inductor current question

-)

In a metal you can estimate the electron thermal velocity (v). mv^2 =3D kT. (m is the electron mass in the metal and k is Boltmanns constant.) And in most cases this turns out to be a lot faster than the drift velocity from any applied electric field.

George H.

Reply to
George Herold
Loading thread data ...

I think m_e is the only thing that changes between materials, of course the electron mass is the *effective* electron mass, because the atomic lattice acts like a magnifying lens. And it has different values in different directions, which may be unexpected to those who think of mass as constant.

Anyway, in silicon, I recall it's on the order of 10^5 m/s. Pretty fast. An important quality of silicon is, when very pure, it has a low carrier density, so for a given current density, the carrier velocity is pretty high. It's possible to get it so high, it would otherwise exceed the thermal velocity. But this doesn't occur, and the resulting characteristic becomes constant-current as velocity saturates!

I don't actually know how easy it is to do this. The electric field must be very close to breakdown. Most devices are designed to avalanche before this occurs (or before other phenomenon, like punch-through).

Gallium arsenide does have a constant current characteristic, in fact negative resistance, which is due to an unrelated effect with the energy levels (a very quick effect, hence making Gunn "diodes" effective in the microwave range, which are not diodes but actually unodes!).

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms
Reply to
Tim Williams

he

or

e

No they dont !!

Reply to
cbarn24050

the

stor

the

Every circuit has capacitance, which is a low impendance to fast voltage edges. It will also form a resonant circuit with the inductor so you get oscillations as well as RF radiation.

Reply to
cbarn24050

More likely, IME, that transistor was already ruined (shorted) and the fuse blew in the normal overcurrent fashion.

Reply to
JosephKK

V_eht * Q_e = 0.5 * M_e V^2

The left half is the work done on the electron, the right half is it's kinetic energy.

--
¡spu?? ou '?? ?oo?
Reply to
Jasen Betts

If you check out the way the one dimensional bandgap curves plot it is closer to a transistor.

Reply to
JosephKK

=3D

Yup and in semiconductors the mass can be negative. Well, we just call those holes and put the negative sign on the charge.

One of the best pieces of data I helped take in grad school showed cyclotron resonance of the electron and the two holes in GaAlAs quantum wells, as well as the 1s-2p, and 1s-3p exciton transistions.

t.

I think that is what they call 'hot electrons', but that is way outside my knowledge comfort zone.

George H.

st

ore

Reply to
George Herold

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.