Low-K process explanation

[Stryyder]

RCdelay= 2ρε(4L²/P²+L²/T²)

The speed of an electrical signal in an IC is governed by two components – the switching time of an individual transistor, known as transistor gate delay, and the signal propagation time between transistors, known as RC delay (R is metal wire resistance, C is interlevel dielectric capacitance):

RCdelay= 2ρε(4L²/P²+L²/T²) ,

where: ρ is metal resistivity, ε – permittivity of the interlevel dielectric (ILD) (ε is referred to as k in this field), L – line length, P – metal pitch, T – metal thickness.

 

[tazdevl]

K is a factor in capacitance, but it is not capacitance.

k is for conductance not capacitance.

Were you responding to me or Russ? Either way, look two posts above yours.

LOL Sorry, I blame two vodka tonics before dinner on my nonsensical response.

 

[jbond04]

Keep in mind that the high resistance materials, while they will reduce heat, will also decrease transistor switching speeds. This is probably not good for high-speed circuits, at least in the near-term, but would be excellent for handheld devices.

Which is why Intel has developed new metal gate materials to combat that effect:

"The second part of the solution is the development of a metal gate material, since the high-k gate dielectric is not compatible with today's transistor gate. The combination of the high-k gate dielectric with the metal gate enables a drastic reduction in current leakage while maintaining very high transistor performance, making it possible to drive Moore's Law and technology innovation well into the next decade."

http://www.intel.com/pressroom/archive/releases/20031105tech.htm

 

[Dio]

Keep in mind that the high resistance materials, while they will reduce heat
...
Edit: fixed, low conductance=high resistance.

Think it might need fixing again. Ohm's law V = IR and P = VI would disagree with you there (power dissipation is proportional to R, so high resistance materials will increase heat).

 

[PC-Engine]

Heh isn't that how electric heaters work?

 

[bloodbob]

Keep in mind that the high resistance materials, while they will reduce heat
...
Edit: fixed, low conductance=high resistance.

Think it might need fixing again. Ohm's law V = IR and P = VI would disagree with you there (power dissipation is proportional to R, so high resistance materials will increase heat).

P = VI

V= IR

P= I^2 X R

lets say I = 1 and R=1 and V =1

P = 1^2 X 1 = 1 Watt

okay lets double R

I = 0.5 and r=2 and V=1

P= 0.5^2x2= 0.25 X 2 = 0.5 Watts

Double the R which halved the heat output.

 

[Simon F]

No. (Well, some of it is correct, but the reasoning is wrong)

Low K is describing the stuff between the interconnects. 'K' is the measurement of electromagnetic conductance.

It's been many many years since university physics, but isn't K (well, Kappa, I think) the "permeability" of the dielectric?

Ahh! Google returns this short description

 

[PC-Engine]

P = 1^2 X 1 = 1 Watt

P=IV
V=IR

I=2 R=1

P=I^2R

P=4

I=2 R=2

P=8

Doubling R doubles P

 

[bloodbob]

P = 1^2 X 1 = 1 Watt

P=IV
V=IR

I=2 R=1

P=I^2R

P=4

I=2 R=2

P=8

Doubling R doubles P

Ehh you not keeping voltage constant so may I ask why your increasing the voltage dude? Transistor switching is generally voltage limiting so their is no reason to be increase the voltage to up the current ( remeber we are increase the resistance which is lowering the current leaking so your not wasting it so their should be no reason to up the voltage ).

 

[PC-Engine]

Isn't P=I^2R a consolidated formula? Meaning you don't have to use V=IR separately because it's already integrated into the formula. Or am I missing something? Isn't the whole point of substituting formulas to minimize multiple factors? In P=I^2R you don't need to know the voltage because it's already accounted for in IR. In the power formula all we need to do is change R. If you know I and R then V is known.

V=IR

I=2 R=1

V=2

P=2*2=4

I=2 R=2

V=4

P=2*4=8

 

[PSarge]

P = 1^2 X 1 = 1 Watt

P=IV
V=IR

I=2 R=1

P=I^2R

P=4

I=2 R=2

P=8

Doubling R doubles P

Ehh you not keeping voltage constant so may I ask why your increasing the voltage dude? Transistor switching is generally voltage limiting so their is no reason to be increase the voltage to up the current ( remeber we are increase the resistance which is lowering the current leaking so your not wasting it so their should be no reason to up the voltage ).

