Just a thought ..........

[gdextreme]

This thought came to my mind recently. Say you have a processor with a TDP of 65 watts. I believe the CPU runs on 1.2 volts so that equates to around 55amps. Assuming the wire from PSU to the motherboard has resistance of 0.1 ohm. that means the wires dissipate 302.5 watts of energy. Something doesn't look right. Where am I wrong?

 

[yh125d]

What math did you use to get the 302.5w number, and what other assumptions have you made?

 

[veri745]

This is why CPUs have multiple pins to supply Vcore and other voltages, and motherboards have a different trace for each of those pins.

 

[veri745]

Oh you're talking about the wires from the PSU to the motherboard? If you redo your math, you're assuming that there is a ~6V drop across the wires from the PSU to the motherboard (302W / 55A = 5.5V) You may be off a little in your assumptions.

 

[yh125d]

Yeah I don't know what the hell you did to figure that number

 

[TuxDave]

Yeah your wire resistance is way too high.

 

[Aluvus]

The motherboard regulates down from 12 V to the voltage needed by the processor. The wires from the power supply carry this higher voltage, which means less current, which means lower losses. The lower voltage only needs to be carried from the motherboard's VRMs to the processor, which means a shorter distance, which means less resistance. And as stated above, multiple parrallel conductors are used to lower resistance.

In your example, the current on the PSU wires would be only 5.5 A. For your given resistance of 0.1 Ohms, this would mean a loss of about 3 W to heat. In reality, these wires are 2 parallel conductors, at a minimum 20 AWG or thicker. Some random sites from Google give the resistance of such wires as about 30 Ohms/kilometer, so for 2 parrallel conductors about 15 Ohms/km. For a realistic length of 50 cm, this gives a resistance of less than 0.01 Ohms, and thus less than 0.3 W lost as heat.

Originally posted by: yh125d
What math did you use to get the 302.5w number, and what other assumptions have you made?

For the current: P_load = V * I (for DC systems)
I = 54 A (OP should have rounded down)

In this case P_load is the power drawn by the processor and V is the voltage measured between the CPU's power and ground pins. I is the total current carried by all of those pins.

For the lost powr: P_heat = I^2 * R (sometimes called "I-squared R losses")
P_heat = 292 W

I is the current found above, R is the resistance that it flows through. OP chose R to be the resistance of the cable, but this is invalid because the cable actually sees a much lower current. Essentially, the OP is treating the processor and cable as two resistive elements in series, which they are not.

Assuming you accepted all of the numbers, this would all be valid if not for the motherboard regulating down the voltage.

 

[yh125d]

Well first of all OP, you can't use TDP to estimate the current flow through a CPU

 

[gdextreme]

I assumed resistance of wire to be 0.1ohm. i^2r gives power dissipated. So 55^2(0.1) = 3025(0.1) = 302.5W


Say even if you are not using TDP as power usage I think its afe to assume a I7 even at stock might use 65 watts when its at load. Also what would be the safe assumption of the resistance of the wire from PSU to the motherboard?

What exactly is a VRM? Could anybody point me to an article of some sort?

@Aluvus

That is very informative. It makes sense for the motherboard to step down from 12v to 1.2V. Could you point me to an article on the voltage regulators for CPU's?

 

[Cookie Monster]

Wiki has your answer

Basically the VRM is the part of the motherboard that regulates the supply voltage to the CPU. Kind of like a DC to DC voltage converter that regulates its output to a wanted figure using PWM (phase width modulation). Simply put, its a buck converter.

A very simple buck converter is made up of the PWM controller (feedback controller + constant current source, a FET) and power stage (inductor + output capacitor + diode). Assuming its a switching regulator of course.

From my understanding, there is little amperes flowing from the PSU. What the PSU does is that it sets up a voltage value, e.g 12V for the constant current source built into the motherboard. The constant current source is required for the PWM controller of the VRM (and this is usually a small value).

I dont think ill explain how PWM works (because there are alot of ways of doing it e.g sawtooth generator), but basically motherboards use ICs that produce square waveforms with varying duty cycle (how long it stays ON/OFF) dependent on the load current. This is paired with an inductor and a capacitor. These components are required so that when the PWM controlled switch is OFF i.e no current is flowing, the current i.e voltage is supplied by the inductor.

So basically in this case, the load at the output of the VRM is the CPU. However your not going to see ~300Ws of power being dissipated across the PSU wire nor see 55.4As flowing through any wire or component. Ive blown up 14W power resistors (these are made of ceramic/turn of wire) before and only had 0.7As going through them!

Its hard to determine the idle power of a CPU unless direct measurements are taken at the output of the VRM.

I think its inaccurate to base of power consumption based on the TDP figure because it is not the maximum power that the processor can dissipate.? Its a number for heatsink designers so that they can design a HS which can dissipate the required TDP value.

