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#1 |
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diyAudio Member
Join Date: Sep 2005
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Have some honking 30volt tranformers, used as a pair would give me 42.x volts rectified. But, I would rather be below 40 volts. Given mass of them would say they are around 600-1000va each.
Question about 2 approaches and looking for better solution. 1 - use 3-4 heavy duty diodes ->| ->| on positive rail and biased matching number of diodes biased in other direction of negative rail doing it after rectifier and before the caps. Would that work? 2 - what's a simple transitor assisted approach? like have seen transitor aiding a 78xx/79xx voltage regulator, for current boost. should I be looking zener ? or am I just burning power squeezing transformer into a different design (figured, "...i just need to drop a few volts"). Extra transformer, like variac or step down of some sort out of question... |
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#2 |
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diyAudio Member
Join Date: Sep 2003
Location: Upstate NY
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What is the end application? Class A or other?
How about unwinding a few turns of the secondary until you get the voltage you want? That way you won't be wasting energy heating the room. Don't forget the output voltage will drop under load. Your theoretical 42 volts does not seem to include diode drops. If you have independent secondaries you can use a bridge for each and drop 1.2 volts. |
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#3 |
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diyAudio Member
Join Date: Sep 2005
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thought about holding onto, or leave be, until i get around to doing a classA...
was thinkinopg about using for bpa200 (bridge-par 225wpc) gainclone. I believe the abolute max supply voltage rating for 3886 chips is 42 volt per rail, or as stated in datasheet 84V differential. But worried that if I pass this amp on (inevitable in DIY), that someone else would have a volt or two extra on their household voltage pushing it over the limit. Plus want to avoid tripping chip protection circuits. |
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#4 |
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diyAudio Member
Join Date: Sep 2003
Location: Upstate NY
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You're going to be way over the limit for chip amps. They won't load it much at idle, so assuming 5% regulation and 5% line overvoltage (common) you are over 46 volt rails.
Better to save those transformers for another project if you don't want to unwind it a bit or use a regulator. See the chipamp forum for a few threads on regulated supply gainclones. |
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#5 |
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diyAudio Member
Join Date: Sep 2005
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Yeah... that's what I was kind of figuring.
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#6 |
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Electrons are yellow and more is better!
diyAudio Member
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30 VAC will probably do without any problems. You are just at the limit but not way over it.
__________________
/Per-Anders (my first name) or P-A as my friends call me |
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#7 |
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diyAudio Member
Join Date: Sep 2003
Location: Upstate NY
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P-A -
If the line voltage is 10% high like it frequently is here in the states and the lightly loaded voltage is 5% higher than rated (fairly typical for big toriods) he's fried his amp. Too close to the rated maximum for my comfort, but everyone has their own limits. And those transformers sound like they ought to go into something more interesting than a BPA200 chip amp (wayyy overkill) How about an aleph 2? Bob |
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#8 |
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diyAudio Member
Join Date: Oct 2003
Location: Ottawa, Canada
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Use a "capacitor multiplier", which is sort of like a regulator, except that it isn't. It's really an active ripple filter. It works by using a transistor to drop a few volts from the rail voltage, and that transistor is driven in such a way as to attempt to stabilise the output voltage as much as it can, given the amount of dropout that it works with. Not only will you bring your voltage into line for your gainclones, but you will also have a cleaner supply (less ripple anyway). Have a look at Rod Elliot's project number 15
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#9 |
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diyAudio Member
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For a BPA-200 , to drive an 4 Ohm load , you would need around +-30V rails.
Assume each channel has a max. power consumption of 500W(output +dissipation). So , around 10 volts must be dropped.That is (40/30)*500 = 125W of additional power dissipation in regulators for each channel ! This goes for a larger heat sink. I would better unwind some windings. Regards, Lukas. |
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#10 |
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diyAudio Member
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Sorry for errors in equation , should be (10/40)*500 = 125W.
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