Just so you have a better understanding of what you are looking for, a 220000uF cap will be the size of a soda can or bigger.
The capacitor in Farnell is the size of a soda can - look at the dimensions!
It's a 130W amplifier (assuming a 4R load), so at total capacitance of 0.66F is overkill, perhaps the designer got the decimal place wrong?
A single 22,000µF capacitor per rail should be more than enough - I'd go for 10,000µF.
No, I used Ohm's law.
I assumed a peak output of 32V and with a 4R load that will be 128W which I rounded to 130W, if it's an 8R load then that can be halved.
Maybe I got the peak outout voltage wrong, perhaps it'll be more like 30V? Even so it'll be 110W into 4R or 55W into 8R
A 30W amplifier driving an 8R load only needs a peak output voltage of 22V.
Yes it says class A at the bottom.You failed to notice it's a class A amplifier, you can't treat it as a class AB one. In any case,
You mean +/- 35V rails?70V HT rail gives you around 100W in to 4 ohms, not 130W - overdriving the amplifier to produce 130W would be grossly distorted, and the amp wouldn't last long.
Yes it says class A at the bottom.
It seems silly to build such a large class A amplifier
You mean +/- 35V rails?
It depends on how much voltage loss the output transistors have and how low the distortion is, for 100W into 4R you need 28.3V peak and at 35V a 6.7V loss sounds reasonable for low distortion.
What voltage drop would you typically allow for loading?
Surely it depends on filter capacitors and the impedance of the transformer?
I imagine it'll be more of a problem for bass frequencies.
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