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First Task: Derive numbers for the mobility µ. |
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First we need typical conductivities and electron densities in metals,
which we can take from the table in the link. |
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At the same time we expand the table a bit |
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| Material |
r [W cm] |
s [W–1 cm–1] |
Density d [103 kg m–3] |
Atomic weight w
[1u = 1.66 · 10–27 kg] |
n = d/w
[m–3] |
| Silver (Ag) |
1.6·10–6 |
6.2·105 | 10.49 |
107.9 | 5.85 · 1028 |
| Copper (Cu) |
1.7·10–6 |
5.9·105 | 8.92 |
63.5 | 8.46 · 1028 |
| Lead (Pb) |
21·10–6 |
4.8·104 | 11.34 |
207.2 | 3.3 · 1028 |
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For the mobility µ we have the
equation |
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With q = elementary charge = 1.60 10–19 C
and the density of electrons = density of atoms n calculated above, we obtain, for example for µAg |
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| µAg = |
6.2 · 105
1.6 · 10–19 · 5.85 · 1028
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m3
C · W · cm |
= | 66.2 | |
cm2
C · W |
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The unit is a bit strange, but rembering that 1 C = 1 A · 1 s and 1
W = 1 V / 1 A, we obtain |
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| µAg | = | 66.2 | |
cm2
Vs | | | |
| | | | µCu | = |
43.6 | | cm2
Vs |
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| µPb | = | 9.1 | |
cm2
Vs |
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Second Task: Derive numbers for the drift velocity vD by
considering a reasonable field strength. |
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The mobility µ
was defined as |
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So what is a reasonable field strength in a metal? |
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Easy. Consider a cube with side length l = 1 cm. Then, the relevant
area is F = 1 cm2, and its resistance R is given by |
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A Cu or Ag cube thus would have a resistance of about 1.5 ·10–6
W.
Applying a voltage of 1 V, or equivalently a field strength of 1 V/cm thus produces a current of I
= U/R
» 650 000 A or a current density ofj = 650 000 A/cm2 |
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That seems to be an awfully large current. Yes, but it is the kind of current
density encountered in integrated circuits! Think about it! |
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Nevertheless, the wires in your house carry at most about 30 A (above
that the fuse blows) with a cross section of about 1 mm2; so a reasonable current density is 3000 A/cm2,
which we will get for about U = 1.5 ·10–6
W · 3000 A = 4.5 mV. |
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For a rough estimate we then take a field strength of 5 mV/cm and a mobility
of 50 cm2/Vs and obtain |
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| vD | = | 50 · 5 | |
mV · cm2
cm · V · s | = 0.25 |
cm
s | = 2.5 | mm
s |
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That should come as some surprise! The electrons only have to move v e r y s
l o w l y on average in the current direction (or rather, due to sign conventions,
against it). |
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Is that true, or did we make a mistake? |
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It is true! However, it does not
mean that electrons will not run around like crazy inside the crystal, at very high speeds. It only means that their net movement in current (anti-)direction is very slow. |
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Think of an single fly in a fly swarm (even better, read
the module that discusses this analogy in detail). The flies are flying around at high speed like crazy – but
the fly swarm is not going anywhere as long as it stays in place. There is then no drift velocity and no net fly current! |
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© H. Föll (MaWi 2 Skript)
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