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/stmicroelectronics/2143.pdf
http://www.ortodoxism.ro/datasheets/stmicroelectronics/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/nationalsemiconductor/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!
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/stmicroelectronics/2143.pdf
http://www.ortodoxism.ro/datasheets/stmicroelectronics/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/nationalsemiconductor/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!
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!
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!
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.
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.
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.
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.
Re: Diodes
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?
I Googled the diode part # IN5818 (or is it 1N5818?). It turns up as a "Schottky Barrier Rectifier".
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/Delta 1010 D-A Board Part 1.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.
richie00boy said:
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?
I Googled the diode part # IN5818 (or is it 1N5818?). It turns up as a "Schottky Barrier Rectifier".
lgreen said:
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/Delta 1010 D-A Board Part 1.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.
Re: Diodes
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!
lgreen said:
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!
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?
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?
Hi Richie, I want to try your idea...
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!
richie00boy said:
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!
I'm trying to understand the details of this. So a 12 VAC source would first be rectified to, say, 15-16 VDC, then the voltage doubler would double that to 30-32 VDC before it goes to the regs? Is that the way it works? If so, why do the regs need 30-32 VDC input if they are only 15 V regs? Again, I apologize if I'm showing my ignorance here.
BTW, what would be a very good brand of 12 V wall wart that you would recommend?
I really appreciate your help. Thank you so much!
BTW, what would be a very good brand of 12 V wall wart that you would recommend?
I really appreciate your help. Thank you so much!
Hi,
can you remove the voltage doubler components?
If so, consider using a 15+15Vac transformer and make a regulated dual polarity 16 or 17Vdc supply. Send this pre-regulated supply into the sound card and then let the internal regulators do their job. If this reg voltage is not high enough then jump to an 18+18Vac transformer and send +-18 or 19Vdc into the sound card.
can you remove the voltage doubler components?
If so, consider using a 15+15Vac transformer and make a regulated dual polarity 16 or 17Vdc supply. Send this pre-regulated supply into the sound card and then let the internal regulators do their job. If this reg voltage is not high enough then jump to an 18+18Vac transformer and send +-18 or 19Vdc into the sound card.
AndrewT said:
I'm afraid I'm not knowledgable enough to do such a seemingly drastic change to the circuit.
From what you're saying, it seems that the 15V regs like to see an input voltage greater than 15VDC (like 16-19VDC). If so, is that the way regulators are (they like input voltages higher than the output that they are rated for)? Just wanting to learn more.
Thanks!Hi,
for the regulator to work properly (=regulate) the input voltage >=output voltage+dropout+ripple peak to peak for ALL operating conditions.
The worst case is usually the highest current demand co-inciding with lowest mains supply voltage.
Similarly the worst case for regulator dissipation is usually highest mains supply voltage tested against a variety of output currents.
If you double regulate then the first REG absorbs all the mains variation. External or first stage regulation also removes the mains ripple from the DC voltage supply to the sound card.
The low space requirements on the sound card make it less able to dissipate significant power so feeding near minimum voltage to meet all curent demands ensures that disspation is kept low and also ensures that reg drop out never occurs (until you short out the output lead).
for the regulator to work properly (=regulate) the input voltage >=output voltage+dropout+ripple peak to peak for ALL operating conditions.
The worst case is usually the highest current demand co-inciding with lowest mains supply voltage.
Similarly the worst case for regulator dissipation is usually highest mains supply voltage tested against a variety of output currents.
If you double regulate then the first REG absorbs all the mains variation. External or first stage regulation also removes the mains ripple from the DC voltage supply to the sound card.
The low space requirements on the sound card make it less able to dissipate significant power so feeding near minimum voltage to meet all curent demands ensures that disspation is kept low and also ensures that reg drop out never occurs (until you short out the output lead).
mtl777 said:I'm trying to understand the details of this. So a 12 VAC source would first be rectified to, say, 15-16 VDC, then the voltage doubler would double that to 30-32 VDC before it goes to the regs? Is that the way it works? If so, why do the regs need 30-32 VDC input if they are only 15 V regs? Again, I apologize if I'm showing my ignorance here.
BTW, what would be a very good brand of 12 V wall wart that you would recommend?
I really appreciate your help. Thank you so much!
You almost have it. The total supply needs to be at least 15V either side of ground, i.e. 30V minimum across everything. This is why the halfwave rectifier voltage doubler is used. As Andrew said a regulator can only work if it has some headroom to buffer the changing input voltage with.
As for brands I have no idea, there is no need for anything fancy.
Thanks for the great explanation, guys! I'm learning a lot and would like to learn more...
What is ripple peak to peak? Is it one of the characteristic properties of a regulator? I can't seem to find it in the regulator's datasheet. How or where can I find it?
I also want to know... Is it a known fact of electronics that when you rectify an AC source, you get a slightly higher DC voltage than the AC source voltage? I'm wondering because the one who suggested the power supply mods in the thread I mentioned said, "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." He seems to be saying that rectifying 9V AC will give 12-13V DC. If that is true, then my next question is, how much DC volts would you get by rectifying 12V AC? (Based on what the guy said, I was guessing 15-16V DC but wasn't sure.)
Thanks again!
AndrewT said:
What is ripple peak to peak? Is it one of the characteristic properties of a regulator? I can't seem to find it in the regulator's datasheet. How or where can I find it?
