fun with laptops

[Shalom]

Most modern CPU's will use 100w alone..

That is something I've always wondered about. If the CPU is dissipating 100 watts of heat, at a supply voltage of 1.25 to 1.4 volts, Ohm's Law says it's gotta be drawing about 70 to 80 amps. How the hell does it do this without melting the PCB traces?

(Just checked the list of P4's at Wikipedia; there are some that dissipate as much as 115 watts at 1.25 volts. That's 92 amps of current. Having read this, I now understand why the P4 needs those four extra red wires from the power supply to the CPU, but I still can't see how the contacts on the chip or the traces on the circuit board can handle that much current without welding the chip into the socket.)

[Hyndis]

That is something I've always wondered about. If the CPU is dissipating 100 watts of heat, at a supply voltage of 1.25 to 1.4 volts, Ohm's Law says it's gotta be drawing about 70 to 80 amps. How the hell does it do this without melting the PCB traces?

(Just checked the list of P4's at Wikipedia; there are some that dissipate as much as 115 watts at 1.25 volts. That's 92 amps of current. Having read this, I now understand why the P4 needs those four extra red wires from the power supply to the CPU, but I still can't see how the contacts on the chip or the traces on the circuit board can handle that much current without welding the chip into the socket.)

They don't!

A mobo without a heat sink will overheat within seconds. Most BIOS programs do have failsafes that will instantly and automatically shut down the machine, but should that fail and the machine not be shut down you'll be cooking chips within a minute.

[Shalom]

It's not the chips I'm talking about. Obviously they'd melt without a heat sink/fan. With the heat sink they can safely dissipate all that heat, but they're still drawing ~90 amps of current through a PC board trace that's less than a millimeter thick. This is what I don't understand; you'd think that level of current would need something as thick as a car battery jumper cable.

[Pedersen]

The only issue is that almost *all* household circuits are significantly less than 90amps. In fact, almost all of the circuits I've ever seen have been 20A or less.

In other words, like it or not, those computers can not be drawing 90A, else the circuit itself would melt.

[lordlundar]

It's not the chips I'm talking about. Obviously they'd melt without a heat sink/fan. With the heat sink they can safely dissipate all that heat, but they're still drawing ~90 amps of current through a PC board trace that's less than a millimeter thick. This is what I don't understand; you'd think that level of current would need something as thick as a car battery jumper cable.

When talking about welding, you have to take voltage into account, and there just isn't enough voltage to do the job.

[Shalom]

The only issue is that almost *all* household circuits are significantly less than 90amps. In fact, almost all of the circuits I've ever seen have been 20A or less.

In other words, like it or not, those computers can not be drawing 90A, else the circuit itself would melt.

Power is defined as voltage * amperage. For power to remain constant, lowering the voltage means you must raise the amperage. For a simplified example, a transformer that has an output of 5 volts at 20 amps and an input of 117 volts would draw about 0.85 amps (20*100 =~ 117*0.85).

Now the maximum rated capacity of the power supply on a typical modern desktop system is 450 watts. At 117 volts, which is standard in the New York area (115 elsewhere in North America, don't ask me why the difference) 450 watts means a draw of 400/117=about 3.8 amps maximum. This is fine on a 15 amp branch circuit.

But the CPU isn't running on 117 volts, it's running on 1.25 volts, just about 1% of the line voltage. All the power that's being drawn by the CPU eventually winds up as heat, as no other form of energy (e.g. light, motion) is being produced. If 115 watts of heat are being dissipated, that heat must be coming only from the electric current being supplied flowing through the resistance within the chip, as there's no other source of energy. To generate that much heat, the chip must therefore be drawing that much electrical power.

So, to get 115 watts of heat on 1.25 volts, you get watts=volts*amps, or 115 = 1.25 * amps. Solving for amps gets you 115/1.25 = 92 amps.

Am I missing something here?

[TheSHAD0W]

Shalom is correct; power is the constant here, and when you change voltages you also change amperage.

[Pedersen]

Am I missing something here?

Actually, yes, you are missing a couple of critical pieces.

First, amperes is the measure of the flow of current. Basically, how much electric current is flowing. If you step down one of the items in the equation, you must step up another in order to maintain the values. However, getting 90A from a 15A circuit will not happen: You're overstepping the rating for that wire by a factor of 6, which is likely to melt the wire, start a housefire, etc. All of that, of course, assumes that you've managed to do this without tripping the circuit breaker for that circuit.

Don't believe me? Start with the wikipedia article on Amperage, and follow the references. You'll see that what you are discussing can not happen.

Second, and more importantly, you're missing the entire rest of the computer. You are making the assumption that the power comes from the wall outlet, reaches the PSU, and gets pumped directly onto the motherboard at the rated wall outlet voltage.

It doesn't. A number of changes happen at the PSU: Current gets converted from AC to DC, connectors are split out amongst the various components in the machine (CPU fan, case fan(s), optical drive(s), hard drive(s) (solid state or mechanical), USB chain(s), etc). By the time power reaches the CPU, it has been stepped down significantly in terms of voltage, and quite significantly.

