Question: How much power is increased ...

Question:

When the Vcore of a CPU (e.g. TB B 1700+ DLT3C) at 1.5V running at 1.5 GHz is raised to 1.65V in order to overclock at 2.4 GHz, what is the increase in power (active power) that the CPU consumes in order to substain the higher clock rate?

(a) 10% (Vcore is increased by 10% from 1.5V to 1.65V)

(b) 21% (since Vcore is increased by 10%, and power is proportional to V^2. So 1.1 x 1.1 = 1.21, hence 21%)

(c) almost 100%

(d) about 50%

(e) none of the above

I'll give my answer after 5 posts. My answer may be right or may be wrong. And we can start some discussion.

I think this will help us to understand some more underlying theory and math in overclocking.

b

you could actually measure the difference in power consumption assuming you have a volt meter and amp meter or amper clamp something hehehehe

d

I'm thinking towards (b)

Power = I^2*R = V*I = V^2/R
Since the resistance of the chip doesn't change with voltage, power varies to the second power.

(1.65^2)/(1.5^2) = 1.21 = 21% increase in power delivered.



(Question for uneducated (me): what is stock Vcore on Tbred B and barton cores?)

the power is increased by the squre of the voltage increase
and liniearly by the frequency.

In your case:
( (1.65/1.5)^2 ) * (2.4/1.5) =
(1.1^2) * 1.6 =
1.21 * 1.6 =
1.936 = 94%

The power increase of rising the voltage while keeping the same frequency is 21%

Rising the frequency while keeping the same voltage is 60%.

Rising both gives 93.6% increase.

The current doesn't flow all the time through the core - it flows on pulses which gives the frequency (Hz means times per second).
Thats why if you run twice the pulses you will get twice the heat.

Albigger is right for the current/voltage increase.



Side note:

Although, this is true only for resistors and not for transistors nor diods.

A diod has property called voltage drop.
If you connect a resistor and a diod in a chain the voltage on the diod pins will be almost constant (usually for leds - 1.7V). So if you have 3V at the ends of the chain 1.7V go to the diod and 1.3 to the resistor. By the resistor resistance you can calculate the current which is the same for both.

If you run less then 1.7V no current flows.
If you run 4.3V the current is double although 4.3 is not 2*3
(4.3-1.7)/R=2.6/R and (3-1.7)/R=1.3/R

Anyway, What I wanted to say is that if you increase the voltage from 1.5 to 1.65 the power may increas more then V^2/R

I don't have idea what is the voltage drop of 130 nm transistors...

Albigger said:

I'm thinking towards (b)

Power = I^2*R = V*I = V^2/R
Since the resistance of the chip doesn't change with voltage, power varies to the second power.

(1.65^2)/(1.5^2) = 1.21 = 21% increase in power delivered.



(Question for uneducated (me): what is stock Vcore on Tbred B and barton cores?)

Click to expand...

For Tbred B (Model 8) 256KB L2
XP1600+ only rated 1.6V coded DUT3C
XP1700+ rated 1.5V coded DLT3C or rated 1.6V coded DUT3C
XP1800+ rated 1.5V coded DLT3C or rated 1.6V coded DUT3C
XP 2000+, 2100+, 2200+ only rated 1.60V coded DUT3C
XP 2400+, 2600+ only rated 1.65V coded DKT3C
XP 2600+, 2700+ (FSB 333) only rated 1.65V coded DKT3D
(the last D means FSB 333)

For Barton (Model 10) 512KB L2
2500+, 2800+, 3000+ only rated 1.65V

The L stands for rated voltage of 1.5V, the U stands for rated voltage of 1.6V, and K stands for rated voltage of 1.65V.

TBred A, B and Barton are based on 0.13 micron silicon technology. TBred A has one less layer of metal.

Wow! Nice work Pla! I completely ignored the frequency part, which you obviously can't do!

So I take it that Pla pretty much got it right on then?


And thanks for the info about all the default core voltages.


Still it seems kind of hard to believe that going from 1.5V @ 1.5Ghz to 1.65V @ 2.4Ghz would increase the Power to almost twice what it was?

