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Old 2nd October 2007, 09:40 AM   #1
mtl777 is offline mtl777  United States
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Default Replacing Voltage Regulators

Hi:

My first post, and sorry if this is so newbie-ish.

I saw a thread somewhere and they were discussing some mods on the Delta 1010 soundcard which I happen to have. One of the mods suggested was replacing the +/-15V regulators with LDO type. I want to try this mod but there is something I'm not sure of. Let me explain.

The stock regs in my 1010 are L7815CV for +15V and L7915CV for -15V. Here are the datasheets for these:

http://www.ortodoxism.ro/datasheets/...onics/2143.pdf

http://www.ortodoxism.ro/datasheets/...onics/2149.pdf

I want to replace the stock regs with LM2940CT-15 for +15V and LM2990T-15 for -15V, which are LDO type. Here are the datasheets for these:

http://www.ortodoxism.ro/datasheets/...r/DS008822.PDF

http://cache.national.com/ds/LM/LM2990.pdf

I thought it would be simple since the stock and replacement regs have the same pin-out. But on closer look I noticed that the stock +15V reg, as installed on my 1010, has a diode across its ground and output pins. The same thing is true with the stock -15V reg. So my questions are:

1. What are these diodes for?

2. When replacing the regs, should I simply attach these same diodes to the new regs in exactly the same manner that they were attached to the old regs? Will these diodes be compatible with the new regs? Do the new regs need these diodes at all?

Thank you so much for any help on these questions!
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Old 2nd October 2007, 10:16 AM   #2
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Are you sure the diodes are across gnd and o/p, not i/p and o/p? What are you expecting to achieve by putting in low dropout versions? I'm not convinced you will notice any gains.
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Old 2nd October 2007, 09:23 PM   #3
mtl777 is offline mtl777  United States
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I double checked and yes, the diodes are across ground and output. Is this not the usual connection? What does it mean then? BTW the diodes are IN5818. Will it benefit to replace these with a better diode such as Schottky? If so, please let me know the replacement part #.

Regarding the improvements to be gained, replacing the regs is just one part of the whole mods on the power supply. I'll quote the poster who made the suggestion:

"The +/- 15V power supply for the op amps in the breakout box is a bit of a joke. If you rectify 9V ac you usually only get around 12-13Vdc so there's a capacitor-based voltage doubler to boost the voltage high enough to feed to a pair of 15V regulators. Unfortunately the boost caps are undersized so the voltage isn't boosted high enough for the regulators to do anything. This means the op amps are being fed a ripply, noisy, unregulated power supply. To make things worse, higher sample rates draw a heavier supply current and pull down the voltage rails even further, so ripple and headroom are both worse at higher sample rates.

My solution was to put in larger, industrial grade boost capacitors, change the rectifiers for Schottky diodes and replace the regs with low dropout types. The +/- 15V regs dissipate more heat when this is done, so I made some big heatsinks (they need cutouts to clear the other components on the PCB). Now the 15V rails are clean and don't droop when the input voltage drops slightly. To keep it all cool I put a 40mm fan next to the left side of each 1010 in the rack, lined up with the ventilation slots. I also replaced the plastic external power supplies (yuk) with a hefty homemade PSU, with separate floating secondaries for each unit, which is bolted into the back of the rack."

Note, this guy has two Delta 1010 units in his rack. In a later part of the thread, he says:

"Regarding the power supply mods, I possibly went a little OTT with those. I used to build industrial control equipment and one of my pet hates is under engineered power supplies in audio gear. You could get a big improvement just by replacing the +/- 15V regulators with low dropout types like the LM2940CT-15 and LM2990T-15. Also replace the rectifier diodes (the four 1N4001s next to each other near the 9Vac input connector) with Schottky types like the MBR1100, for a lower voltage drop. You'll get a cleaner supply without the extra heat that bigger booster caps would cause."

So that's how I got into this modding thing.

There are many other mods (caps, opamps, clock, etc.) but let's just focus on this for now.

Thanks!
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Old 2nd October 2007, 10:30 PM   #4
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I've not seen the unit, but I would have thought a halfwave voltage doubler was used which is a perfectly fine way of doing it if it's done correctly. Does the unit have an ac wallwart? If so I think you might get good results simply by using a 12v wallwart instead of 9v.