Hope this helps. (I tried to simplify things, but probably threw in too many buzz words!)

 

[gdextreme]

Originally posted by: Cookie Monster
Wiki has your answer

Basically the VRM is the part of the motherboard that regulates the supply voltage to the CPU. Kind of like a DC to DC voltage converter that regulates its output to a wanted figure using PWM (phase width modulation). Simply put, its a buck converter.

A very simple buck converter is made up of the PWM controller (feedback controller + constant current source, a FET) and power stage (inductor + output capacitor + diode). Assuming its a switching regulator of course.

From my understanding, there is little amperes flowing from the PSU. What the PSU does is that it sets up a voltage value, e.g 12V for the constant current source built into the motherboard. The constant current source is required for the PWM controller of the VRM (and this is usually a small value).

I dont think ill explain how PWM works (because there are alot of ways of doing it e.g sawtooth generator), but basically motherboards use ICs that produce square waveforms with varying duty cycle (how long it stays ON/OFF) dependent on the load current. This is paired with an inductor and a capacitor. These components are required so that when the PWM controlled switch is OFF i.e no current is flowing, the current i.e voltage is supplied by the inductor.

So basically in this case, the load at the output of the VRM is the CPU. However your not going to see ~300Ws of power being dissipated across the PSU wire nor see 55.4As flowing through any wire or component. Ive blown up 14W power resistors (these are made of ceramic/turn of wire) before and only had 0.7As going through them!

Its hard to determine the idle power of a CPU unless direct measurements are taken at the output of the VRM.

I think its inaccurate to base of power consumption based on the TDP figure because it is not the maximum power that the processor can dissipate.? Its a number for heatsink designers so that they can design a HS which can dissipate the required TDP value.

Hope this helps. (I tried to simplify things, but probably threw in too many buzz words!)

Thanks for your reply.

Basically in the bolded part you are saying that the CPU is powered in burst of current by using a capacitor and inductor.

PS: Are you an engineer?

 

[Cookie Monster]

The term burst would be inaccurate because current will always be supplied to the load. Ill re-clarify myself. The source of this current (to set up the correct output voltage i.e supply) comes directly from the PSU or the inductor. They take turns based on the PWM. Remember that the PWM is simply a square wave, i.e it can be either a set voltage value (true/ON) or zero voltage (false/OFF), and the time its ON per period depends on the duty cycle. This is the idea behind switching regulators. The switch that connects the PSU to the CPU is ON/OFF (controlled by the PWM), where when its ON, the current is flowing directly from the PSU and when its OFF, the inductor feeds the load because the inductor has an energy storing element which allows you to do this.

However given the complexity of motherboard VRM designs, Im probably dumbing alot of things down. But in essence this is what it does primarily.

 

[soccerballtux]

Originally posted by: veri745
This is why CPUs have multiple pins to supply Vcore and other voltages, and motherboards have a different trace for each of those pins.

And gold pin contacts (very thin layer). Gold is very important to electronics.

 

[soccerballtux]

Originally posted by: Cookie Monster
The term burst would be inaccurate because current will always be supplied to the load. Ill re-clarify myself. The source of this current (to set up the correct output voltage i.e supply) comes directly from the PSU or the inductor. They take turns based on the PWM. Remember that the PWM is simply a square wave, i.e it can be either a set voltage value (true/ON) or zero voltage (false/OFF), and the time its ON per period depends on the duty cycle. This is the idea behind switching regulators. The switch that connects the PSU to the CPU is ON/OFF (controlled by the PWM), where when its ON, the current is flowing directly from the PSU and when its OFF, the inductor feeds the load because the inductor has an energy storing element which allows you to do this.

However given the complexity of motherboard VRM designs, Im probably dumbing alot of things down. But in essence this is what it does primarily.

So, I believe input to the inductor is the PWM'd voltage; the inductor+capacitor regulate the current flow from this PWM'd signal-- without this inductor, the capacitor would charge as fast as it could and the current through the traces from the PWM'd voltage to the capacitor would be very high.

The capacitor supplies the current to the load (the CPU) and keeps it steady in the off part of the duty cycle.

Good picture

Off the top of my head there's nothing wrong with using TDP to calculate the current into the CPU-- the TDP will be the product of the current into, times the voltage drop across, the CPU. But I may be wrong. Those who have said otherwise should explain why.

Anybody know about how the CPU is supplied? Is the power plane under the CPU separate from the one in the rest of the board, with a bunch of vias and short wires straight to the location of the pin on the socket? Or is it wires directly from the VRM with no power plane... I'll have to check out some design docs on Monday that we have at work.

