If a MB dies, is it always worth replacing the PSU as well?

[giorgio652]

I run a computer repair company, and yesterday was given a piece of advice from a fellow computer professional; which was to always replace the PSU at the same time as replacing a dead motherboard.

Theory being, most MB's die due to the PSU sending unstable voltage so it is fairly likely to happen again, so for the price of an extra £20 you and the customer will have piece of mind.

Can any other professionals comment on this? I'm unsure whether to take it as gospel.

Thanks

[street1 (RIP)]

Based on the unpredictable flow of electricity,that same
theory would hold true to replace the whole computer.

[homer]

Based on the unpredictable flow of electricity,that same
theory would hold true to replace the whole computer.

the flow of electricity is quite predictable, i believe you meant to say due to the unpredictable voltages.

personally, i would check the voltages with a voltmeter before simply throwing out the power supply because it might have irregular voltages.

[WillyW]

...
the flow of electricity is quite predictable, i believe you meant to say due to the unpredictable voltages.

personally, i would check the voltages with a voltmeter before simply throwing out the power supply because it might have irregular voltages.

The hard part there is -  how do you check the voltage(s) with a proper load?
If the ps is sketchy,  it could show correct voltage until a load is placed upon it.

By the way -  here's a question:    Does anyone know for sure if it is ok to switch on a modern ps with absolutely nothing plugged into it as a load?

[squall_01]

Thats not nessaryly true as the board could die from ESD.  How does he really know there bad with out testing them?

[patio]

Thats not nessaryly true as the board could die from ESD.  How does he really know there bad with out testing them?

What does this mean ? ?

[homer]

...
the flow of electricity is quite predictable, i believe you meant to say due to the unpredictable voltages.

personally, i would check the voltages with a voltmeter before simply throwing out the power supply because it might have irregular voltages.

The hard part there is -  how do you check the voltage(s) with a proper load?
If the ps is sketchy,  it could show correct voltage until a load is placed upon it.

the voltage at the supply will stay the same regardless if a load is placed on it or not. the voltage after the load may, however, experience a drop, but that is not the fault of the supply, it is the fault of the load.

By the way -  here's a question:    Does anyone know for sure if it is ok to switch on a modern ps with absolutely nothing plugged into it as a load?

its ok.

[squall_01]

Thats not nessaryly true as the board could die from ESD.  How does he really know there bad with out testing them?

What does this mean ? ?

patio? ESD Electro Static Discharge.  USe an volt meter and see if it works with a load

[street1 (RIP)]

So far ,so fun. I'm always learning folks.

[WillyW]

...
the voltage at the supply will stay the same regardless if a load is placed on it or not.

This is not correct.

the voltage after the load may, however, experience a drop,

I don't know what you mean.  Can you re-phrase?

but that is not the fault of the supply, it is the fault of the load.

By the way -  here's a question:    Does anyone know for sure if it is ok to switch on a modern ps with absolutely nothing plugged into it as a load?

its ok.

Ok.
I didn't know if modern switching supplies were ok with no load or not.

[homer]

...
the voltage at the supply will stay the same regardless if a load is placed on it or not.

This is not correct.

what do you mean its not correct? i am sorry if this sounds rude, but what education do you have in electronic circuitry?

I don't know what you mean.  Can you re-phrase?

http://i195.photobucket.com/albums/z55/anonymous1375/sdafsadf.jpg

in this picture you can see that the voltage of the 5V DC supply remains at 5V until it passes through the first load. you can see that after the first load, however, there is a voltage drop. there is also a second voltage drop after the second load.

[WillyW]

...
the voltage at the supply will stay the same regardless if a load is placed on it or not.

This is not correct.

what do you mean its not correct?

I mean - in the context of our conversation -  that it is wrong.

A power supply is designed for a certain voltage at a max given current.

The condition of  "no load"  is infinite resistance.   Therefore, no current.

A short would be be zero resistance.  Therefore, in theory,  unlimited current.
No supply or battery can deliver this.   Thus,  the voltage must fall.   In this extreme example, the voltage would fall to zero.

Try it.    Get yourself a 12 volt car battery.   Hook your voltmeter leads across it.    Measure the voltage.       Now use a big, heavy set of jumper cables.   Short the battery terminals and watch the voltmeter.  It will fall to almost zero.     Of course,  the cables might melt  and/or the battery might blow.

So if you operate in between an infinite resistance and a perfect short,  then you can have varying results.    When the impedance of the load is less than the internal impedence of the supply,  you've probably exceeded the design of the supply, and the voltage will fall.... the supply just cannot keep up with the demand for amperage.   

If you have a defective power supply....  suppose for some reason it can only supply 10 watts and it is a 300 watt supply,   then it doesn't take much of a load before the voltage begins to fall off.   However,  with zero load, the  voltage will be just fine.
 

i am sorry if this sounds rude, but what education do you have in electronic circuitry?

So far, it is not rude.   It is just not relevant.

