3 ms·

How can changing the voltage change the TDP? Isn't the TDP a property of the physical design, not the operating conditions?

pixl97 4y ago
I think in this case it's being used as a measure of thermal performance at "100%" usage at a particular voltage. See the Wikipedia page on this and you'll see in modern processors that TDP is configurable. >TDP specifications for some processors may allow them to work under multiple different power levels, depending on the usage scenario, available cooling capacities and desired power consumption. Technologies that provide such variable TDPs include Intel's configurable TDP (cTDP) and scenario design power (SDP), and AMD's TDP power cap.
haunter 4y ago
Yeah maybe I'm using the wrong term here sorry for that. As far as I know the maximum wattage that the CPU can draw doesn't change by undervolting it. But the power consumption in watts at any given momement drops significantly. Changed the wording.
- deleted 4y ago
  [deleted]
klodolph 4y ago
There’s some terminology being mixed up here. The "D" in TDP stands for design. It's the maximum amount of power that the device is designed to dissipate. The actual power draw is what the comment is talking about. The actual power draw will be different from TDP. The power draw is affected by operating conditions, like voltage and frequency. In a typical CMOS design, gates have capacitance and carry charge. When you apply a positive voltage to a gate, you deposit some charge into that gate from the positive rail. When you apply a negative voltage to that gate, the positive charge you previously deposited (from the positive rail) is dumped into the negative rail. This current, flowing from the positive to the negative rail, is proportional to the voltage (q = cv), proportional to the gate capacitance, and proportional to the switching frequency. The power is current multiplied by voltage. This means that power is proportional to voltage squared in these systems. (Or at least, it has a voltage squared component—this is a simplified model, and there are other things which contribute to power draw.) The switching frequency is also related to voltage. At higher voltages, the transistors are driven harder, and the gates switch faster, so you can rase the clock speed. The end result is that when you lower the voltage and frequency a little bit, you can get a large reduction in power draw.
- haunter 4y ago
  Yeah thanks I mixed it up but your comment is more than insightful :O I'm just a lay person but I wish CPUs, or at least Intel CPUs were not tuned over that much. Or at least they should make undervolting much more accessible and open (not that it's very difficult now but it certainly should be more normal)
  - klodolph 4y ago
    The problem is that you can’t play around with undervolting without consequences. When you reduce the voltage, the gate voltages are lower, and the drain-source resistance of FETs will be higher, which means higher output impedances, and slower switching times. This will not affect every part of the circuit equally—because the circuit has all sorts of different gates and cells in it. Digital ICs, for one thing, are not made out of transistors which are all the same size. Some transistors are made larger than others. So when you undervolt things, you’re changing the performance of the circuit in very difficult to predict ways. If you end up with some gate delay that is too large, you’ll get errors. If you end up with some driver transistor that can no longer drive what it is supposed to drive, you’ll get errors. At this point in history, we have done an immense amount of fine-tuning to our IC designs, and there is not much room for fiddling with voltages. Chips are also so large and complicated that you cannot really know that you are testing the entire chip with a given workload, unless you have a lot of insight into the chip’s architecture. Finally, any circuit which is used for some kind of trusted computing environment needs to block undervolting because undervolting has been used in various exploits to extract keys from hardware, or bypass certain protections. It’s a weird, weird world.
    - haunter 4y ago
      Why does the out of the box voltage is so high though? And as far as I see it is something set by the motherboard manufacturer. Like it varies to brand to brand and betwee chipsets too (Z/H/B). Chasing that +5-10% performance doesn't seem to worth that much when the power consumption and temps go so high (so more noise from the fans, more CPU fan power consumption too etc) Same with the P1/P2 limits. By default on my motherboard it was 253w/unlimited. Which is crazy when the Intel defaults are 125w/181w (at least that's publicly available on the data sheet) I at least wish Intel would show somewhere what are the default voltage numbers. Or the motherboard manufacturer's should be more "law abiding", a simple "Enforce Intel factory defaults" should be more than enough.
burnte 4y ago
Voltage in equals heat out. More voltage is more energy into the chip, more energy dissipated as heat.

