So nothing is wrong?

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My previous thread:

Its a SIMPLE psu one to get things going for this amp. Transformer with centre tap ( that is grounded ) to bridge rectifier to soothing caps.

I was able to get two incandence bulb holders with pig tails and hook it up to my PSU. Hooked up as in positive rail to bulb and bulb to ground, same idea for negative rail. It turns out +28 and -27 volts DC at the 40w bulbs with both connected. Same brightness for both bulbs and the positive and negative voltages are on the correct rails. Measurements taken with a Fluke 115 digital volt meter rounded to nearest volt.

Seems right, does it to you?

I used this transformer because it was handy. Since I am only really looking for 10 or so watts would a 12 volt VCT transformer with a 36 or 24 VA rating be better? Since to me it would logically run cooler and therefore easier to cool. I have read that heat is the main problem with building these amps. As well I would never have a over-voltage situation.
I don't know offhand but you can figure it out. You cannot use any voltage you want. The voltage will determine the max rated output power. For a particular load resistance, there is only one supply voltage that will give a particular desired max rated output power level.

Take the transformer secondary's rated RMS output voltage times sqrt(2), i.e. x 1.414, then subtract 1.5 V or so for rectifiers, and that's your nominal rail voltage, Vrail, when under load.

Then subtract the "clipping voltage" from the graph in the LM1875 datasheet (for your loaded-down rail voltage value, Vrail). If there is no graph, use 2.5 to 3 Volts for Vclipping.

Then subtract 10% of Vrail, for the ripple voltage, Vripple. Call the result Vpeak, since it will be the peak signal voltage you can have, when running at max output power.

Then peak output power = Vpeak x Vpeak / Rload. And the maximum continuous RMS output power will be (Vpeak / 1.414 ) x (Vpeak / 1.414 ) / Rload.

For the reservoir capacitance, use

C = [ (1000000 * Vpeak) / (Rload x (Vripple) ) ] * [ (1/(2 * fmains)) + (0.02 / Cap_voltage_rating) ] microfarads,

where Vripple = Vrail - Vclipping - Vpeak, or, Vripple = 0.10 x Vrail .

Since an amplifier is usually only a 1/2 to 2/3 efficient, typically 55% to 65%, use a transformer VA rating that is at least 2X the max RMS output power that you calculated (actually, use 4X, since you calculated for only one channel).
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Thanks for the information, though I did do this already half of the calculations myself :p.

For 12 volts AC the Vrail voltage under load would be 15.49 VDC. According to the graph that would give me a 10w output. I did this part already.

I didn't do the Vclip, Vripple or Vpeak, since the soothing caps I am using are 0.12F, I figured that I didn't need to do those since the caps are so big there wouldn't be any problem. Yes I am being very careful around those caps. As of right now I can not read any AC voltage using my multimeter on the rails under load, it reads 0.022VAC which is background noise to my meter and no 60hz measured.

Thank you for advising me on the VA rating for the transformer. Since my desired output would be 10 watts than I would need a 12v VCT with a VA of 40 at least.

Did I miss something?
12 * sqrt(2) = 16.97

16.97 - 1.5 = 15.47 (Vrail)

15.47 - 3 = 12.47 (= Vrail - Vclip) [Important to not skip this step]

12.47 - 0.022 = 12.45 V (Vpeak = Vrail - Vclip - Vripple) [to "calibrate out" the multimeter, short leads and subtract that reading]

12.45 Vpk / sqrt(2) = 8.8 V RMS

8.8 * 8.8 / 8 Ohms = 9.69 Watts RMS

LT-Spice, with everything the same except Schottky diode bridges (and a different discrete transistor amplifier) gives 10.27 Watts RMS and 37 mV ripple.
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