 

[PM650]

Originally posted by: soccerballtux

Originally posted by: Cookie Monster
The term burst would be inaccurate because current will always be supplied to the load. Ill re-clarify myself. The source of this current (to set up the correct output voltage i.e supply) comes directly from the PSU or the inductor. They take turns based on the PWM. Remember that the PWM is simply a square wave, i.e it can be either a set voltage value (true/ON) or zero voltage (false/OFF), and the time its ON per period depends on the duty cycle. This is the idea behind switching regulators. The switch that connects the PSU to the CPU is ON/OFF (controlled by the PWM), where when its ON, the current is flowing directly from the PSU and when its OFF, the inductor feeds the load because the inductor has an energy storing element which allows you to do this.

However given the complexity of motherboard VRM designs, Im probably dumbing alot of things down. But in essence this is what it does primarily.

So, I believe input to the inductor is the PWM'd voltage; the inductor+capacitor regulate the current flow from this PWM'd signal-- without this inductor, the capacitor would charge as fast as it could and the current through the traces from the PWM'd voltage to the capacitor would be very high.

The capacitor supplies the current to the load (the CPU) and keeps it steady in the off part of the duty cycle.

Good picture

Off the top of my head there's nothing wrong with using TDP to calculate the current into the CPU-- the TDP will be the product of the current into, times the voltage drop across, the CPU. But I may be wrong. Those who have said otherwise should explain why.

Anybody know about how the CPU is supplied? Is the power plane under the CPU separate from the one in the rest of the board, with a bunch of vias and short wires straight to the location of the pin on the socket? Or is it wires directly from the VRM with no power plane... I'll have to check out some design docs on Monday that we have at work.

Everything uses the same ground eventually on the pcb, but it may have a semi-isolated ground plane to prevent large currents present in the vrm area from flowing in grounds used for low-current things. The proximity of components used in the buck converters should however be as close as possible to the load to reduce ripple & losses - 100A through a 35µm-thick copper plane ('1oz copper') is not a trivial matter (70µm = 2oz copper). Placing capacitance closest to the load is typically the highest priority - this minimizes overall ripple and transients due to a load variation.

I would think they have to use vias through all layers to get the Vcore to the cpu - most sockets are surface mount/bga and do not have pins to penetrate the layers...The inner 2 layers are specified by the original ATX spec to be dedicated to power planes but often all 4 are used in small areas for the vrm. Connecting all 4 layers beneath the mosfets with vias also helps in disipating the heat (fiberglass is not a great heat conductor).

 

[Cookie Monster]

Originally posted by: soccerballtux
So, I believe input to the inductor is the PWM'd voltage; the inductor+capacitor regulate the current flow from this PWM'd signal-- without this inductor, the capacitor would charge as fast as it could and the current through the traces from the PWM'd voltage to the capacitor would be very high.

The capacitor supplies the current to the load (the CPU) and keeps it steady in the off part of the duty cycle.

Good picture

The "input" to the inductor is normally connected to the source of a FET (switch seen in that picture) and a diode. The PWM is simply the gate voltage of the FET which either connects the drain and the source, or disconnects it for a lack of a better term.

The capacitor job in this case isnt to supply the load by discharging. From my understanding, the output capacitor merely sets the required output voltage (because the load is parallel with the capacitor), reduce the ripple at the output (most of the AC component created by noise/switching frequency and so forth goes down the cap instead of at the output), and has largely to with the buck converter operating in the continuous mode (delta Ion = delta Ioff). Regarding the third statement, when the FET switches off, the inductor reacts because something has changed. The voltage difference across the inductor is the diodes FB voltage. Since v(t) = L*di/dt, the delta Ioff is very slow compared to delta Ion. This is highly undesirable. What the output cap does is it increases this voltage difference across the inductor, i.e increasing delta Ioff. This should hopefully create a triangular waveform representing the output voltage (and showing that the inductor is operating in continuous mode) at the output which can be seen by an oscilloscope.

I also have to mention that the PWM signal is basically to turn the switch on or off. You dont want any high current flowing through your FET all the time (unless its properly cooled but these get hot pretty fast then pop), because it could be damaging to the component.

Heres a good picture to show you what the PWM signal is doing and what the supposed "switch" looks like. After all transistors are switches!

 

[aigomorla]

Originally posted by: Cookie Monster
Wiki has your answer

Basically the VRM is the part of the motherboard that regulates the supply voltage to the CPU. Kind of like a DC to DC voltage converter that regulates its output to a wanted figure using PWM (phase width modulation). Simply put, its a buck converter.

this is why i love my classified.

800W dilvery to the cpu via 12 phase voltera...

what more could i possibly want when overclocking this beast?