I don't know what you mean.  Can you re-phrase?

http://i195.photobucket.com/albums/z55/anonymous1375/sdafsadf.jpg

in this picture you can see that the voltage of the 5V DC supply remains at 5V until it passes through the first load. you can see that after the first load, however, there is a voltage drop. there is also a second voltage drop after the second load.

Ah.   An example of a voltage divider.     Earlier it seemed we were talking about just one load,  or the sum total of them all,  as were were talking about measuring the voltage delivered by the supply.

[homer]

Thus,  the voltage must fall.   In this extreme example, the voltage would fall to zero.

you are incorrect. current and resistance are directly proportional to voltage.

V (voltage)=I (current) x R (resistance)

if what you are saying is true then, according to this mathematical formula, electricity would cease to flow once a short circuit is introduced. this would mean it would be impossible to destroy a battery by shorting the two terminals.

If you have a defective power supply....  suppose for some reason it can only supply 10 watts and it is a 300 watt supply,   then it doesn't take much of a load before the voltage begins to fall off.   However,  with zero load, the  voltage will be just fine.

the voltage will not drop unless there is a problem with the voltage regulators.

i will go more in depth later, but i am suffering from the flu right now and i can not hold a thought for more then 10 seconds.

Ah.   An example of a voltage divider.

that was not an example of a voltage divider, it did not have a voltage out, that was an example of a simple series circuit. one must have the resistor there to drop the voltage from 5V to roughly 1.5V, the voltage most common LEDs use. i also included 2 loads to show you that voltage drops happen after the load.

[WillyW]

Thus,  the voltage must fall.   In this extreme example, the voltage would fall to zero.

you are incorrect.

Nope.  I'm exactly correct.

current and resistance are directly proportional to voltage.

V (voltage)=I (current) x R (resistance)

Right.   

All three are related.
Now plug in some numbers.   Remember!... you don't have unlimited current available.

Here you go:
Suppose you have a 12V supply that can provide 30 amps.
If you load it with .8 ohms,  what it be supplying?      15 amps,  right?
If you load it with .4 ohms,   now it is supplying 30 amps.

What happens if you load it with .2 ohms?   
I=E/R    ,   so would current be  60 amps?      Of course not - the ps can't do it.   
So, we now we have R and I (fixed at 30 max).    E=I*R    ,  thus voltage is
6V.

The voltage must fall.

You were correct in quoting Ohm's Law, and it is the reason the voltage will fall.

if what you are saying is true then, according to this mathematical formula, electricity would cease to flow once a short circuit is introduced.

In theory,  that is exactly correct.   And it is not what I'm saying,  it is what Ohm said. 
Try it.   Plug numbers into Ohm's Law. 

Pick a voltage.    12v.   
0  ohms  - that's your theorectical short circuit ohm value
I=ER
I= 12 * 0
I= 0

this would mean it would be impossible to destroy a battery by shorting the two terminals.

Because there would no longer be any potential difference in the two terminals.   

But.... that's not what really happens when you "short" them.     

If you have a defective power supply....  suppose for some reason it can only supply 10 watts and it is a 300 watt supply,   then it doesn't take much of a load before the voltage begins to fall off.   However,  with zero load, the  voltage will be just fine.

the voltage will not drop unless there is a problem

Bingo!
Now you've got it.    The voltage may drop if there is a problem.
And this is not what you said earlier:
"the voltage at the supply will stay the same regardless if a load is placed on it or not."   , that I pointed out was incorrect. 

with the voltage regulators.

i will go more in depth later, but i am suffering from the flu right now and i can not hold a thought for more then 10 seconds.

Ah.   An example of a voltage divider.

that was not an example of a voltage divider, it did not have a voltage out, that was an example of a simple series circuit. one must have the resistor there to drop the voltage from 5V to roughly 1.5V, the voltage most common LEDs use. i also included 2 loads

??
Each will have its own voltage drop across it.    Simply measure that, and you'll see.   That's a voltage divider.

to show you that voltage drops happen after the load.

I don't know what you mean here... voltage drops happen across loads.
But that is ok....  you seem to want to argue with me,  and I'm not interested in that.     You made a rather blanket statement that was not correct,  and in the interest of the forum readers,  and also to do you a bit of a favor,   I pointed it out.      With Ohm's Law that you are obviously familiar with,  and some consideration about both the theoretical and the real,  you'll grasp it.

[patio]

Thats not nessaryly true as the board could die from ESD.  How does he really know there bad with out testing them?

What does this mean ? ?

patio? ESD Electro Static Discharge.  USe an volt meter and see if it works with a load

I'm aware of what ESD is...is was asking what this has to do with a PSU issue ? ?
 

[homer]

upon doing some research, we are both correct. the example you provided...

Here you go:
Suppose you have a 12V supply that can provide 30 amps.
If you load it with .8 ohms,  what it be supplying?      15 amps,  right?
If you load it with .4 ohms,   now it is supplying 30 amps.

