I'm building an active tube-powered DI box for my studio, and I'm looking for a little clarification on the PS, specifically what happens to voltages during rectification and why I should use one arrangement or another. (There is a LOT of information on the forum about this, I know, but I'm trying to winnow this down to easy bites!)
For starters, here's a schematic a tech provided me; it's slightly modded from the original (based on the Altec 1566) to accomodate a new power transformer:
[IMGDEAD]http://thegoldenbears.net/images/AlternatePS.jpg[/IMGDEAD]
The diodes are 1N4007; I don't know what the bridge rectifier for the filament supply is supposed to be. Here are my questions:
-If the filament winding produces 6.3VAC, what will the DC output be? Another poster writes elsewhere:
In addition, I know from my last PS adventure that there will be losses across the filtering caps. A helpful poster wrote:
-Now to the B+. This is less confusing to me; if I spend enough time scratching my head, I think I can figure out voltage gains/losses across the diodes, caps and resistors. Question is: Do the voltage gains called out for a bridge rectifier above also hold true for the arrangement used for the B+? Again, I think I have all the tools I need, perhaps minus a few brain cells; what I'm asking for is an experienced pair of eyes to tell me: "You're on the right track, just do the math and adjust values accordingly," or, "This is the worst idea since the Edsel."
As always, replies and wisdom are most appreciated.
-Seth
For starters, here's a schematic a tech provided me; it's slightly modded from the original (based on the Altec 1566) to accomodate a new power transformer:
[IMGDEAD]http://thegoldenbears.net/images/AlternatePS.jpg[/IMGDEAD]
The diodes are 1N4007; I don't know what the bridge rectifier for the filament supply is supposed to be. Here are my questions:
-If the filament winding produces 6.3VAC, what will the DC output be? Another poster writes elsewhere:
"When using a full-wave bridge rectifier arrangement, the peak DC voltage output will be 1.414 times the AC voltage input to the rectifiers, but this does need to be reduced by 2x the rectifier 'losses' to give the final voltage level for the circuit. This will depend upon the rectifiers used, but as another generalisation, 0.6V would be average here for each diode, so the final output from such an arrangement would be approx. 1.2 volts less than the calculated 1.4-ish times the AC from the transformer's secondaries."
In addition, I know from my last PS adventure that there will be losses across the filtering caps. A helpful poster wrote:
You are discharging those capacitors and the change of voltage is: V = It/C = 0.3 x 0.00833/1000 x 10-6 = nearly 3V each
-Now to the B+. This is less confusing to me; if I spend enough time scratching my head, I think I can figure out voltage gains/losses across the diodes, caps and resistors. Question is: Do the voltage gains called out for a bridge rectifier above also hold true for the arrangement used for the B+? Again, I think I have all the tools I need, perhaps minus a few brain cells; what I'm asking for is an experienced pair of eyes to tell me: "You're on the right track, just do the math and adjust values accordingly," or, "This is the worst idea since the Edsel."
As always, replies and wisdom are most appreciated.
-Seth