# Thread: Some Important information on measuring your cpus voltage.

1. ## Some Important information on measuring your cpus voltage.

I've decided to write this after seeing countless posts on this forum showing incorrect measurement points for taking cpu core voltages. It seems that there has been a mix up somewhere along the line about where exactly is the correct place to measure this voltage and various people from mods to n00bs are getting it wrong.

So with the old adage “Give a man a fish; you have fed him for today. Teach a man to fish; and you have fed him for a lifetime” I'm going to explain how the power to your cpu is derived from the 12V line and in doing so show you where the actual vcore should be measured from.

I'm gonna attempt to keep this simple for those of you without an electronics background while trying not to alienate those who do so we'll start with a simple diagram. I will be borrowing heavily from answers.com as they have decent diagrams and easy to understand words here:

The simplest way to reduce a DC voltage is to use a voltage divider circuit, but voltage dividers waste energy, since they operate by bleeding off excess voltage as heat. A buck converter, on the other hand, can be remarkably efficient (easily up to 95&#37; for integrated circuits) and self-regulating, making it useful for tasks such as converting the 12-24v typical battery voltage in a laptop down to the several volts needed by the processor.

The operation of the buck converter is fairly simple, with an inductor (the squiggle to the right) and two switches (usually a transistor and a diode) that control the inductor. It alternates between connecting the inductor to source voltage to store energy in the inductor and discharging the inductor into the load.
OK so got that? basically you turn on the circuit and the inductor energises itself from the supply, you turn off the switch and the inductor de-energises into the load. On-off-on-off very quickly and you can control the voltage that the load "sees" by the rate that you keep it on and off. Magic.

So lets make it a little more complicated why not...

A synchronous buck converter is a modified version of the basic buck converter circuit topology in which the diode, D, is replaced by a second switch, S2. This modification is a tradeoff between increased cost and improved efficiency.

In a standard buck converter, the freewheeling diode turns on, on its own, shortly after the switch turns off, as a result of the rising voltage across the diode. This voltage drop across the diode results in a power loss....

...by replacing diode D with switch S2, which is advantageously selected for low losses, the converter efficiency can be improved. For example, a MOSFET with very low RDSON might be selected for S2, providing power loss on switch 2
So here we have a circuit that's a little more like what we find on a motherboard these days, with the two MOSFETs acting together to help keep down losses and allow for a wider range of voltage control due to the switching regulation being more flexible. This particular circuit can be used for applications that require low(ish) power demands like a nvidia 860i SPP:

you can match up the major components there and see its just like the circuit diagram above.

OK so lets scale things up a little:

So here we have a 3 phase circuit and a real world version of it on a motherboard. Again the reason here is losses and also the current draw from the load is now split between each phase, plus you can improve on voltage ripple by adding phases *cough* Gigabyte *cough* (although a well designed power supply will be fine with less!!!)

OK so we kinda understand whats going on here I hope, MOSFETs turn on and off very quickly regulating the 12volt line into something alot smaller by energising an inductor and allowing it to de-energise into the load (cpu) via some smoothing capacitors (to reduce the ripples).

Sorted.

Skim readers stop here!

So here's the important bit, where do you measure this derived voltage from?

Using the slightly modified image from answers I propose two spots, red and blue.

Red is either at the legs of the MOSFET or one leg of the inductor. I have seen alot of people point this out as the correct place to measure the vcore. This is incorrect and I'll explain why:

If you measure from this point you will get a voltage that is roughly vcore depending on how good you multimeter is. The reason why? because its a voltage that's being turned on and off in the order of several hundred times a second. So depending on how well you multimeter copes with that it'll basically attempt to average it giving an incorrect voltage. This is why alot, and I mean ALOT of people report different vcore's for the same board and complain about incorrectly reported voltage via software programs. Sure they're not 100% but they may be closer than you think...

The place you DO want to take your measurements is "after" the inductor (blue spot). Now things have been made a little more difficult with the onset of newer "cube" inductors, normally it you probed the leg of the inductor that was pointing towards the cpu socket you were sorted, the best way now is to find either an empty socket for the smoothing capacitors and use the positive terminal or even the positive side of a SMD capacitor which are becoming more popular. Failing that make a jumper lead and solder it to the underside of the board at the corresponding leg of the inductor/smoothing capacitor.

I hope that this will help alot of you not make this easy mistake, and will save countless hours of headaches about inaccurate vcore measurements.

With any luck this could be made a sticky for future reference

John

2. This is so helpful I would like to post it on my local forum with your permission please?

3. No worries mate, the important thing here is that people know, and understand why

4. Good read, thanks for taking the time to put it together for the community

5. So when we do voltmods we are changing the way the controller signals the switching of the FETS? Which makes a FET heat up more, faster or slower switching? What makes higher/lower voltage - faster or slower switching? What do the SMD's (the one's we often mod) typically do in the circut?

Excellent post Johnny

6. So basically by vmodding you actually underclock the mosfets, is that correct? And that should mean higher voltage = more ripple?

7. Originally Posted by G H Z
So when we do voltmods we are changing the way the controller signals the switching of the FETS? Which makes a FET heat up more, faster or slower switching? What makes higher/lower voltage - faster or slower switching? What do the SMD's (the one's we often mod) typically do in the circut?

Excellent post Johnny
Thanks G H Z :cheers:

When you vmod for more voltage or control over the voltage you are simply changing the feedback voltage. This is a very simple circuit, just two resistors in series between the actual output voltage and ground. The values of the two resistors defines the setpoint voltage that is measured between the resistors (see the site below)

http://www.electronics2000.co.uk/calc/calcdiv.htm

The vmod basically soldering in parallel with the lower resistor, which means that the lower that variable resistor goes the lower the voltage at V2 (for the image in the link) is. Now if the perceived voltage is lower then the control circuit will attempt to increase the output voltage to compensate, for example:

we set 1.5volts in the bios -> controller set 1.5volts and feedback loop shows i.5volts

we add a vmod so that the feedback loop now sees 1.35volts (10% reduction) so 1.5v in BIOS controller attempts to set 1.5volts but feed back says its 1.35 volts so controller ups the output to 1.65 volts so that the feedback loop now shows 1.5volts, as is the requirement in the BIOS.

Now that is various simplifications in there, things like OVP and feedback stability have been ignored but I hope you get a general feel for whats going on

Technically, in regards to your comments on switching its to do with how long the FETs are either on or off for a given time, called the duty cycle. Ideally you want to be switching as fast as you can to limit ripple somewhat ( although the act of switching will inherently create spikes in the output that will be dealt with according to how good your smoothing circuit is tuned) but the bigger the voltage the longer the switches will have to be "ON" for a given cycle.

While this will have something to do with the increase in heat I'd personally say its because the cpu or whatever you're supplying is drawing more current because you're running it faster.

8. Great post. I'd be screwed without it!

9. Thank you for taking the time to create a well written and informative post!

Anyone have more resources for explanations of this type?

10. Nice information.....

11. Can you upload some pictures to ease understanding?

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