.e. Vrms^2/R = 100W, then you would need output Vrms=sqrt(100R), which would be 20 Vrms for 4 Ohms or 28.28 Vrms for 8 Ohms. And Vpeak = sqrt(2) x Vrms. So, for 100W, you would need 28.28 Vpeak for 4 Ohms, or 40 Vpeak for 8 Ohms. The DC supply rails would need to be more than Vpeak, to account for the chipamp's max output voltage versus power supply voltage spec, per the chipamp's datasheet.

For good, concise information about what DC voltage and ripple to expect from a given linear power supply, look here:

http://www.zen22142.zen.co.uk/Design/dcpsu.htm

But don't forget to calculate for worst-case mains-voltage excursions, and include 'ripple average versus load current' and transformer regulation effects.

For information about how close to the power rails a particular chipamp's output signal can swing, see the chipamp's datasheet.
 
KettermanJ said:
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


I believe he asked where. A likely spot is www.alliedelec.com.
They're very nice over the phone.
Check out the Stancor/White Rogers, and the Hammond transformer lineups.
Check out this http://www.alliedelec.com/Search/ProductDetail.asp?SKU=266-0047 bridge rectifier unit.
Have a look at the quality Mallory and Nichicon caps available at 50v and 63v, in sufficient capacitance values for your power supply.
They have one heatsink large enough for amplifier projects.

Here is another power supply option:
http://www.electronics123.com/s.nl/it.A/id.1715/.f

Check out the basic diagram at the bottom of this file:
http://electronics123.net/amazon/datasheet/k114.pdf

For more elegant power supplies, you can find them with a search here at diyaudio.com.
Edit: One of them is available in kit form at www.chipamp.com
 
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