Probably to get the same amount of charge flowing in the same time (i.e. current is charge / time). To get a transistor to switch you have to charge the capacitor on its "gate" to a certain voltage. This takes a set amount of charge. If you double the resistance without increasing the voltage, you will half the rate at which you charge that capacitor. Your circuit now runs slower. Up the voltage and your cap is charging fast again, but you've doubled the power.

Low-K processes have small capacitances (K is the dielectric constant of the insulator between layers). If you have a small cap, it takes less time to charge (fast circuit) or you can run at a lower voltage (low power).

High resistance and Large Capacitors = slow circuit or high power. Pick one.
Low resistance and Small Capactitors = fast circuit or low power. Pick one.

P.S. I've probably over simplified this, as I've just kept all the relationships linear (i.e. cap's don't charge in linear time), but it'll do for this discusion

 

[RussSchultz]

No. (Well, some of it is correct, but the reasoning is wrong)

Low K is describing the stuff between the interconnects. 'K' is the measurement of electromagnetic conductance.

It's been many many years since university physics, but isn't K (well, Kappa, I think) the "permeability" of the dielectric?

Ahh! Google returns this short description

Yes, I should have put "conductance" in quotes, or used the more appropriate word of permeability.

 

[Bahadir]

Keep in mind that the high resistance materials, while they will reduce heat
...
Edit: fixed, low conductance=high resistance.

Think it might need fixing again. Ohm's law V = IR and P = VI would disagree with you there (power dissipation is proportional to R, so high resistance materials will increase heat).

P = VI

V= IR

P= I^2 X R

lets say I = 1 and R=1 and V =1

P = 1^2 X 1 = 1 Watt

okay lets double R

I = 0.5 and r=2 and V=1

P= 0.5^2x2= 0.25 X 2 = 0.5 Watts

Double the R which halved the heat output.

Someone's definitely got their calculations wrong!
In the first example, I = 1, R = 1, V =1

In the second example if R = 2 ohms, and current (I) is still 1 Amp, then
power dissipated would be:
P = I^2 * R = 1*2 = 2 Watts

From this formula:
P = I^2 * R you can see that Power is directly proportional to Resistance.
But then again, P = V^2 / R and P = V*I.

 

[Daliden]

You gotta keep the voltage the same, otherwise the calculations are meaningless when talking about computer parts -- right?

 

[KimB]

Keep in mind that the high resistance materials, while they will reduce heat
...
Edit: fixed, low conductance=high resistance.

Think it might need fixing again. Ohm's law V = IR and P = VI would disagree with you there (power dissipation is proportional to R, so high resistance materials will increase heat).

No. Current is reduced while voltage differential is kept the same. You have to use the equation:
P=V^2/R.

So power goes as 1/R.

 

[KimB]

You gotta keep the voltage the same, otherwise the calculations are meaningless when talking about computer parts -- right?

Well, there are effects due to high frequency operation, but those factors aren't necessary to just get the basic idea of what's going on.

 

[arjan de lumens]

V = I * R
P = V * I
= I^2 * R
= V^2 / R

To sum it up:
If you want to keep the CURRENT constant with a doubling of the resistance, then the power draw, and the needed voltage, will be doubled.
If you want to to keep the VOLTAGE constant with a doubling of the resistance, then the power draw and the current will be halved.

 

[VtC]

Current is reduced while voltage differential is kept the same. You have to use the equation:
P=V^2/R.

P=V^2/R is the preferable equation, IMO. In electronics, you control V and R. I is the thing that changes on you.

So, when calculating stuff like power you have to use P=V^2/R or else, as bloodbob did, remember that when you vary R you'll need to recalculate I rather than V. For me, it's easier to use an equation which only includes the independent variables in the first place. Hence, P=V^2/R.

 

[KimB]

P=V^2/R is the preferable equation, IMO. In electronics, you control V and R. I is the thing that changes on you.

Well, not always. In this case current is the dependent variable. There are electronic systems where current is roughly constant (such as some of the currents in transistors....).

 

[vandersl]

I think the best way to think of these circuits is the need to move little bits of charge around.

If you need to charge a gate through a resistance R with a voltage V, then your charge current is I=V/R.

For the sake of argument, say you need to move a charge C on or off a gate. The time it takes to move the charge is t=C/I.

This can be written as t=C/(V/R), or t=C*R/V.

If you double V you can cut the switching time in half.

If you cut R in half, you can cut the switching time in half.

If you reduce the required charge, you reduce the switching time (for a given R and V).

Note that the reality is much more complicated than just R, V, I and C.