Powering Two TDA7294 Amplifiers Using a Single +65V Supply—Is It Possible?

Need some help fellow electrical engineers:
 I have this dual power supply circuit pictured here, with the only difference being that my voltage rails are at +/- 65 volts instead of +/- 40.  Ultimately, I need to power two TDA7294 power amplifiers from this.  So, in essence, I need two dual power supplies from this setup.    Now, my logic is that since I have the +/- 65 volts, I should be able to divide that into two +30 volt rails, two -30 volt rails, and my existing ground.  I've experimented with a simple two resistor voltage divider, and that does give me the desired +30 volts, but dividing 2 different voltages at the same time using this method is exceedingly inefficient, especially for a power amplifier that is very power hungry. 
I hope I've explained this clearly enough, and hopefully someone out there has some advice, I've spent a lot of time researching this but only get so far as a single dual power supply.  Again, my final need is two separate dual power supplies from this single dual power supply, or perhaps some other alternative to my existing architecture.  Any ideas are welcome.  Thanks!Part 2: My second, less important question, is why am I getting +/- 65 volts from this setup?  The secondary of my transformer yields 95 volts AC with a center ground.  I have hooked this up to the exact circuit pictured.  When I measure the outputs after the capacitors,  the positive lead to ground, and the negative lead to ground, I get +/- 65 volts DC respectively.  When I measure the DC output directly from the bridge rectifier, I get about 85 volts DC.  Why am i dropping 10 volts across the rectifier? And then why am I losing another 20 volts across the capacitors?  Is it because it's a dual power supply?

Hi Rick-I think the more important question is "what's wrong with the power supply?"  It's normal for the secondary voltage to be a little high when there's no load, but 95 volts out of a secondary that's supposed to be running 60 volts is just too much.  What is the part number of the transformer?As for your question about "losing voltage across that capacitors"- you're not losing voltage at all.  Take a closer look.  The rectifier looks like a full-wave bridge but it's not.  It's actually TWO full-wave center tap rectifiers- one providing a positive voltage and the other providing a negative voltage.  If the transformer was actually putting out 30-0-30, then I would expect the output voltage across each capacitor to be 30 X 1.414 =  42 volts, minus about a volt for the forward voltage drop in the rectifiers, giving about 41 volts.
The reason I multiplied by 1.414 is that the AC voltage is measured in RMS (root-mean-square), but the rectifiers cause the voltage across the capacitors to be the peak voltage of the AC input.  To convert from the RMS voltage of a sine wave to the peak value of the sine wave we multiply by 1.414.Now-back to your first question about running one amplifier on one side of the power supply and the other amplifier on the other side- This has a number of problems. You want the power supplies to each amplifier to see equal voltage each side of ground, and it would not work to do this with two dividers.  Also- these amplifiers draw a lot of current during peak volume levels, making dividers impractical.Let's fix the power supply first.  Got that transformer number?-Rick  

Rick,
I think there is a disconnect between his second question and his diagram.
In his question, he said he has a 95V center-tap transformer which is giving him +/- 65VSo the transformer should be 42.5 - 0 - 42.5 and we'd expect the output to be42.5 * 1.414 = 60V ( minus voltage drop across the diodes so really about 59V when loaded)If the transformer is specified for 95V with the expected load from the amplifiers, the no-load voltage could be higher since you don't have the resistance loss in the transformer winding. In that case 65V is a bit high but not unreasonable. I'd be very careful not to plan on having that much at full load though.
How well regulated does the supply to these amplifiers need to be? The lowest power way to power these would be to just put one on each side of the ground but the regulation wouldn't be stable at all.