That means that the heat being put into our waterblocks/heatsinks scales with the power input (assuming the secondary losses are rather constant or at least negligible) and varies as V^2 and frequency^1.

Could we test this reliably in any way to see the effects of voltage and frequency????

So when I boost my XP1700+ DLT3C to 2800MHz and 2.2vcore then the power consumption goes up over 4 fold!!

From your calculations I used (2.2/1.5)^2*(2800/1466) = 4.11.

How does this correlate to the power output? If the power output at stock is 50Watts (http://users.erols.com/chare/elec.htm ) does this mean that the power output for the overclock is over 200Watts No wonder my phaser was having trouble with this overclock!

Here is a link to a program called NewWatt3 (scroll down to find the program). It will calculate the wattage of your overclocked/overvolted CPU. The program already has all of the default wattage information for most CPU's, both Intel and AMD. All you have to do is input the overclocked CPU speed and CPU voltage.

Russell_hq said:

So when I boost my XP1700+ DLT3C to 2800MHz and 2.2vcore then the power consumption goes up over 4 fold!!

From your calculations I used (2.2/1.5)^2*(2800/1466) = 4.11.

How does this correlate to the power output? If the power output at stock is 50Watts (http://users.erols.com/chare/elec.htm ) does this mean that the power output for the overclock is over 200Watts No wonder my phaser was having trouble with this overclock!

Click to expand...

I think the active power will go up to about 4 times the active power at 1.5V 1466 MHz.

BTW, and the active current Icore will be 2.8 times (=2.2x2.8/(1.5x1.466)). The rated 1700+ at 1.5V Vcore takes about 29.9A (nominal from spec sheet). So at 2.2V 2800 MHz oc, the active current will be 2.8 times the active current at 1.5V = 2.8 x 29.9A = 83.7A !!!!

Things are a bit more complicated than that. About 10-18%, take it as 13%, of the rated current is the dc bias or leakage current. Take that into account, the Icore at 2.2V 2800 MHz would be about

Icore = 29.9 * 0.87 * 2.8 + 29.9 * 0.13 * (2.2/1.5) = 72.8 + 5.7 = 78.5A !!!!

About 2.5+ times that of a PSU spec to run normally (with some oc), but not at this level. That are ways to estimate the current needed by the 5V/12V lines due to Icore for this level of oc.

E.g. Assume Vcore is derived on the 12V line, assume regulator effeiciency of 80%, converting 12V to 2.2V Vcore, the current would be

I_12V = 1.25 (2.2 * 78.5) / 12 = 18.0 A

This is the active current on the 12V to maintain the huge oc power. Adding other current for fans, harddrives, dvd, cdrw, ... you will see what PSU you will need.

Other big 12V current consumption:
Fan = 0.2 - 0.7 A (e.g. TT SFII)
Hard Drive ~ 1.3 A (e.g. WD 120G 8MB SE)
DVD, CDRW ~ 1 A
...

If the Vcore is derived on the 5V line, assume regulator effeiciency of 80%, converting 5V to 2.2V Vcore, the current would be

I_5V = 1.25 (2.2 * 78.5) / 5 = 43 A

This is the active current on the 5V to maintain the huge oc power. Adding other current harddrives, dvd, cdrw, ... you will see what PSU you will need.

I think the answer to the question is (c).

The power consumed by a CPU is

Power = Power_static + Power_active

The Power_active is proportional to C V^2 f, where V is the Vcore and f is the clock freqency, C is the equivalence capacitor respresenting the chip. The active power is for doing the computation by charging (and discharging) the 100's millions of small capactors (from coupling among transistors, metal wires, silicon substrate) inside the chips through the 10's of millions of fet transistors.

The Power_static is about 10-18% of the total power. It is for biasing the millions of transistors as leakage current. It changes with V^2 where V is the Vcore. Also this current is sensitive to temperatue, it increases as die temperature and Vcore increase during oc.

Using a simplifed picture, the CPU can be viewed macroscopically as a resistor and a capacitor in parallel for power and current estimate. (There is also inductor in series, but skip it for this discussion).


