1000 whats!

[c_hegge]

Re: 1000 whats!

Only two molex plugs? Even a $10 unit I found on fleabay had four.

[PCBONEZ]

Re: 1000 whats!

Total dissipated watts for both would be 10-20% less. And even that due to lower diode drop voltage because of current.

Diodes are not mosfets.

Really?
Did you take a special class for that?

They are both 'silicon'.

And it's FAR more than 10-20%.
Go read some data sheets and do the math.
.

[ratdude747]

Re: 1000 whats!

Only two molex plugs? Even a $10 unit I found on fleabay had four.

six

3 per wire, with a berg for a floppy on the end

[Rulycat]

Re: 1000 whats!

Why do manufacturers like to use a jap lytic for the primary and crap lytics for the sceondaries?

[Pyr0Beast]

Re: 1000 whats!

Really?
Did you take a special class for that?

They are both 'silicon'.

And it's FAR more than 10-20%.
Go read some data sheets and do the math.
.

No, it is common knowledge.

So is sand - 'silicon', but it doesn't conduct anything.

How much 'far' more it is ?
Take 1N5818 and 1A passing current.
With single diode, forward voltage is 0.5V @ 1A, 0.5W dissipation
Halve the passing current and drop is 0.45V-0.43 @ 0.5A x 2 = 0.45W, which is 10%-15% less, like I have said.

[PCBONEZ]

Re: 1000 whats!

You are full of shit as usual.

Long detailed thread on this topic, oh, 2-3 years ago complete with charts and graphs and data sheet excerpts as well and a link to an article that covers in in detail that [unlike yourself] understands how things work.

I suggest you go read it.
- You might even learn something. [If that's even possible.]
.

[Pyr0Beast]

Re: 1000 whats!

Well. You are full of shit as well.

Give me the link and stop whining.

Also. Your rants do not prove you are right in any point you've made in previous posts.

[PCBONEZ]

Re: 1000 whats!

ANYBODY can confirm what I said all on their own.
- Including you..

You have a totally twisted mind when it comes to how things work.
You don't even understand the basics.
And arguing with you is a waste of time.
.

[Pyr0Beast]

Re: 1000 whats!

Can you point to that 'anybody' that will confirm what you said all on their own ?
Excluding me.

Yes, Arguing with me is a waste of time. Especially when your arguments don't hold water and are nothing short of insulting anyone who even dares to question your 'knowledge'.

[Trinite]

Re: 1000 whats!

Studied into it a bit more, and still think the heatsinks are a bit on the small side for a 950W unit, as you are dealing with 150-200W of heat dissipation at full load. Not that it is insufficient to cool the MOSFETs and rectifiers, but it could throw out enough radiant heat to bake any adjacent cheap electrolytics. Just another good reason to recap. Ultimately depends on the CFM of the fan, I guess. Some heatsinks are absolutely massive and are coupled with a silent, low CFM fan. If it is a higher CFM fan, then the heatsinks would be appropriately sized.

Pyro, PCBONES is correct. If you have two resistive loads connected in parallel, total resistance is halved, and thus total heat output is halved.

[Pyr0Beast]

Re: 1000 whats!

Studied into it a bit more, and still think the heatsinks are a bit on the small side for a 950W unit, as you are dealing with 150-200W of heat dissipation at full load.

Well, yes and no. Such heatsink can dissipate about 50-100W tops. I really doubt that this unit can do 950W. Perhaps peak, but not constant power. (Those transformers can shift about 250-300W)

As things scale in power, so does their efficiency. Bigger electric transformers have much better efficiency than small transformers.
But bigger SMPS's have troubles dealing with light loads due to very short on-times, so pulse skipping is needed to deal with that. If there is a logic to do that in this PSU it should be fairly cool at moderate loads.

Pyro, PCBONES is correct. If you have two resistive loads connected in parallel, total resistance is halved, and thus total heat output is halved.

No he is not.
Diodes are NOT pure resistive loads. They have their forward voltage at rated current and there is no way to avoid that.

[ratdude747]

Re: 1000 whats!

Studied into it a bit more, and still think the heatsinks are a bit on the small side for a 950W unit, as you are dealing with 150-200W of heat dissipation at full load. Not that it is insufficient to cool the MOSFETs and rectifiers, but it could throw out enough radiant heat to bake any adjacent cheap electrolytics. Just another good reason to recap. Ultimately depends on the CFM of the fan, I guess. Some heatsinks are absolutely massive and are coupled with a silent, low CFM fan. If it is a higher CFM fan, then the heatsinks would be appropriately sized.

Pyro, PCBONES is correct. If you have two resistive loads connected in parallel, total resistance is halved, and thus total heat output is halved.

it has a huge 135mm fan. it is quiet but i have barely given it any load. my guess is that it can make it is the twin large bridge rectifiers... if the goal was to skimp, there would only be one w/o a heatsink. the teapos had to go but any big ones are gone.