Again, if you don't believe me, check out a pinout for the ATX connectors, and you'll see. Nowhere there does the CPU stand a chance at drawing the full 120V AC.

Shalom is correct; power is the constant here, and when you change voltages you also change amperage.

Regrettably, no, power is not the constant here. The power reaching all of the individual devices is indeed a constant, but the power that's reaching them is not the constant you and he are saying they are.

[Shalom]

First, amperes is the measure of the flow of current. Basically, how much electric current is flowing. If you step down one of the items in the equation, you must step up another in order to maintain the values. However, getting 90A from a 15A circuit will not happen: You're overstepping the rating for that wire by a factor of 6, which is likely to melt the wire, start a housefire, etc. All of that, of course, assumes that you've managed to do this without tripping the circuit breaker for that circuit.

What you're missing here is that the 92 amps I'm wondering about are at the output end of the computer's power distribution chain, which is at a much lower voltage; the input end, at the wall, is drawing less than 4 amps. You aren't going to overheat a 15 amp circuit with only four amps. I get the feeling you didn't entirely understand what I wrote.

By the time power reaches the CPU, it has been stepped down significantly in terms of voltage, and quite significantly.

Which is exactly my point. As the voltage goes down, the amperage goes up. It's simple mathematics. P = I * E (watts = amps * volts); if P is held constant, then E(in) * I(in) == E(out) * I(out). As one goes down, the other goes up.

Let's pretend that the only thing being powered is the CPU; ignoring fans, hard drives, video cards, memory and the rest of the junk on the motherboard and in the case. That CPU by itself is dissipating 115 watts of heat; as there's no other source of energy to it but electric, that means that it's drawing 115 watts of power from the power supply and ultimately from the wall socket. (The rest of the junk in the computer is pulling maybe another 200-300 watts depending on what's in there, but it's irrelevant for this discussion.)

If E(in)=115 volts (or 117 in NYC), which is what's coming in from the wall, then I(in), which is the CPU's share of what the ATX power supply is drawing from the wall socket, is one amp. Again, this is fine on a household 15 amp branch circuit. Now let's look at what the power supply does with this 115 watts:

E(in) * I(in) = E(out) * I(out)
E(in) = 115V
I(in) = 1A
E(out) = 1.25V
I(out) = (115)(1) = (1.25)x; x=(115)(1)/(1.25) = 92 amps.

115 volts at 1 amp goes in; 1¼ volt at 92 amps comes out. 115 watts either way you look at it. It's that simple.

(I'm oversimplifying here, obviously. Modern CPUs are powered from the 12 volt rail; actual output from the ATX supply destined for the CPU is 12 volts at about 9.6 amps, which is then further stepped down to 1.25 volts by circuitry on the motherboard itself. Again, voltage down, amperage up. Also there must be some heat losses in the power supply itself, or else it wouldn't be fitted with 2 fans.)

Again, if you don't believe me, check out a pinout for the ATX connectors, and you'll see. Nowhere there does the CPU stand a chance at drawing the full 120V AC.

Where did I ever say it did? I said 1.25 volts, not 125 volts. That decimal point is important.

[Pedersen]

What you're missing here is that the 92 amps I'm wondering about are at the output end of the computer's power distribution chain, which is at a much lower voltage; the input end, at the wall, is drawing less than 4 amps. You aren't going to overheat a 15 amp circuit with only four amps. I get the feeling you didn't entirely understand what I wrote.

To some degree, you are correct. I am not an electrician, nor an EE, and it's been a while since I've looked at the formulas. However, you still have a critical flaw in your thinking. I've just pulled out a PSU that I have that went bad on me (fan started making a clicking sound, otherwise it would still be in use). BTW, it's a 680W PSU, so it's numbers are even higher than your 450W numbers that you've been spouting.

What I found interesting about it is that it lists the AC input, from the wall, as 120V 10A. And then it lists the DC output as follows:

+3.3V at 38A, +5V @ 40A, +12V1 @ 24A, +12V2 @ 22A, -5V @0.3A, -12V @0.8A, +5VSB @ 2A

As I stated earlier, the power is converted from AC to DC and then distributed. The motherboard gets considerably less power than you have been claiming, as you can see from the outputs above.

Do I know where the extra goes? No. Nor do I particularly care. But I did know that you were wrong in what you were stating, and that was the point of my original note on this topic.

Oh, and by the way:

Where did I ever say it did? I said 1.25 volts, not 125 volts. That decimal point is important.

Just like I said DC current reaches the motherboard, not AC. That D is significant. As is my having said 120 volts, not 125 volts. That 5 instead of a 0 is something else that's rather important.

[Pimento]

Hmmm...my Toshiba Quasmio runs warm to the touch but no excessively hot. Even when playing games or what have you.

Mind you the thing is huuuge......ive yet to find a case that it fits in properly.