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# Overclocking "theory" - Page 2

Quote:

I went through the E8000 series technical data sheets, understood a 'bit' of it, but it didn't contain anything pertaining to what I was looking for.
Kal's example made sense to me, but I'm not able to relate it to Blameless's explination:
Quote:
Anyway, the faster you try to switch a transistor the less time you have for a change of state, and the more voltage typically needed to make the difference between on and off reliably distinguishable.
I'm not able to relate the 2 statements. Firstly the faster you switch a transistor the less time you have for a change of state from on to off, OK, got that!
Now, how does increasing voltage help in distinguishing the change in state and how is that related to the time required for change of states.

It takes more energy and hence more voltage to switch faster.
This is how I look at it from an EE's point of view. Please excuse the paint diagrams

An ideal transistor's base has no capacitance, meaning that when a voltage is applied instantaneously, it instantaneously switches on, and vice versa for switching off, creating a perfect square wave.

In reality, the transistor's base has a certain amount of capacitance. This means that the square waves rising and falling edges actually follow an exponential curve determined by the capacitance. For explanation purpose, I will ignore the falling edge.

As the working frequency increases, the base of the transistor doesn't have time to reach the voltage applied (vcore in this case) before the voltage starts to fall again.

If the frequency is increased further, not only does the voltage not reach the desired level, but it does not even reach the base threshold voltage of the transistor, meaning that it doesn't even turn on.

By increasing the voltage applied to the transistor, we decrease the charging time of the base capacitor restoring the wave's shape to a more square one and the base voltage successfully crosses the threshold again and switching the transistor on.

Note, this may not be an accurate representation, it is purely on theory. And yes, this is a very simple explanation and there's stuff that I've purposefully omitted.
Edited by ARandomOWL - 4/2/12 at 5:22pm
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Quote:
Originally Posted by Kal777

I think in order for the OP to understand all this he may need to get a good understanding on what voltage and current actually is and how it works and how current and voltage are different but linked.

I always remember it as current is what powers the circuit and is measured in Amps and voltage is what pushes the current through the circuit. So in respects one of the causes related to increasing voltage is that the current moves from A to B faster.

Voltage is also related to current by its increase, as voltage increases the supplied current also increases respectivly. If you produce 1A at 2v and increase the voltage to 4v the amps would also increase to 2A, what this means is that not only does increasing the voltage increase the speed current moves it also increases its charge.

Dude, I'm an Electronics Engineer, that stuff is like our abc's. The problem was your explanations were too simplistic for me, a layman could probably understand them better, but it was just confusing me further because I couldn't relate it to what I knew about transistors.
Quote:
Originally Posted by GeneO

It takes more energy and hence more voltage to switch faster.

Yea, that I understood, I was looking into something more detailed at the transistor level.
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Quote:
Originally Posted by ARandomOWl

This is how I look at it from an EE's point of view. Please excuse the paint diagrams
An ideal transistor's base has no capacitance, meaning that when a voltage is applied instantaneously, it instantaneously switches on, and vice versa for switching off, creating a perfect square wave.
In reality, the transistor's base has a certain amount of capacitance. This means that the square waves rising and falling edges actually follow an exponential curve determined by the capacitance. For explanation purpose, I will ignore the falling edge.
As the working frequency increases, the base of the transistor doesn't have time to reach the voltage applied (vcore in this case) before the voltage starts to fall again.
If the frequency is increased further, not only does the voltage not reach the desired level, but it does not even reach the base threshold voltage of the transistor, meaning that it doesn't even turn on.
By increasing the voltage applied to the transistor, we decrease the charging time of the base capacitor restoring the wave's shape to a more square one and the base voltage successfully crosses the threshold again and switching the transistor on.
Note, this may not be an accurate representation, it is purely on theory. And yes, this is a very simple explanation and there's stuff that I've purposefully omitted.

Gm = dIDS / dVGS
Output current divided by input voltage of a transistor, looked it up in an old textbook of mine, further derivations of the above got me to power consumption of a transistor:
Power = CPD x VDD^2 x Frequency, Capacitance times Voltage squared times Frequency.

I think I kind of get the whole picture now, but ofcourse this applies to single or small groups of transistors in ICs. Its probably much more complicated in VLSI ICs like the modern day processors.

+Rep x 2

Thanks!

Also a thank you to everyone else who replied.
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