CPU消耗的功率

我认为，对于电流的CPU电源我和电压ü是我·U 。

我想知道如何从维基百科得出以下结论？

CPU消耗的功率大约与CPU频率成正比，并且与CPU电压的平方成正比：

P = CV ² f

（其中C是电容，f是频率，V是电压）。

MSalters的回答是80％正确。估算值来自通过电阻器以恒定电压对电容器充电和放电所需的平均功率。这是因为CPU以及每个集成电路都是大量的开关，每个开关驱动另一个开关。

基本上，您可以将一个阶段建模为MOS逆变器（它可以更复杂，但是功率保持不变）为下一个阶段的输入栅极电容充电。因此，这全都归结为一个电阻为电容充电，而另一个电阻对其放电（当然不是同时：）。

我要显示的公式来自数字集成电路 -Rabaey，Chakandrasan和Nikolic的设计观点。

考虑一个由MOS充电的电容器：

从供应中获取的能量将是

$$E_{VDD} = \int_0^\infty i_{VDD}(t)V_{DD}dt=V_{DD}\int_0^\infty C_L \frac{dv_{out}}{dt}dt = C_L V_{DD} \int_0^{VDD} dv_{out} = C_L {V_{DD}}^2$$

而最终存储在电容器中的能量将是

$$E_{C} = \int_0^\infty i_{VDD}(t)v_{out}dt = ... = \frac{C_L {V_{DD}}^2}{2}$$

_{当然，正如史蒂文指出的那样，我们不会等待无限的时间来对电容器充电和放电。但这甚至不取决于电阻，因为它的影响取决于电容器的最终电压。除此之外，在考虑瞬态过渡之前，我们希望在下一个栅极上施加一定的电压。因此，假设它的Vdd为95％，我们可以将其排除在外。}

因此，与MOS的输出电阻无关，它将消耗存储在电容器中的能量的一半，以恒定电压对其进行充电。在放电阶段，存储在电容器中的能量将在pMOS上耗散。

如果您认为在一个开关周期中存在一个L-> H和H-> L过渡，并定义了该逆变器完成一个周期的频率，则该简单门的功耗为：$f_S$

$$P = \frac{E_{VDD}}{t} = E_{VDD} \cdot f_S = C_L {V_{DD}}^2 f_S$$

请注意，如果您有N个门，将功率乘以N就足够了。现在，对于复杂的电路，情况会稍微复杂一些，因为并非所有门都将以相同的频率进行通断。您可以将参数定义为每个周期通行的门的平均分数。$\alpha<1$

所以公式变成

$$P_{TOT} = \alpha N C_L {V_{DD}}^2 f_S$$

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由于R引起的原因的小幅证明：正如史蒂文所写，电容器中的能量为：

$$E_C = \dfrac{V_{DD}^2 \cdot C}{2} \left(1 - e^{\dfrac{-2T_{charge}}{RC}}\right)$$

因此很明显，由于有限的充电时间，R是存储在电容器中能量的一个因素。但是，如果说必须将栅极充电至90％Vdd才能完成过渡，那么Tcharge和RC之间的固定比率为：

$$T_{charge} = \frac{-log(0.1)RC}{2} = kRC$$

选择它后，我们又有了一个独立于R的能量。

请注意，从0到kRC进行积分而不是无限进行积分，但是计算变得稍微复杂些。

我之前发布了另一个答案，但这不是很好，语言也不正确，我想对冒犯者表示歉意。

我一直在考虑这个问题，我认为这里的问题是对我来说，引用的文字表明电容是造成功耗的原因。事实并非如此。有抵抗力。

Voilà和成对的互补式MOS。MOSFET与电容器一起形成电荷泵。当输出变高时，P-MOSFET导通，它将通过对电容器充电；当输出变低时，电容器将通过N-MOSFET 放电至V S S。两个MOSFET都有导通电阻，这使得它们在充电/放电时会消耗功率。现在本建议电阻值无关紧要，而我说相反。好吧，我们都是对的，也都是错的。 $V_{DD}$$V_{SS}$

第一本：电容器电压和电流在充电过程中呈指数变化。目前

$I = \dfrac{V_{DD}}{R} e^{\dfrac{-t}{RC}}$

$P = I^2 R = \dfrac{V_{DD}^2}{R} e^{\dfrac{-2t}{RC}}$

and integrating over time gives us energy dissipated in the resistor:

$U = \dfrac{V_{DD}^2}{R} \displaystyle \int_{t=0}^{\infty} e^{\dfrac{-2t}{RC}} \mathrm{d}t = \dfrac{V_{DD}^2}{R} \dfrac{RC}{2} = \dfrac{V_{DD}^2 \cdot C}{2}$

which is indeed independent of $R$. So it looks like Ben is right.

Now me. "Infinity!? Are you out of your mind? This job has to be done in 0.3ns!" At school we seemed to have ages to charge a capacitor. If $t$ is finite we get

$U = \dfrac{V_{DD}^2}{R} \displaystyle \int_{t=0}^{t1} e^{\dfrac{-2t}{RC}}\mathrm{d}t = \dfrac{V_{DD}^2 \cdot C}{2} \left(1 - e^{\dfrac{-2t}{RC}}\right)$

and then $R$ is still a factor.
In practice it won't matter however since $RC \ll T_{CLOCK}$.

I cut some corners here assuming that $R$ is constant. But it's not easy. $R(t)$ depends on the gate's voltage, which depends on the gate's capacitance's charge curve, which depends on $R$. Easy if it's a linear system, but this isn't, so I chose for the exponential as an approximation.

Conclusion: while the dissipation is expressed in terms of $C$ it happens in $R$, which on first sight seems to have nothing to do with it.

What can be done about it? Lowering $R$ is no use. Can we decrease $C$? It would help to decrease the charge being drained from $V_{DD}$ to $V_{SS}$, but we need $C$. The gate capacitance is what makes a MOSFET work!

What if $R$ were zero, absolute zero? Then we wouldn't have dissipation, right? In that case switching would give an infinite $di/dt$, which would cause the switching energy to be radiated instead of dissipated, but the amount of energy would be the same. Your CPU would get less hot, but would be a wideband 100W RF noise transmitter.

The main power draw in CPU's is caused by the charging and discharging of capacitors during calculations. These electrical charges are dissipated in resistors, turning the associated electrical energy into heat.

The amount of energy in each capacitor is C_i/2 · V². If this capacitor is charged and discharged f times per second, the energy going in and out is C_i/2 · V² · f. Sum for all switching capacitors and substituting C = ΣC_i/2, you get C · V² · f

Capacitance is measured in Farads, which is Coulombs per Volt.

Frequency is measured in Hertz, which is units per second.

Reducing we get Coulomb-Volts per second, more commonly known as Watts, a unit of power.

Generally the current consumed by a device is proportional to the voltage. Since the power is voltage*current, the power becomes proportional to the square of the voltage.

Your equation is correct for the power drawn at any particular instant. But the current drawn by the CPU is not constant. The CPU is running at some frequency and changing states on a regular basis. It uses a certain amount of power for each state change.

If you understand I as the RMS current (the square root of the average of the square of the current) then your equation is correct. Putting these together, you get:

V · I(Rms) = C · V^2 · F
I(Rms) = C · V · F

So the average current varies linearly with the voltage, frequency, and capacitance. The power varies with the square of the DC supply voltage .