From the air cooling example, going from 1.5V 1.5 GHz to 1.65V 2.4GHz,

OC Pactive = (1.65 * 1.65 * 2.4)/(1.5 * 1.5 * 1.5) = 1.93 (93 % increase) <---- answer (c) on active power, which is almost 100%

Assuming 10% of total power is static powe.
OC Pstatic = 1.1 * 1.1 = 1.21 (21 % increase)

Going into more detail about total power, weighing with both static and active power,
Total OC power = 1.21 * 0.1 + 1.93 * 0.9 = 1.86 (86% increase)

My estimate could be wrong, there may be better model and estiamtes. Comments welcome.

We really have to adjust our thinking for these high current CPU oc to run at high clock rate. Especially picking up the right PSU, so the 5V/12V line have sufficient curent to maintain steady Vcore. It is very differnet than 6-12 months ago, when these things are at 1.7-1.8V 2GHz range.

Edward2 said:

Here is a link to a program called NewWatt3 (scroll down to find the program). It will calculate the wattage of your overclocked/overvolted CPU. The program already has all of the default wattage information for most CPU's, both Intel and AMD. All you have to do is input the overclocked CPU speed and CPU voltage.

Click to expand...

Thanks. It is a very nice program and it confirms what I think.

Would be nice if it does current estimate too.

I'm certainly no expert, but it doesn't seem logical to me that power consumption would scale linearly with frequency. You are correct in that there are twice as many pulses with twice the clock speed, but if this is the case, are these pulses necessarily the same length as the non-oc'd pulses?
What I'm asking is, if you have (for simplicity) 5 pulses a second and then you oc to 10 pulses a second, are each of those 10 pulses the same length (in time) as the 5 pulses from the original chip? or are they shorter and therefore create less heat per pulse?
The way I see it, the pulses would be more rapid but shorter as the chip must generate more of them per second therefore the full current is not flowing for as long during each pulse so the increase in heat/power dissipated is not truly linear. I'm assuming this because once you overclock too far, the chip simply can't keep up as the pulses are much too quick (short) and the transistors' bases cannot fully charge and or discharge.

As I said, I'm no expert. I've never had a class where this kind of detail was covered, but I thought I'd at least throw my ideas out.

Lord_MiL and hitechjb1 - very good points!!!

I don't know anymore...
I guess the pulses should be shorter. But if they are reciprocal to the frequency then there would be no power increase with the frequency increase. So I guess it is somewhere in between.

If anybody knows please say.

Another point is that 1700+ and 2400+ have less then proportioanal power ratings.

Also for shorter pulses don't you need quicker gates?

Also when you run a pulse after sometime it has run through the cpu. The smaller the chip the less time it needs. If you run a second pulse before the first one has finished it will not run as expected. Thats why they keep making smaller the chips.

Anyway, anybody who knows?

What is the active power of a CPU at frequency f and voltage V

When a capacitor C is charged to charge Q within a time T by a current I. Let V be the voltage across the capacitor

Q = C V
I = Q / T
I = C V / T

If the capacitor is charged repetitively by a clock of frequency f of period T, since f = 1/T. So

I = C V f

Hence the current is proportional to the clock frequency f.

So the shorter the period T (faster clock), the bigger the current I. Keeping V constant and all the capacitance inside the chip remain constant (1st order).

When the clock is double, the shape of the pulse (described by the aspect ratio or duty cyle) remains the same, i.e. the high (logic 1) and low (logic 0) intervals remain the same (1st order), generated by the internal clock and pulse generators.

It is correct that when the clock is double (frequency is twice), the pulse width is half, as a consequence, it takes half the time to charge up the same capacitance with the same voltage V, HENCE the current is DOUBLE (and not halved), because I = C V / T. (In this paragraph, V is kept constant. It will be more current if V is also increase.)

Further for power,

P = V I = C V^2 f

That explains why going from 1.5V 1.5 GHz to 1.65V 2.4GHz, the active power is almost double.

And when going from 1.5V 1.47MHz to 2.2V 2.8GHz as Russell_hq is doing (see earlier post), the active power would be about 4x !! and the active current would be around 2.8x !! (CHECK the PSU !!!!)


