When I finished my dual mono Aleph J last year, I suspected that this was going to be my next project. Well... now the parts are all coming in over the next few weeks, so I can finally connect up homeless F5 and F6 boards. Before I dive in, though, I wanted to be certain that I'm getting my grounding scheme correct (and/or not making any other boneheaded wiring errors). I think the attached jives with what you're saying wirewiggler, but please let me know if it doesn't. (The bridge rectifiers will likely be mounted on the side heatsinks, but shown on the baseplate for clarity.) Also, any thoughts on whether the shield on the umbilical cables should be grounded on one, both, or neither ends?
Hi guys,
I'm planning to build an external power supply for my future FW and ZM amplifiers. Taking ZM's guidance into account, an external PSU requires the installation of an extra capacitor bank in the amplifier chassis.
Since I'm not absolutely sure how to make the wiring/electrical connections between the two 'sections' of the PSU, I have made a sketch (depicting only one channel) that shows my blueprint for these connections.
I would appreciate any corrections or comments to my sketch.
My external PSU will have 120000uF per channel (CRC - 60mF/0.1Ω/60mF). Will 20000uF per channel be enough in the extra capacitor bank inside the amplifier chassis (RC - 0.1Ω/20mF)?
Or, should I go higher? 40mF? 60mF?
As a disclaimer note I have to say that my sketch was based on Octopus Technicus brilliant diagram for Aleph J's PSU.
Thanks.
I'm planning to build an external power supply for my future FW and ZM amplifiers. Taking ZM's guidance into account, an external PSU requires the installation of an extra capacitor bank in the amplifier chassis.
Since I'm not absolutely sure how to make the wiring/electrical connections between the two 'sections' of the PSU, I have made a sketch (depicting only one channel) that shows my blueprint for these connections.
I would appreciate any corrections or comments to my sketch.
My external PSU will have 120000uF per channel (CRC - 60mF/0.1Ω/60mF). Will 20000uF per channel be enough in the extra capacitor bank inside the amplifier chassis (RC - 0.1Ω/20mF)?
Or, should I go higher? 40mF? 60mF?
As a disclaimer note I have to say that my sketch was based on Octopus Technicus brilliant diagram for Aleph J's PSU.
Thanks.
Hi Analogico,
that looks good to me - except that I think you don't need the extra R in the amplifier chassis. If I follow Zen Mods reasoning here correctly, the wire between PSU chassis and amp chassis would act as "R" between PSU out and extra capacitor bank in the amplifier chassis.
Best regards, Claas
that looks good to me - except that I think you don't need the extra R in the amplifier chassis. If I follow Zen Mods reasoning here correctly, the wire between PSU chassis and amp chassis would act as "R" between PSU out and extra capacitor bank in the amplifier chassis.
Best regards, Claas
If you're already buying 60mF caps, get the volume discount and fill the amp chassis up with more of those.
Well, that is one of the reasons why I wanted to know if anyone saw any drawbacks in using only 20mF (2 x 10mF per channel) in the amplifier chassis.
I had already bought all the components for my 'internal' PSU when I decided I wanted to build an 'external' PSU instead. And so, now I need some additional capacitors for the amplifier chassis.
But, since I have 4 x Nichicon KG 10000uF caps, left over from another project, I wanted to make sure I could get away with only 20mF (per channel) in the amplifier chassis, given that I have 120mF (per channel) in the external PSU chassis.
Can you guys share your opinions on this?
Thanks for your feedback.
This would be cool. And then take wcwc's idea of the swappable sink plates and incorporate that into the monos. Maybe do them like four half width modules that can stack to be the dims of a conventional 8U combo, 4U psu combo with separate monos, or dual 4U mono/psu combo.
I'm looking for 0R1 wirewound resistors, to be used in the CRC PSU of FW clone amps.
However, as far as the resistors I've found are concerned, most of their manufacturers indicate the following formula to calculate the resistor's Rated Continuous Working Voltage:
V=√(P X R)
From what I understand, this means that for a 0R1/10W resistor we have the following calculation:
√(10 x 0.1) = 1V
Only 1volt of maximum continuous working voltage!