Apart from the above advice - which is very valid - I just had a look at the TDA7294 Data Sheet (nice IC!).  It advises that the absolute maximum power supply voltage with no signal must be +/- 50V.  The specified maximum is +/-40V.  If you want a reliable amplifier then you'd be well advised to stick to those limits.Now if you just use a basic rectified / smoothed AC for your power it is going to go up and down depending on the load.  Amplifiers can handle this but you can find that the power will go above the +/- 50V limit with no output, and yet sink below the +/- 40V specification during full output.  You want to power two amps - there is no objection to running them off the same supply rails, but again if one is working at full output and the other is not putting out much power then you may get interference between the two.The datasheet lists 10A as the maximum peak output current but this will be the maximum current that the output transistors can take.  100W at +/-40V (ie 80V) would be 1.25A.  To be comfortable I'd double that and then double again for 2 amplifierss - that's 5A.  The rule of thumb for power supply caps is 2000uF per amp - so that means you should look at 10000uF in each side rather than your 4700uF.However, bearing in mind the small difference between the specified and the absolute maximum power rail voltage, I'd suggest you look at some form of regulation.  This could be as simple as a 40-45V zener diode and a darlington transistor on each side, but look at your power dissipations and be sure to heatsink adequately..  Alternatively you could get two commercial 48V switching power supplies - you may be able to tweak the output voltage down to nearer 40V, or use a few diodes in each leg (1N540x  rated at 3A continuous should be adequate) to drop the voltage, or use the Zener/Darlington arrangement above - 8V is adequate headroom to do this and would minimise the dissipation in the regulator transistors.If you need more detail on any of the above, yell out

Rick, just reread your original question and there's no easy answer to it.  Your transformer gives 95 Volts Centre tapped - that is 47.5 - 0 - 47.5 (not 42.5 as Elizabeth used).  When you rectify AC and use capacitors for smoothing, as in your diagram, the capaictors charge to the peak value of the AC waveform which is 1.414 times the RMS value - in this case 47.5 x 1.414 = 67 volts.  However in between the AC peaks, if current is being drawn by a load, the voltage will reduce (the capacitors are discharging into the load) until the next peak.   Here's how this works in diagrammatic form:
The bigger the capacitor, the less the droop between half cycles of the rectified AC waveform.  Conversely, the bigger the load, the bigger the droop, or ripple as it is often called.You really need to think about using a transformer with the correct voltage to avoid stressing or even destroying your amplifier ICs.  The TDA7294 datasheet does say it has good power supply rejection (ie power supply changes getting through to the output).  But my previous comments about using a regulator on your power supply lines is still valid.  If you did this you'd be able to use  relatively small capacitors in the power supply, as the droop could go down to around 45-50 volts before the regulator would not be able to regulate properly.Dividing your supply lines with resistors is not really feasible, it would as you say be extremely inefficient.

Thank you all so much for your expeditious replies!!  Your advice is extremely helpful!
I decided to post a new question so that i could upload pictures of my actual transformer and keep things organized. 
Thanks again guys, you're awesome!

This power supply is apparently intended to put out +-40V at 5A. This is 40V*5A = 200W from each output, or 400W total. The schematic shows no series resistance between rectifiers and transformer, so the turn-on surge will be huge: At first turn-on the 220V line will see a short circuit until the capacitors charge initially during the first half-cycle of the 220VAC. The surge could melt the rectifiers, giving a hard short across the transformer secondary. You will need an over-rated on/off switch, as the switch will usually arc each time the switch is turned on. The surge might be big enough to trip the 220V circuit breaker. Also, put a slow-blow fuse in series with the on/off switch. The series inductance and series resistance of the transformer will reduce the surge to some extent.
An old-fashioned way of improving the situation, is to put an inductor (called a choke) in series with each rectifier output, before the filter capacitor. This changes the filter from capacitor-input to choke-input. Each inductor is essentially the core and secondary winding of a transformer similar to the transformer already in the circuit. A choke-input filter eliminates the turn-on surge. A choke-input filter has constant output of the average of the AC input (90% of the open-circuit AC RMS voltage) minus the resistive drops in the transformer and choke and minus the rectifier drop, provided the load current exceeds a critical value. To calculate the critical current, see power-supply design articles from before 1970 for details. The Radio Amateur's Handbook is one reference. The original papers are by O.H. Schade in the Proceedings of the IEEE from the 1930s (then called the Proceedings of the IRE).
I'd recommend buying a pair of switching power supplies rated to output 40V @ 5A each.

Very fascinating discussion, Now I know this is an old thread but my question to petertraneus-anderson is are you saying it is possble to convert +40v to -40v as all off the shelf SMPS are plus volts.