[Pyr0Beast]

Re: 1000 whats!

It may also use H-bridge for driving a transformer and a bit higher driving voltage helps as well.

[370forlife]

Re: 1000 whats!

http://www.hardwaresecrets.com/artic...y-Review/871/7

This is the 850W version. Considering the exhaust temp was 53.2C at full load with an intake temp of 47.5C, I think those heatsinks do their job just fine.

Even drawing 950W from this unit only increased the intake temperature 6.4C.

If you guys REALLY want to have a discussion about skimpy heatsinks, look at some of sirtec/High Power's and FSP's offerings. Even so, most of those units do just fine.

[Pyr0Beast]

Re: 1000 whats!

Exhaust temp does not relate perfectly to temperature of heatsinks and even less to junction temperature. Burning the unit up at mere 10% overload proves this nicely.
53°C at exhaust (article says only PSU temperature) can be easily 80°C+ on junction.

[Trinite]

Re: 1000 whats!

No he is not.
Diodes are NOT pure resistive loads. They have their forward voltage at rated current and there is no way to avoid that.

Silicon is a semiconductor, which makes it a resistive load. There is no energy being dissipated by means of inductance or capacitance, so all heat dissipation is created by resistance. Transistors can get around this by switching, but they still dissipate some heat from leakage current and from internal resistance.

[Pyr0Beast]

Re: 1000 whats!

Silicon is a semiconductor, which makes it a resistive load. There is no energy being dissipated by means of inductance or capacitance, so all heat dissipation is created by resistance. Transistors can get around this by switching, but they still dissipate some heat from leakage current and from internal resistance.

Yes, silicon is a semiconductor. But that does not make it a resistor.
Diodes exhibit a _dynamic_ resistance, which is quite different from static resistance.

Check diode on ohm meter, then put it into circuit, load it with different loads and calculate its resistance again. Surprised ?

Transistors, bipolar and mosfets, dissipate quite a lot of heat due to capacitance, esp. at frequencies measured in GHz

[Trinite]

Re: 1000 whats!

Yes, silicon is a semiconductor. But that does not make it a resistor.
Diodes exhibit a _dynamic_ resistance, which is quite different from static resistance.

Check diode on ohm meter, then put it into circuit, load it with different loads and calculate its resistance again. Surprised ?

Transistors, bipolar and mosfets, dissipate quite a lot of heat due to capacitance, esp. at frequencies measured in GHz

Yes, diodes have a nonlinear relationship (curve) between voltage and current, and do not follow Ohm's Law. However, they are still resistive loads; they simply do not behave in a similar fashion as resistors do.

If you have two diodes in parallel, then the derivative of the resistance curve will be half that of the curve of a single diode. Thus, resistance will be halved. Regardless of load, any point along the two diode curve will be at half-value compared to a single diode.

[PCBONEZ]

Re: 1000 whats!

Yes, diodes have a nonlinear relationship (curve) between voltage and current, and do not follow Ohm's Law. However, they are still resistive loads; they simply do not behave in a similar fashion as resistors do.

If you have two diodes in parallel, then the derivative of the resistance curve will be half that of the curve of a single diode. Thus, resistance will be halved. Regardless of load, any point along the two diode curve will be at half-value compared to a single diode.

Exactly.

But your time will be less wasted teaching a box of rocks how to sing in harmony than trying to teach Pyr0Beast ANYTHING at all.
.

[Pyr0Beast]

Re: 1000 whats!

Exactly.

Exactly as wrong as previous times.

But your time will be less wasted teaching a box of rocks how to sing in harmony than trying to teach Pyr0Beast ANYTHING at all.

Dropping out of 'school' really does not waste your time with doing homework. However, it does not mean you won't benefit from it.

(Also, rocks can be taught how to sing; They exhibit resonant behaviour given the right amplitude and frequency)

Yes, diodes have a nonlinear relationship (curve) between voltage and current, and do not follow Ohm's Law.

Correct: Yes, diodes have a nonlinear relationship (curve) between voltage and current
False: and do not follow Ohm's Law

Diodes follow Ohm's law _perfectly_. At a given temperature and current, you will get exactly the same forward voltage drop, from which you can calculate resistance.
So do lightbulbs.

False: If you have two diodes in parallel, then the derivative of the resistance curve will be half that of the curve of a single diode. Thus, resistance will be halved.

Two diodes in paralel - halving operating current -> says nothing about halving resistance, you are jumping to (wrong) conclusions.
Diodes do not exhibit a !! Linear !! resistance, like ordinary resistors do.
Therefore halving current does not halve voltage drop !

Regardless of load, any point along the two diode curve will be at half-value compared to a single diode.

FALSE: If you have two diodes in parallel, then the derivative of the resistance curve will be half that of the curve of a single diode. Thus, resistance will be halved. Regardless of load, any point along the two diode curve will be at half-value compared to a single diode.

Nope. It won't be halved because that isn't a resistor, it is a diode.