pixl97 4y ago

I think in this case it's being used as a measure of thermal performance at "100%" usage at a particular voltage. See the Wikipedia page on this and you'll see in modern processors that TDP is configurable. >TDP specifications for some processors may allow them to work under multiple different power levels, depending on the usage scenario, available cooling capacities and desired power consumption. Technologies that provide such variable TDPs include Intel's configurable TDP (cTDP) and scenario design power (SDP), and AMD's TDP power cap.

haunter 4y ago

Yeah maybe I'm using the wrong term here sorry for that. As far as I know the maximum wattage that the CPU can draw doesn't change by undervolting it. But the power consumption in watts at any given momement drops significantly. Changed the wording.

deleted 4y ago
[deleted]

klodolph 4y ago

There’s some terminology being mixed up here. The "D" in TDP stands for design. It's the maximum amount of power that the device is designed to dissipate. The actual power draw is what the comment is talking about. The actual power draw will be different from TDP. The power draw is affected by operating conditions, like voltage and frequency. In a typical CMOS design, gates have capacitance and carry charge. When you apply a positive voltage to a gate, you deposit some charge into that gate from the positive rail. When you apply a negative voltage to that gate, the positive charge you previously deposited (from the positive rail) is dumped into the negative rail. This current, flowing from the positive to the negative rail, is proportional to the voltage (q = cv), proportional to the gate capacitance, and proportional to the switching frequency. The power is current multiplied by voltage. This means that power is proportional to voltage squared in these systems. (Or at least, it has a voltage squared component—this is a simplified model, and there are other things which contribute to power draw.) The switching frequency is also related to voltage. At higher voltages, the transistors are driven harder, and the gates switch faster, so you can rase the clock speed. The end result is that when you lower the voltage and frequency a little bit, you can get a large reduction in power draw.

haunter 4y ago
Yeah thanks I mixed it up but your comment is more than insightful :O I'm just a lay person but I wish CPUs, or at least Intel CPUs were not tuned over that much. Or at least they should make undervolting much more accessible and open (not that it's very difficult now but it certainly should be more normal)
- klodolph 4y ago
  The problem is that you can’t play around with undervolting without consequences. When you reduce the voltage, the gate voltages are lower, and the drain-source resistance of FETs will be higher, which means higher output impedances, and slower switching times. This will not affect every part of the circuit equally—because the circuit has all sorts of different gates and cells in it. Digital ICs, for one thing, are not made out of transistors which are all the same size. Some transistors are made larger than others. So when you undervolt things, you’re changing the performance of the circuit in very difficult to predict ways. If you end up with some gate delay that is too large, you’ll get errors. If you end up with some driver transistor that can no longer drive what it is supposed to drive, you’ll get errors. At this point in history, we have done an immense amount of fine-tuning to our IC designs, and there is not much room for fiddling with voltages. Chips are also so large and complicated that you cannot really know that you are testing the entire chip with a given workload, unless you have a lot of insight into the chip’s architecture. Finally, any circuit which is used for some kind of trusted computing environment needs to block undervolting because undervolting has been used in various exploits to extract keys from hardware, or bypass certain protections. It’s a weird, weird world.
  - haunter 4y ago
    Why does the out of the box voltage is so high though? And as far as I see it is something set by the motherboard manufacturer. Like it varies to brand to brand and betwee chipsets too (Z/H/B). Chasing that +5-10% performance doesn't seem to worth that much when the power consumption and temps go so high (so more noise from the fans, more CPU fan power consumption too etc) Same with the P1/P2 limits. By default on my motherboard it was 253w/unlimited. Which is crazy when the Intel defaults are 125w/181w (at least that's publicly available on the data sheet) I at least wish Intel would show somewhere what are the default voltage numbers. Or the motherboard manufacturer's should be more "law abiding", a simple "Enforce Intel factory defaults" should be more than enough.

burnte 4y ago

Voltage in equals heat out. More voltage is more energy into the chip, more energy dissipated as heat.