Mastero MK-4 ATX PSU with features. - Page 5

MrPerfectionest · 05-03-2004, 08:21 AM

Can you give some information on why a different regulator for the 5V standby rail? The ATX specification mentions this is only used for wake-on-lan and other applications where the other power rails are powered down. Also, the specification calls for at most 2A on this line. If the 5V standby rail is only connected to the 5VSB pins of the motherboard then this 5A capability is largely going to go to waste. Did I miss some discussion in the MK 3.5 thread as to why this is like this?

Sinorm · 05-03-2004, 07:13 PM

There is no computer-related reason to have 5v on the stanby rail, but there are other possible reasons. You could use the 5v rail to run other components for your carputer setup that you wanted on before the computer, or something along those lines. Just a guess.

MrPerfectionest · 05-04-2004, 02:51 AM

In this application, is the 5V Stb truly a stand by rail? That is, will it be on without having to turn the rest of the supply on? I can't quite make it out from the component placement if it is or not. It appears to be but I wanted to make sure.

Also, I am wondering what kind of heat sinking is going to be on that regulator so that it can really meet a 5A spec. If someone decides to hang something off that line that only draws an amp of current then you are looking at (13.8V-5V)*1 = ~9W of heat that has to be dissipated, 3A and you are in the 30W range which would pretty much take up half 1/4 of the PSU case with a heat sink and a necessary fan to keep it cool.

Mastero · 05-04-2004, 08:12 AM

Shall post full details and a manual of the PSU soon...

Mastero

Mastero · 05-05-2004, 02:36 PM

mk-4 partlist

Enjoy guys
Mastero

no1knows · 05-05-2004, 03:06 PM

Mastero, when dya think the kits will be ready to order?

Anyone in the UK know an easy way to source all these parts?

Cheers

clemen · 05-05-2004, 03:37 PM

Please tell how the shutdown controler will work?

MatrixPC · 05-05-2004, 04:28 PM

Quote: Originally Posted by clemen

Please tell how the shutdown controler will work?

When the ACC line has power, the SDC turn on the PSU to STAND BY state. If the mobo has a "resume on power loss" feature or similar, the computer will boot.
When the ACC line loss power, the SDC go to its shutdown sequence.
The computer has 0 ~ 12 minutes run time before the "power button press" signal is generate. This is good for short stop. After the time expired, power button is press and the computer will shutdown/hibernate (may be not if it hang). 0~3 minutes after the "power button press" signal, the SDC cut power to the PSU
When the ACC line loss power, and you press the "Kill Switch", the SDC send the "power button press" and cut power after 0~3 minutes.
This SDC doesn't turn on your computer. If your mobo doesn't have the "resume on power loss", you have to find another way to turn the computer on.

xBrady · 05-05-2004, 04:48 PM

I am kind of a noob when it comes to the electronics but I was wondering what features the OPUS Power Supplies have that this one wouldn't? I really hate OPUS Sales right now so I am definitely looking for a better (faster) alternative. My brother and father know plenty about electronics to help me put this together and install it so that won't be a problem.
Thanks

rando · 05-05-2004, 07:44 PM

Quote: Originally Posted by MrPerfectionest

Also, I am wondering what kind of heat sinking is going to be on that regulator so that it can really meet a 5A spec. If someone decides to hang something off that line that only draws an amp of current then you are looking at (13.8V-5V)*1 = ~9W of heat that has to be dissipated, 3A and you are in the 30W range which would pretty much take up half 1/4 of the PSU case with a heat sink and a necessary fan to keep it cool.

I have no idea of this is correct.

Pd = (Vin - Vout)Iload + VinIg

Eff = Pout/Pin => Pin = Pout/Eff
Pd = Pout/Eff - Pout = Pout(1/Eff - 1) = VoutIout(1/Eff - 1)

Assume Vout = 5V; Iout = 5A; Eff = 80% @ operating Vin and Temp
=> Pd = 5*5(1/0.8 - 1) = 25(1.25 - 1) = 6.25W

BraGa · 05-05-2004, 07:50 PM

Hi all,
In the MK4 - component layout file, there's a 4700uF/35V capacitor that I presume should be 2200uF/35V (right?)!

Joao

Mastero · 05-06-2004, 02:25 AM

yes braga its 2200uf /35v shall change it soon ... !

Mastero

MrPerfectionest · 05-06-2004, 05:10 AM

Quote:

I have no idea of this is correct.

Pd = (Vin - Vout)Iload + VinIg

Eff = Pout/Pin => Pin = Pout/Eff
Pd = Pout/Eff - Pout = Pout(1/Eff - 1) = VoutIout(1/Eff - 1)

Assume Vout = 5V; Iout = 5A; Eff = 80% @ operating Vin and Temp
=> Pd = 5*5(1/0.8 - 1) = 25(1.25 - 1) = 6.25W

This is incorrect. The first equation you listed is THE one that matters. A linear regulator gets its name because the internal transistor is operating in its linear region which looks like a resistor in the circuit. An extremely simplified thevenin equivalent circuit looks like a voltage divider, one resistor being the load and the other resistor being the regulator. If you want 5V out with 14V in, then the 5V gets dropped across the load and 9V gets dropped across the regulator. The same current that flows to the load flows through the regulator so you will get 9V * Iload = Pd for regulator and 5V * Iload for the load. The regulator itself also consumes power at the input voltage to ground (the VinIg part of the equation) but this should be pretty small compared to the other factor so I neglected it in my post above.

To truly determine at what current and Vin the regulator will survive, the thermal resistance of the regulator package, combined with that of whatever heatsink is attached to it, combined with the ambient air temp will tell how much the internal temperatures in the regulator will rise and if this is in its operating range. Maybe when I get some time in the next day or so I will do this calculation. Since I don't have much experience with thermal resistance values of different heat sinks I will have to do a little bit of research to get these values and then I will perform the calculations to see what is possible.

But, using common sense calculations like I did above and comparing to other applications should be enough to give a general idea of what is possible. For reference, an Athlon processor consumes about 60W of heat and you should know how large a heatsink (in addition to forced cooling) is necessary to keep that operating at ~50C where ambient temps are ~25-30C. This regulator consuming ~25W (3A load with 14V Vin) in ambient air already at ~50C (trunk temperatures on a hot summer day) with a dinky TO-220 equivalent heat sink isn't going to survive.

Sorry for the long post and maybe this isn't the place, but figured since so many people are interested in how some of this technology works I would try to enlighten them.

rando · 05-06-2004, 10:29 AM

Ah, I was assuming that Ig could move in the opposite direction of Iload and thus have a reducing effect on the total power dissipation equation. It's been a few years since I completed my EE work.

Can you tell me what's wrong with my efficiency equations? Efficiency isn't entirely dependent upon the Vin-Vout delta, is it? If so, does that mean that the efficiency of the regulator is closer to 50-60% at 5v/3A and 30-40% at 5v/5a, assuming a 14v input?

MrPerfectionest · 05-06-2004, 10:57 AM

Your equations look fine, but in the case of a linear regulator it doesn't have a set efficiency, and you assumed 80% to make your equation work. Have a look in the data sheets and see if you can see anywhere that it quotes a specific efficiency?