So, I am fascinated by the different size filter/resevoir caps people use in their PSU designs. From what I have learned in school, and from what I've seen in practice in other areas of electronics, there seems to be a rule of thumb about the "size" these caps should be.
Well, to be fair, it is a rule of thumb about the Vr(pp), voltage ripple peak to peak, which should not exceed 10% of the Vp(rect), or peak rectified voltage.
The approximate equation is:
Vr(pp) = (1/fRC)*Vp(rect)
where f is 120hz (for anyone with 60hz mains), R is the load resistance, and C is the total capacitance of your filter caps.
Since a transformer with 25 Vrms secondaries puts out a Vp(rect) of 35V, our target Vr(pp) should be 10% of that, or 3.5V. If then solve Vr(pp) for C we get:
C = [ Vp(rect)/fR ] / Vr(pp)
substituting our numbers and assuming an 8ohm load **...
C = [ 35/(120*8) ] / 3.5 = 10416uF
So, it looks as though at least 10,000uF is needed for each rail in a PSU with 35V rails just to come close to reducing the voltage ripple to around 3.5V peak to peak.
But wait! From what I have read in multiple sources, it is ideal to shoot for a ripple of 2V peak to peak. Of course, what really matters is the PSRR of the amp, but, anyway.
A Vr(pp) of 2V requires about 18411uF
But, there are seemingly few people who are putting that kind of capacitance into their designs. Some even argue that it makes certain amps, like the GCs, sound poor in the mids and highs.
So, if I had a question it would be two-fold: (a) is there any reason you wouldn't really want to reduce Vr(pp) to 2V, and (b) is there empirical data to support the claim that such high capacitance has a negative effect on amplifier performance.
Well, those are my thoughts.🙂
**It is true that a PSU won't directly "see" a speaker as its load under most circumstances; however, if you calculate the Iopeak of the LM3886 at 50W output you get 3.5Amps draw. Using Ohms Law, and knowing the rail voltage, you can determine that the PSU will "see" a load of about 10ohms at its maximum current output to the chip.
Well, to be fair, it is a rule of thumb about the Vr(pp), voltage ripple peak to peak, which should not exceed 10% of the Vp(rect), or peak rectified voltage.
The approximate equation is:
Vr(pp) = (1/fRC)*Vp(rect)
where f is 120hz (for anyone with 60hz mains), R is the load resistance, and C is the total capacitance of your filter caps.
Since a transformer with 25 Vrms secondaries puts out a Vp(rect) of 35V, our target Vr(pp) should be 10% of that, or 3.5V. If then solve Vr(pp) for C we get:
C = [ Vp(rect)/fR ] / Vr(pp)
substituting our numbers and assuming an 8ohm load **...
C = [ 35/(120*8) ] / 3.5 = 10416uF
So, it looks as though at least 10,000uF is needed for each rail in a PSU with 35V rails just to come close to reducing the voltage ripple to around 3.5V peak to peak.
But wait! From what I have read in multiple sources, it is ideal to shoot for a ripple of 2V peak to peak. Of course, what really matters is the PSRR of the amp, but, anyway.
A Vr(pp) of 2V requires about 18411uF
But, there are seemingly few people who are putting that kind of capacitance into their designs. Some even argue that it makes certain amps, like the GCs, sound poor in the mids and highs.
So, if I had a question it would be two-fold: (a) is there any reason you wouldn't really want to reduce Vr(pp) to 2V, and (b) is there empirical data to support the claim that such high capacitance has a negative effect on amplifier performance.
Well, those are my thoughts.🙂
**It is true that a PSU won't directly "see" a speaker as its load under most circumstances; however, if you calculate the Iopeak of the LM3886 at 50W output you get 3.5Amps draw. Using Ohms Law, and knowing the rail voltage, you can determine that the PSU will "see" a load of about 10ohms at its maximum current output to the chip.