I also want to know... Is it a known fact of electronics that when you rectify an AC source, you get a slightly higher DC voltage than the AC source voltage? I'm wondering because the one who suggested the power supply mods in the thread I mentioned said, "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." He seems to be saying that rectifying 9V AC will give 12-13V DC. If that is true, then my next question is, how much DC volts would you get by rectifying 12V AC? (Based on what the guy said, I was guessing 15-16V DC but wasn't sure.)
Thanks again!
AC voltage is specified as the average or RMS value. When you rectify and smooth AC into DC the capacitors store charge almost up to the peaks of the waveform, so you get a seemingly higher voltage. There is no magic gain. The peak of an AC waveform is sqrt(2) times the RMS. About 1.414 times.
Hi Mt,
a rough rule of thumb for selecting a transformer for a regulated supply is.
Vac = regulated Vdc.
At low DC i.e. <10Vdc you need slightly higher Vac.
At high DC i.e. >30Vdc you can select slightly lower Vac.
If you need dual polarity then you need two supplies each capable of a single polarity supply.
For +-15Vdc you need a 15+15Vac transformer. If you require 100mA of DC output then the transformer must be rated for double the DC requirement. 200mAac would do.
If the DC output is continuous then the transformer will get warm or even hot. I would recommend you double the current rating for continuous DC output.
15+15Vac @ 200mAac needs 6VA transformer for intermittant duty and 12VA for continuous duty of 100mAdc output from the regulators.
The difference in cost between 6VA and 12VA is quite small, even 18VA and 25VA only cost a little extra and there are benefits in lower regulation with the higher VA transformers, which in turn, reduces the regulator heatsink requirement.
a rough rule of thumb for selecting a transformer for a regulated supply is.
Vac = regulated Vdc.
At low DC i.e. <10Vdc you need slightly higher Vac.
At high DC i.e. >30Vdc you can select slightly lower Vac.
If you need dual polarity then you need two supplies each capable of a single polarity supply.
For +-15Vdc you need a 15+15Vac transformer. If you require 100mA of DC output then the transformer must be rated for double the DC requirement. 200mAac would do.
If the DC output is continuous then the transformer will get warm or even hot. I would recommend you double the current rating for continuous DC output.
15+15Vac @ 200mAac needs 6VA transformer for intermittant duty and 12VA for continuous duty of 100mAdc output from the regulators.
The difference in cost between 6VA and 12VA is quite small, even 18VA and 25VA only cost a little extra and there are benefits in lower regulation with the higher VA transformers, which in turn, reduces the regulator heatsink requirement.
Thanks, guys! Now let's see if i got this right. What if I combine the 12V wall wart suggestion with the other suggestion of "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."? Here's my analysis of this:
Rectifying 12V AC would give about 17V DC peak (12 x 1.414). Then the voltage doubler turns that into 34V, which is then the peak input voltage going into the regulators. From the LM2990 datasheet (page 3), I'm seeing that the safe input voltage range for the negative voltage regulator is -26V to +0.3V. Since -34V is outside this range, then this would be unsafe, right?
On the other hand, if a 9V AC wall wart is used instead of 12V, rectifying 9V AC would give about 12.7V DC peak (9 x 1.414). Then the voltage doubler turns that into 25.4V, which becomes the peak input voltage to the regulators. Since -25.4V is still within the safe range for the negative voltage regulator, then this would be fine, right?
Please let me know if my analysis is correct. I'm looking into the possibility of combining the 12V wall wart suggestion with the LDO replacement regs suggestion, but it seems from the above analysis that this combination would be unsafe. If my analysis is correct, then the safe approach would be either: (a) to use the 12V wall wart without changing the regs; or, (b) to use the existing 9V wall wart and change the regs (and also the rectifier diodes).
I would appreciate your comments. Thanks again!
Rectifying 12V AC would give about 17V DC peak (12 x 1.414). Then the voltage doubler turns that into 34V, which is then the peak input voltage going into the regulators. From the LM2990 datasheet (page 3), I'm seeing that the safe input voltage range for the negative voltage regulator is -26V to +0.3V. Since -34V is outside this range, then this would be unsafe, right?
On the other hand, if a 9V AC wall wart is used instead of 12V, rectifying 9V AC would give about 12.7V DC peak (9 x 1.414). Then the voltage doubler turns that into 25.4V, which becomes the peak input voltage to the regulators. Since -25.4V is still within the safe range for the negative voltage regulator, then this would be fine, right?
Please let me know if my analysis is correct. I'm looking into the possibility of combining the 12V wall wart suggestion with the LDO replacement regs suggestion, but it seems from the above analysis that this combination would be unsafe. If my analysis is correct, then the safe approach would be either: (a) to use the 12V wall wart without changing the regs; or, (b) to use the existing 9V wall wart and change the regs (and also the rectifier diodes).
I would appreciate your comments. Thanks again!
No remember the voltage doubler produces the voltage that is needed across BOTH regulators to make a positive and a negative rail pair. So if you end up with 34 volts then each regulator will see 17 volts. If you went with the 12 wall wart there would be no need whatsoever to use low dropout regs.
Although looking at it again, to allow for mains tolerance and other losses, a 15V transformer would be better. In fact if it has 15 volt regulators inside I can't even see how it worked at all (as a regulated supply) with a 9V wall wart.
Although looking at it again, to allow for mains tolerance and other losses, a 15V transformer would be better. In fact if it has 15 volt regulators inside I can't even see how it worked at all (as a regulated supply) with a 9V wall wart.
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