 

[PM650]

Originally posted by: Cookie Monster

Originally posted by: soccerballtux
So, I believe input to the inductor is the PWM'd voltage; the inductor+capacitor regulate the current flow from this PWM'd signal-- without this inductor, the capacitor would charge as fast as it could and the current through the traces from the PWM'd voltage to the capacitor would be very high.

The capacitor supplies the current to the load (the CPU) and keeps it steady in the off part of the duty cycle.

Good picture

The "input" to the inductor is normally connected to the source of a FET (switch seen in that picture) and a diode. The PWM is simply the gate voltage of the FET which either connects the drain and the source, or disconnects it for a lack of a better term.

The capacitor job in this case isnt to supply the load by discharging. From my understanding, the output capacitor merely sets the required output voltage (because the load is parallel with the capacitor), reduce the ripple at the output (most of the AC component created by noise/switching frequency and so forth goes down the cap instead of at the output), and has largely to with the buck converter operating in the continuous mode (delta Ion = delta Ioff). Regarding the third statement, when the FET switches off, the inductor reacts because something has changed. The voltage difference across the inductor is the diodes FB voltage. Since v(t) = L*di/dt, the delta Ioff is very slow compared to delta Ion. This is highly undesirable. What the output cap does is it increases this voltage difference across the inductor, i.e increasing delta Ioff. This should hopefully create a triangular waveform representing the output voltage (and showing that the inductor is operating in continuous mode) at the output which can be seen by an oscilloscope.

I also have to mention that the PWM signal is basically to turn the switch on or off. You dont want any high current flowing through your FET all the time (unless its properly cooled but these get hot pretty fast then pop), because it could be damaging to the component.

Heres a good picture to show you what the PWM signal is doing and what the supposed "switch" looks like. After all transistors are switches!

The capacitor's job is to smooth out the ripple, however it has nothing to do with setting the voltage; output voltage is a function of the duty cycle. The voltage difference across the inductor after the switch turns off is related only to the input voltage and the duty cycle. Moving from continuous mode to discontinuous mode is only a problem when synchronous rectification is not used, not the case for computer VRMs. Maintaining continuous mode when the inductor current is allowed to go negative is forced continuous conduction mode and can be allowed to happen with sync. rectification. Transitioning to discontinuous mode is purely a function of the load anyways; if the load never requires a current below a known value, the converter can be designed to never enter discontinuous.

 

[aigomorla]

okey u guys all lost me. :T

my understanding was the VRM's control'd CPU power dilvery... this is one reason why i cant look at MATX boards anymore because the VRM's are sub par compared to the full sized gaming boards.

Also the Caps i though were banks of energy so they could supply them to the vrms when needed. Which is why the whole raze on solid caps won. (more effficency, better durability).

now where do the mosfets come in then? Or is that another word for VRM?

 

[PM650]

Originally posted by: aigomorla
okey u guys all lost me. :T

my understanding was the VRM's control'd CPU power dilvery... this is one reason why i cant look at MATX boards anymore because the VRM's are sub par compared to the full sized gaming boards.

Also the Caps i though were banks of energy so they could supply them to the vrms when needed. Which is why the whole raze on solid caps won. (more effficency, better durability).

now where do the mosfets come in then? Or is that another word for VRM?

The term VRM refers to the switching fet(s), inductor and associated capacitors. VRM for a mobo is simply a buck converter, a form of SMPS (switch mode power supply). The buck converter topology always converts a single voltage to another voltage equal to or below it (typ. 12V -> vcore). Multiple identical phases can be used to adjust the load on each phase to the desired level based on efficiency or some other factor.

The VRM is a form of switching/PWM voltage regulator. You already know this, as you set the vcore in the bios and subsequently expect this voltage to be maintained regardless of how much power the cpu is drawing. Yes, the VRM is ultimately providing power to the cpu, but in a form (i.e. voltage) that the cpu can run on. Capacitors are used on both sides (input & output) of any power supply, switch-mode types especially. The input capacitors provide a small reserve of energy for the VRM to use due to the main PSU capacitors being far away - however its not the distance that necessitates this but the resulting resistance/inductance of the wires & connectors that separate the psu from the VRM input. Capacitors used for a VRM will typically be 16V (input side) and 2.5/6.3V for output side (solid polymers are available at 2.5V but wet electrolytics are usually not available below 6.3V and thus are not used). Solid polymer capacitors exhibit lower ESR and can thus provide better performance in smoothing the voltage waveform with less capacitance than would otherwise be required from wet electrolytics. I also think the solid--cap trend is to somewhat get beyond the capacitor plague.

 

[Rubycon]

Originally posted by: aigomorla

this is why i love my classified.

800W dilvery to the cpu via 12 phase voltera...

what more could i possibly want when overclocking this beast?

The Classified has 600W capacity to the CPU if both 8 pin headers are in use.