 |
This will be a short statement of
very elementary facts from basic physics or electrical engineering. |
|
 |
Any
"wire" used to carry electricity has a resistance R (we
will not consider superconductors here) - just measure it via R =
U/I. Its resistance measured in Ohms (O) might be small, but it is never zero. |
|
|
Any
"wire" that serves its purpose is isolated from the environment by
some dielectric (air or some insulation material) having some dielectric
constant er |
|
|
At some distance from any wire is always another conductor at some
arbitrary potential. Even if our hypothetical device consist of a wire only,
there is always the "earth" somewhere at (by definition) zero
potential. It follows: Any wire has some
capacitance C to something else. Again, this unintentional
parasitic capacity might be small, but
it is never zero. Moreover, whatever value it has, it is proportional to the
dielectric constant er of the
dielectric in question. |
|
Any wire (at
not too high frequencies) thus can be described by an equivalent circuit
diagram that looks like this. |
|
|
|
|
|
|
|
|
|
What happens if we put a digital signal on one end
of the wire, which we call the "input" end? Ideally, that means that
the voltage or better potential on the wire (the not grounded upper
"wire" in the picture above) jumps from 0V to, say, 5V
instantaneously. What happens at the output? |
|
|
Ideally, the potential would also go up suddenly
after a certain time t0 which is dictated by the speed
of light c, because nothing can move faster than that. we thus
have t0 = l/c with l =
length of the wire, and c = speed of light "in" the
wire, whatever that means. |
|
|
However, as soon as we raise the potential at the
input, we have to put charge in the parasitic capacitor C. The
voltage at this capacitor is the output voltage, and for a voltage
U we need to have the charge Q = C ·
U stored in the capacitor. |
|
|
For that a current has to flow into the
capacitor, and that current will be restricted by the resistance
R. Note that for just transmitting information as jumps in the
potential, current flow would not really be necessary, but our parasitic
capacitor needs charges flowing in (and later out) of it, if its potential is
to change. |
|
It is easy to describe quantitatively what
happens. For the current I flowing during the charging of the
capacitor we have two equations. |
|
|
|
|
|
| I = C · |
dUout
dt |
= |
Uin –
Uout
R |
|
|
|
|
|
|
|
This is an easy differential equation. We will
just look at the solutions for the two cases of Uin
going "up" to Uin-on at t = 0
and at Uin going "down from " to
Uin-on to 0; again at t = 0. |
|
|
For the "going up" case we obtain |
|
|
|
|
|
| Uout =
Uin-on · {1 – exp–
(t/RC)} |
|
|
|
|
|
|
|
This describes an output voltage that increases
as a function of time with the time constant RC; i.e.
after the time RC the output voltage is
Uout = Uin-on · (1 –
1/e) = 0.63 · Uin. |
|
If we look at what happens if the input voltage is
"going down", i.e.switched to zero (after it has been on long enough
to make sure the output voltage is equal to the input voltage), we obtain in
the same way: |
|
|
|
|
|
| Uout =
Uin-on · exp– (t/RC)} |
|
|
|
|
|
|
|
The output voltage thus decays with the time
constant RC from the "on" value to zero. |
|
Putting all of this together for some input
signals gives the following picture: |
|
|
|
|
|
|
|
|
|
|
What we have now is that at the
output end of a "wire" with length l, the voltage will
start to go up after a time t0 = l/c with
c = speed of light, i.e. the speed of electromagnetic wave propagation
in the wire (c might not be exactly the speed of light in vacuum, but
that is not the important part here). |
|
|
The voltage then will start to increase, after
the time constant RC it will be at about 2/3 of the
input voltage. After a few time constant it is identical to the input
voltage. |