What happens if you load it with .2 ohms?   
I=E/R    ,   so would current be  60 amps?      Of course not - the ps can't do it.   
So, we now we have R and I (fixed at 30 max).    E=I*R    ,  thus voltage is
6V.

...would be correct if the power supply had a current limiter set at 30 amps. the voltage would drop to compensate for the increased amperage. however, if it did not have a current limiter, the voltage would remain constant and it would draw 60A, thereby either blowing the power supplies fuse or blowing the power supply itself.

so it seems our argument is based on if power supplies have a current limiter or not. if they do, then yes, there will be voltage drops under load if a rated 300 watt faulty PSU can only supply 10 watts. if they dont have current limiters, then, using the same faulty PSU scenario there will be no voltage drop even under load, the PSU will simply blow a fuse. personally, i would rather have a dead power supply then a dead mothboard, graphics card, RAM, etc, caused by an undervoltage.

Pick a voltage.    12v.   
0  ohms  - that's your theorectical short circuit ohm value
I=ER
I= 12 * 0
I= 0

your formula is incorrect. I=E/R, not E x R

I= E/R
I= 12/0
I= ∞

anyways i believe, if you wish to continue this discussion, we should do so via PM to avoid cluttering up the thread. i wish to continue the discussion of the 12V battery example later, but as i have stated before, i am not really in the best health and it does have an effect on my ability to concentrate.

[Dias de verano]

Theory being, most MB's die due to the PSU sending unstable voltage so it is fairly likely to happen again, so for the price of an extra £20 you and the customer will have piece of mind.

I don't know where your pal got that theory from. Anyway, rather than giving customers peace of mind, you're more likely to have them coming back in 6-12 months to give you a piece of their mind, if you only spend £20 on the power supply. Unless you're getting them in bulk, I suppose. Good ones cost £40 - £60 I have found.

By the way, when you short a battery, the current that flows is determined by the (non-zero) resistance of the short and the internal resistance of the battery.

[homer]

By the way, when you short a battery, the current that flows is determined by the (non-zero) resistance of the short and the internal resistance of the battery.

...we know.

[Dias de verano]

By the way, when you short a battery, the current that flows is determined by the (non-zero) resistance of the short and the internal resistance of the battery.

...we know.

Homer, I know that you know.

[WillyW]

...
your formula is incorrect. I=E/R, not E x R

Right.  Probably a typo as I re-composed.

anyways i believe, if you wish to continue this discussion,

No.   I already told you that you appear to want to simply argue with me, and I'm not interested in that.

You've lost the point.  All these things are examples - trying to explain the original point to you.    And it still stands:  When testing a power supply, simply checking for proper voltage in a no-load condition is certainly better than nothing,  it is not conclusive.   To see if a source of power is good, it should be checked while operating under a load.

I've tried to explain it to you.   Maybe I'm not doing a good job of that. 

[WillyW]

By the way, when you short a battery, the current that flows is determined by the (non-zero) resistance of the short and the internal resistance of the battery.

...we know.

Homer, I know that you know.

Do you have a point,  or anything worthwhile to add?

[homer]

I already told you that you appear to want to simply argue with me

i am not arguing for the sake of arguing, i am arguing because alot of your examples you used go against what i have been taught. this is why i asked what your education is in this. i would like to know if i am having a discussion with someone who has been professionally taught at a university or college or someone who simply looked it up on the internet.

simply checking for proper voltage in a no-load condition is certainly better than nothing,  it is not conclusive.

as i stated before, it depends if the supply has a current limiter or not. if it does, then the voltage will drop once a load is placed on it that exceeds the current limit. if it does not, then the voltage will remain constant regardless if a load is placed on it or not.

anyways, im feeling a little healthier today and i would like to bring up the battery example that you provided. you stated that voltage would fall to zero in the event of a short. i stated that was incorrect because then it would be impossible to destroy a battery. you rebuttled by saying in theory i was correct, so i assume you mean in practicality i was incorrect. however, these formulas work in theory and in practicality.

you have a 12v battery with a dead short. in theory a dead short is 0 ohms, however 0 ohms will never be achieved due to the resistance of the battery and the resistance of the material used to short the terminals. now lets use your assumption that voltage falls to zero in the case of a short circuit. im going to use 0.001 ohms as the cumulative resistance of the battery and the material shorting the terminals

V=IR
I=V/R
I=0/0.001
I=0

if there is no current, a battery will not catch fire or explode if the terminals are shorted.

lets try it again with 12 volts being maintained.

V=IR
I=V/R
I=12/0.001
I=12000

12000A will destroy a battery, 12000A will melt leads, 12000A will start the battery on fire and that is exactly what happens when you short a battery in reality. this proves that ohms law works both in theory and in practicality.

to sum it up...no voltage=no current. no current=no short. your battery example is incorrect.

[Dias de verano]

Do you have a point,  or anything worthwhile to add?

Well, now! If we're getting snippy, what point, or additional information did your post add, WillyW?