Therefore, I conclude that resistors with these characteristics are not suitable for the CRC PSU of a FW amp.
Am I drawing the wrong conclusion?
However, as far as the resistors I've found are concerned, most of their manufacturers indicate the following formula to calculate the resistor's Rated Continuous Working Voltage:
V=√(P X R)
From what I understand, this means that for a 0R1/10W resistor we have the following calculation:
√(10 x 0.1) = 1V
Only 1volt of maximum continuous working voltage!
Therefore, I conclude that resistors with these characteristics are not suitable for the CRC PSU of a FW amp.
Am I drawing the wrong conclusion?
The equation is correct.
The 0.1R resistor will drop 1V at 10W dissipation.
The current through the resistor at a 1V drop is 1V/0.1R=10A, quite a large amount.
Note that the resistor drops 1V only if your power supply is supplying 10A.
Generally most builders try to keep the power dissipation to 1/2 or less of maximum design power dissipation of the resistor for longevity and lower heat (power dissipated as heat).
So you are drawing the wrong conclusion.
By the way understanding Ohm's Law is very handy for electronics diy.
The 0.1R resistor will drop 1V at 10W dissipation.
The current through the resistor at a 1V drop is 1V/0.1R=10A, quite a large amount.
Note that the resistor drops 1V only if your power supply is supplying 10A.
Generally most builders try to keep the power dissipation to 1/2 or less of maximum design power dissipation of the resistor for longevity and lower heat (power dissipated as heat).
So you are drawing the wrong conclusion.
By the way understanding Ohm's Law is very handy for electronics diy.
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If you still need to purchase resistors for your power supply, don't use a single 0R1 resistor. Get a bunch (25-50, so you catch the price break) of 3w 0R47 resistors and put them in parallel with one another. Use 5 or 6 of them. Resistors in parallel will ADD the wattage rating (so you'll have a 15-18w equivalent power rating) and you'll end up with 0R078 to 0R094 in overall resistance.
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Eric,
I have a bunch of Panasonic ERX-3SJR47 resistors, but I was thinking about trying some wirewound resistors. But it is a possibility to be considered.
ranshdow,
This PSU will power amplifiers like M2x, F4, SissySit, LuDEF, etc.
I also have boards for the F3, but for this amp I plan to use another PSU.
Ben Mah,
At first I wasn't understanding your explanation, but after some thought I think I get it.
I will use a different example to make sure that I do indeed understand.
According to the manufacturer's formula [ V = √(P X R) - which is basically the application of Ohm's law], for a 0R2/5W resistor we would only achieve a voltage drop of 1V [ √(5A X 0R2) =1V ] if the current supplied by the PSU was 5A (1V / 0R2 = 5A).
And, if I used two 0R2/5W resistors in parallel, not only could they dissipate a total of 10W (P = V X I = V^2/R = 1V/0R1 = 10W), but they could also pass a total current of 10A (I = I1 + I2 = 5A + 5A = 10A).
Thank you all.
I have a bunch of Panasonic ERX-3SJR47 resistors, but I was thinking about trying some wirewound resistors. But it is a possibility to be considered.
ranshdow,
This PSU will power amplifiers like M2x, F4, SissySit, LuDEF, etc.
I also have boards for the F3, but for this amp I plan to use another PSU.
Ben Mah,
At first I wasn't understanding your explanation, but after some thought I think I get it.
I will use a different example to make sure that I do indeed understand.
According to the manufacturer's formula [ V = √(P X R) - which is basically the application of Ohm's law], for a 0R2/5W resistor we would only achieve a voltage drop of 1V [ √(5A X 0R2) =1V ] if the current supplied by the PSU was 5A (1V / 0R2 = 5A).
And, if I used two 0R2/5W resistors in parallel, not only could they dissipate a total of 10W (P = V X I = V^2/R = 1V/0R1 = 10W), but they could also pass a total current of 10A (I = I1 + I2 = 5A + 5A = 10A).
Thank you all.
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