|
|
If the input signal goes "down" again,
the output signal will follow, again with a delay of
t0 and then with a decay time again given by the time
constant RC. |
|
Now consider that we are discussing digital
applications. The output voltage will be interpreted as either
"0" or "1"; let's say the low potential =
0, the high potential = 1. |
|
|
During the transients of the voltage, we are
neither here nor there, but our electronic circuitry typically will interpret a
potential < U/2 as 0, > U/2 as
1. That simply means that the time it takes for a signal to travel
through a wire with the length l and then to be recognized by
what ever follows as a "1" is t0 plus
about one time constant RC of the wire. That means we have an
additional delay time given by the "undesirable" properties of the
wire. |
|
|
Even worth! The picture makes clear, that as soon
as you start to transmit signals shorter than a few time constant
RC, your output signal can't even follow the input any more. In
other words, at input frequencies much larger than about 1/RC,
you run into big problems. |
|
In plain words: You cannot run your circuitry at
frequencies > 1/RC - give or take a factor of 2 -
3.! |
| |
Where does that leave us? How large
is RC typically for the "wires" in an integrated
circuit. Let's look at a simplified situation: |
|
|
|
|
|
|
|
|
|
|
Let's look at a typical Al or Cu
conductor on a chip with a length of 1 cm, a cross-sectional area of
0.5 µm2 and the specific resistivity of
a typical metal of
about 2 mWcm.
The total resistance of our "wire" then is |
|
|
|
|
|
| R = |
r · l
A |
= |
2 · 10–6
Wcm2
0.5 · 10 –8cm2 |
= 400 W |
|
|
|
|
|
|
|
The capacitor in the picture is formed by the two
longish "plates", each with an area l · w =
l · A/dOx (w is the
lateral extension of the conductor strip; i.e w =
A/dOx). With typical numbers like
dOx = 300 nm, dielectric constant e = 3.7 (the value for SiO2), the
capacity of the wires then is |
|
|
|
|
|
| C = ee0 |
A · l
d2Ox |
= |
3.7 · 8.85 10–12 A
· s · 0.5 · 10–12 m2 ·
10–2 m
9 · 104 · 10–18 V · m ·
m2 |
= 1.82 ·
10–12 |
A · s
V |
= 1.82 pF |
|
|
|
|
|
|
For our time constant RC we obtain
|
|
|
|
|
|
| R · C = 400
· 1.82 · 10–12 |
A · s · V
V · A |
= 7.28 ·
10–10 s |
|
|
|
|
|
|
The maximal frequency we can transmit through this
"wire " then would be nmax =
1/RC = 1.36 GHz. This may appear a bit "handwaving" or
just a rough estimation; nevertheless, the problem should be clear. |
|
|
On a real chip, signals travel through many wires; possibly
far shorter then 1 cm, and possibly not all the way atop another wire,
so the capacity could be somewhat smaller. On the other hand, on a real chip,
theer are also neighboring wires to the left and the rigth, which increases the
capacity. |
|
|
Nowadys, we always use a multilevel metalization scheme and
being smart, we will keep the small wires on the lower levels quite short, and
give the long wires (i.e. for power supply) on the upper levels a large
cross-section - keeping R down in both cases. Nevertheless, a
clock frequency of 4 GHz, standard nowadays on many mass-produced
microprocessors, is an amazing feat considering the number from above. |
|
|
Essentially, the specific resistivity r of the conductor and the dielectric constant
e of the "intermetal dielectric" limit the
maximal frequency of the chip after all tricks of an optimized geometry have
been exhausted. |
|
Decreasing r by
a factor of about 1.6 by switching from Al to Cu started
to make a lot of sense around the year 2000. |
|
|
Replacing the ubiquitous SiO2 with an
e » 3.7 by a so-called "low k" dielectric with an e of 1.5 or so would make a lot of sense right
now (2005), unfortunately, nobody knows exactly how to do it (despite a
Billion $ or so, that have been spent on the search for a low-k material
up to now). |
|
|
|
© H. Föll
5.1.3 Basic Concepts of Connecting Transistors 
To index