It would be a bad idea to put schottky diodes in the location you suggest as they are to protect against reverse current when the unit is turned off.
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Old 3rd October 2007, 05:56 AM   #5
lgreen is offline lgreen  United States
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Default Diodes

I hate to disagree with you, but are you sure those diodes are not like D1 and D2 on the LM317 datasheet (Figure 3, page 9) Typically there to prevent the regulator from blowing up if the output drops more slowly than the input, and thus could exceed the input (a bad thing) when the device is turned off. If so, no need to change. they are hopefully never used.
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Old 3rd October 2007, 09:12 AM   #6
mtl777 is offline mtl777  United States
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Default Re: Diodes

Quote:
Originally posted by richie00boy
I've not seen the unit, but I would have thought a halfwave voltage doubler was used which is a perfectly fine way of doing it if it's done correctly. Does the unit have an ac wallwart? If so I think you might get good results simply by using a 12v wallwart instead of 9v.
Yes it has a 120 VAC (input) to 9 VAC (output) wall wart. Are you suggesting a 12 V wall wart that outputs 12 V under load or a 12 V one that outputs 9 V under load?

Quote:
It would be a bad idea to put schottky diodes in the location you suggest as they are to protect against reverse current when the unit is turned off.
I Googled the diode part # IN5818 (or is it 1N5818?). It turns up as a "Schottky Barrier Rectifier".


Quote:
Originally posted by lgreen
I hate to disagree with you, but are you sure those diodes are not like D1 and D2 on the LM317 datasheet (Figure 3, page 9)
Those are two diodes in that LM317 figure. Mine has only one diode attached to each reg. The diode is soldered directly to the ground and output pins of the reg. Here's a picture of my 1010's D-A board where the power supply section is located at the top portion of the board:

http://home.ca.rr.com/mtl777/Files1/...20Part%201.jpg

The picture is not so clear as I only used a scanner, but I hope you'll be able to see it better with a little zooming-in. The regs I want to replace are those two near each other on the upper center of the board. The one with the green paint on its screw is the -15V reg. To its right is the +15V reg. To the right of the +15V reg are the four IN4001 diodes that were suggested for replacement with Schottky's in the discussion thread that I mentioned.
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Old 3rd October 2007, 11:12 PM   #7
mtl777 is offline mtl777  United States
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Default Re: Diodes

Quote:
Originally posted by lgreen
I hate to disagree with you, but are you sure those diodes are not like D1 and D2 on the LM317 datasheet (Figure 3, page 9)
I'm sorry, I think I know now what you mean. All the while I was thinking about two diodes directly attached to the pins. But now I realize there's another diode near the -15V reg, and it's soldered to the board, not the pins. It looks like this might be the D1 you referred to. I'll have to open my 1010 again and confirm with a tester if this diode is connected to the input and output pins of the -15V reg. I'll do that later. If this is what it looks like, then this diode must be D1, and the diode that is soldered directly to the ground and output pins of the -15V reg must be D2.

What is strange though is that the +15V reg does not seem to have a similar board-mounted diode in close proximity to it. Or maybe there is one and it's just hidden. I'll have to verify this later when I open my 1010.

Thanks for the heads up on the D1 and D2 diodes!
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Old 4th October 2007, 01:54 PM   #8
mtl777 is offline mtl777  United States
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I opened my 1010, looked more closely and did continuity tests. I found out that the diode near the -15V reg is not connected to it but to the +5V reg on the upper left area of the board. So I traced the connection of the input pin of the -15V reg and saw that it first hit a capacitor, then it hit one of the four 1N4001 diodes next to each other on the upper right portion of the board. I also checked the connection of the output pin of the -15V reg, but I could not find a connection to any diode other than the one that's directly soldered to it and the ground pin. I did the same investigations on the +15V reg and found the same things as with the -15V reg (input pin first connects to a capacitor, then to another one of the four 1N4001 diodes next to each other, etc.). To my dismay, this doesn't look to me like the LM317 diagram of protection diodes. So the mystery still remains. Or maybe I am missing something. Could anybody please shed light on this?

This is just getting complicated. Can't I just approach it by the simple logic that replacing the existing regs with LDO type should follow exactly the same connections as before, since the only major difference of the replacement regs is that they are LDO type and are of better quality/performance than the existing regs? The replacement regs have exactly the same pin-out and are fixed type just as the existing ones are, so why should they need to be installed any differently?
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Old 5th October 2007, 06:43 AM   #9
mtl777 is offline mtl777  United States
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Hi Richie, I want to try your idea...

Quote:
Originally posted by richie00boy
... I think you might get good results simply by using a 12v wallwart instead of 9v.
I guess a regulated 12V wall wart to replace my unregulated 9V would be good, right?

What's the principle behind this? I'd like to know how this works. How many volts DC do you usually get when you rectify 12VAC? How much voltage is ideally needed to feed the pair of 15V regs?

Please bear with me for my newbie questions. I really am new to this and just starting to learn about electonics, so would greatly appreciate your explanation and advice.

Thanks!
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Old 5th October 2007, 05:33 PM   #10
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My logic is simply that if the 9v wallwart causes a lot of voltage sag and the voltage doubler barely scrapes enough to work the regulators, bumping up the input voltage should alleviate all that.

You can't get 12v regulated AC wall warts. All regulated supplies are DC.
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