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Old 1st December 2011, 02:56 PM   #1
jamesdb is offline jamesdb  United States
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Default PSU Model for a 9V PSU, parts question

I have modeled a power supply in PSUD II that should produce about 9V, will be used to supply a voltage regulator that will output a steady 5V. The PSU model is attached.
The transformer is a Signal flatpack transformer 12.6V 200ma in series hookup (Digikey part#595-1313-ND )here Digi-Key - 595-1313-ND (Manufacturer - LP-12-200)
I used 12ohms for the DCR in the PSU model since the flatpack will be hooked up in series connection - does anyone know if that is correct for the flatpack series connection?
The 27A 600V IXYS FRED Bridge Rectifier is here http://www.partsconnexion.com/product4794.html
Will it work well for this power supply? I also used the LN4005 in the PSU model since it has a 600V PIV and seemed closest to the 600V IXYS FRED Bridge Rectifier, is that okay as well? Thanks in advance for any comments.
The values for the parts in the model are
T1=12.6V 12ohms DCR
Bridge=LN4005
C1=4uF 1ohm
L1=300mH 5ohms
C2=940uF 1 ohm
R1=9.1 ohm
C3=2mF 1 ohm
I1=50mA
Attached Images
File Type: jpg PSU_9V_for_5VRegulator.JPG (176.1 KB, 126 views)

Last edited by jamesdb; 1st December 2011 at 03:00 PM. Reason: To add extra details
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Old 1st December 2011, 05:05 PM   #2
Elvee is offline Elvee  Belgium
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Quote:
Originally Posted by jamesdb View Post
The 27A 600V IXYS FRED Bridge Rectifier is here http://www.partsconnexion.com/product4794.html
Will it work well for this power supply? Thanks in advance for any comments.
That's sheer madness: the bridge is more than 100x too large.
Overdesign to such an extent can only be counter-productive.
Quote:
The values for the parts in the model are
T1=12.6V 12ohms DCR
Bridge=LN4005
C1=4uF 1ohm
L1=300mH 5ohms
C2=940uF 1 ohm
R1=9.1 ohm
C3=2mF 1 ohm
I1=50mA
Why do you have 1ohm resistance for all capacitors? That is extremely poor. Are they from a specialist "audiophile" series?
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Old 1st December 2011, 11:54 PM   #3
Minion is offline Minion  Canada
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I wouldn"t use that transformer ..... if you use it at 12.6v AC that will give you about 17V DC so you will have to drop 12v DC in the regulator which depending on the current draw can mean a lot of heat for the regulator to dissapate ......

If you use it at 6.3v AC you might not have enough voltage overhead for the regulator ....6.3v X 1.41V=8.8v - 1.5v diode drop =7.3 - 5v=2.3v overhead for the regulator , that might be enough depending on the regulator but it doesn"t leave any voltage left if the mains voltage fluctuates .....


I would look for one that does 8v AC or 9v AC ...... Or you can find a 9v DC wallmart power adapter and open it up and take the transformer out and use that which would give you 9v DC which means the Regulator will have enough overhead and will only have to drop 4v and you can probably find one for a buck at the thrift store .....

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Old 2nd December 2011, 12:03 AM   #4
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Quote:
Originally Posted by Minion View Post
I would look for one that does 8v AC or 9v AC ...... Or you can find a 9v DC wallmart power adapter and open it up and take the transformer out and use that which would give you 9v DC which means the Regulator will have enough overhead and will only have to drop 4v and you can probably find one for a buck at the thrift store .....

Or just use the power adaptors 9VAC straight and that will be safer than having mains in your enclosure.
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Old 2nd December 2011, 04:16 AM   #5
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Hi James dB,

I have no idea what is the answer to your question. However, I was delighted to find your mention of PSU Desinger! I've been designing and building my first linear unregulated PS using the free student edition of SPICE. I thought "wouldn't it be great if someone made a design program specifically for linear, unregulated power supplies?" But I figured, no one is going to bother with that. And here it is! in your post!

It just goes to show the advantage of wandering aimlessly through interesting-looking posts.

Good luck with your build James, and I hope you find the answer to your question, whatever it is.
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Old 2nd December 2011, 11:43 AM   #6
AndrewT is offline AndrewT  Scotland
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report states:
simulate for 900ms after a reporting delay of 0s.
This deals with start up and shortly thereafter.

Change that line to 100ms after 1000ms (1s)
and look at what your various voltages and currents are doing in the continuous mode.

BTW,
you have created and modeled an rCLCRC filter. Mains hum on the output should be very low.
Most of the performance will be controlled by the last C. That's where you put the money in high quality audio applications.
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Last edited by AndrewT; 2nd December 2011 at 11:45 AM.
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Old 2nd December 2011, 11:56 AM   #7
jamesdb is offline jamesdb  United States
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Elvee, I used 1ohm for caps as they are small and the PSUD default starts with a 100ohm cap at 2 ohms. Many websites do not show the resistance for caps on sale so thats just a guess. I have the particular bridge on hand, but Imay be better off buying some individual diodes having more suitable ratings, and making a bridge, thanks.
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Old 3rd December 2011, 01:41 AM   #8
! is offline !  United States
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What are your ripple goals? It seems you are going to a lot of trouble for a very basic circuit, if all the CLCRC filtering is needed it is probably only because you used a poor regulator IC.

With a 50mA current requirement you can get away with using many regulators. Parallel secondary for the transformer is about 6.3VAC turned into 7.6V. A 2.6V margin above 5.0V output should be fine for 50mA current using a typical LM7805, LM317, etc., let alone low dropout regulators if you don't have the L & R before the regulator. For example with LM317, at 50mA and 50C operating temperature, dropout is around 1.5V. Yes AC mains variations could give less margin but the mains shouldn't be changing that much.

However, you could go the other route and use 12VAC in from series secondary transformer windings and have 16VDC to regulator, since with only 50mA current that's only 0.05A * 11V drop = 0.6W heat dissipation. You won't win any efficiency awards but it is only half a watt meaning you don't even need a heatsink on a TO220 sized regulator in typical room temperature environments.

So, I'm suggesting all you probably need is the transformer, some cheap 1N4001 or better diode bridge, a 330uF+ capacitor, 0.1uF film or ceramic capacitor before the regulator, typical LM317 datasheet components to reach 5V out. Filtering before a regulator isn't all that important unless you have a lot of high frequency noise then a couple ferrite beads on the regulator input and output leads might help. I mean within the assumption that you at least have enough capacitance that the supply voltage prior to the regulator never drops below the minimum threshold (1.5V above 5V output @ 50mA/50C) in the case of LM317 mentioned above).

Last edited by !; 3rd December 2011 at 01:53 AM.
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Old 3rd December 2011, 07:08 AM   #9
Elvee is offline Elvee  Belgium
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Quote:
Originally Posted by jamesdb View Post
Elvee, I used 1ohm for caps as they are small and the PSUD default starts with a 100ohm cap at 2 ohms. Many websites do not show the resistance for caps on sale so thats just a guess. I have the particular bridge on hand, but Imay be better off buying some individual diodes having more suitable ratings, and making a bridge, thanks.
Your 4µ is most likely some plastic type, meaning it will have a resistance in the 0.1Ω range or less, and the e-lytics will normally have some tens of millohm.
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Old 4th December 2011, 02:31 AM   #10
PRR is offline PRR  United States
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> a power supply ... produce about 9V... to supply a voltage regulator .... 5V

Click the image to open in full size.

Anything more is excessive.

(The "400V" bridge IS excessive... but the same price as 100V so you may as well buy only 400V jobs by the 10-pack.)
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