As far as whether the transistors can keep up when the clock is increased. It always won't and will fail at certain freq. This is a separate issue. But as long as it can run at that freqency with a certain Vcore increase, the above current and power estimate hold.

Thank you, hitechjb1!
This enlightens us.

Another question:
When you increase the voltage and frequency so that the cpu still works, f.e. Barton from 1.65 1826 to 1.825 2300 and it still works stable at 49C full load,
Is there any chance the cpu can burn/short circuit or get damaged?
Or does the damage point comes after it stops functioning?

As for max Vcore, there is a link in the earlier post by Russell_hq to a CPU table. It agrees with my estimate (differs by 50 mV in Vcore).

hitechjb1 said:

According to the AMD datasheet for model 8: datasheet link

In Chapter 8,

Quoted:
"The AMD Athlon XP processor model 8 should not be subjected to conditions exceeding the absolute ratings, as such conditions can adversely affect long-term reliability or result in functional damage."

The absolute rating for Vcore = Vcc_core_dc_max + 0.5 V

Vcc_core_dc_max = Vcc_core_nominal + 0.05 V
Vcc_core_ac_max = Vcc_core_nominal + 0.15 V

Vcc_core_nominal = 1.50 V for DLT3C
Vcc_core_nominal = 1.60 V for DUT3C
Vcc_core_nominal = 1.65 V for DKT3C

E.g. 1700 DLT3C
Vcc_core_dc_max = 1.5 + 0.05 = 1.55 V
The absolute rating for Vcore = 1.55 + 0.5 = 2.05 V

The above is based on my interpretation. Pls read the datasheet, and make your judgement.

Click to expand...

I think don't treat the temperature 40 or 45 or 50 and Vcore 1.9, 2.0 or 2.1 measured by the mb or any programs as absolute. Since there are measurement errors, calibration errors, ... For example, I think if one put that max Vcore on the 1700 DLT3C on air without paying attention to the die tempeature sensitivity over voltage, most likely will kill the chip (because the sensor can be off by 10C).

Not like system manufacturers working at nominal, for overclocking, we are working at the edge of the spec and a particular part, so each CPU and part have to be treated differently. In particular, the sensitivity of die temperature (which also relects internal leakage current) against voltage.

When Vcore is increased to a point that the clock can nolonger respond to the rate of clock per Vcore increase earlier, say it is reaching 1/8 (just an example) of the original rate of clock/Vcore. It is reaching the limit.

E.g. at the beginning of oc the TB B 1700+ DLT3C, I can get 150 MHz per 25 mV Vcore increase. At the last 100 MHz, I only get 25 MHz for 50 mV Vcore increase (a rate of less than 1/10 !!!).

There is no point of pushing any more, since the Vcore increase is not for adding the active power to run the CPU faster, it is consumed in the static leakage current (which is positively sensitive to temperature), and in turn heats up the chip more, and hence drawing more leakage current, ... and heats up more, ... and getting into run away, positive feedback. Don't overclock to the point that the temperature (and current) begins to run away from a small increase of voltage.

If voltage is increased few steps instead of smallest increment, it may pass that run away point for current and temperature and kill the chip.

Thanks a lot!

Pla said:

Thank you, hitechjb1!
This enlightens us.

Another question:
When you increase the voltage and frequency so that the cpu still works, f.e. Barton from 1.65 1826 to 1.825 2300 and it still works stable at 49C full load,
Is there any chance the cpu can burn/short circuit or get damaged?
Or does the damage point comes after it stops functioning?

Click to expand...

Short answer:

The system will run into all kinds problems, such as cannot post, cannot boot windows completely, BSOD after getting into windows, Prime95 crash, 3DMark crash, .... first, before damaging the chips. And one would know to stop.

When certain progrm crashes during CPU oc, it means there are certain CPU commands not being executed correctly. At this early sign, it would indicate that some circuits in the CPU (may be just one transistor out of millions) is not fast enough to meet the overall timing requirement set by the faster clock. I.e. the higher Vcore is not enough to make all the transistors involved to run fast enough, ... so crashes.

Unless you increase Vcore at big multiple steps instead of smallest step of 25 mV given by the bios, and hence without knowing passing the run away point and so damaging the chip.