2
$\begingroup$

Can any device be made in such a way that the current through it is independent of carrier mobility?

In a usual semiconductor or a device which is made of semiconductor material the current due to the charge carriers (both thermally generated and carriers due to doping) is a function of mobility which in turn is a function of temperature, doping levels, electric field applied. My question is:

Is there any device which can be made in a way that the conduction through it is independent of the carrier mobility?

$\endgroup$

2 Answers 2

1
$\begingroup$

In an absolute sense, the answer is no - if we could set the carrier mobility to zero than no current will flow (of course one could argue that isn't a semiconductor anymore - I agree but like to stake out the extreme limits).

In a volume of like material, Ohms law will prevail, and the resistivity will be dependent on carrier mobility.

In a single device (diode, transistor) similar issues would impact performance.

In a linear IC, you can certainly provide feedback to compensate for carrier mobility differences over some range, dependent on the design. This is needed since there exist die-to-die, wafer-to-wafer, and lot-to-lot variations in mobility. How to design appropriately is more a question for the EE StackExchange site perhaps.

$\endgroup$
0
$\begingroup$

I agree with @JonCuster. I would, however, like to supplement his answer by mentioning a “trick” which can be used to circumvent the mobility-dependence problem. There is a way in which the “mobility,” as per the conventional definition, does not affect the current through the device. This is possible if the device is “ballistic.” With a ballistic device, however, we would be cheating, since mobility is undefined for a ballistic device. Before elaborating on these statements, let me first describe mobility from a microscopic point of view.

The microscopic interpretation of a phenomenological parameter such as mobility is connected to the rate of carrier scattering during carrier transport. In other words, mobility is a measure of the rate at which a current-carrying particle scatters off of phonons, electrons, impurities, lattice defects, etc. Mathematically this can be expressed as$$\mu = \frac{e \tau}{m_{\rm eff}}$$where $e$ is the unit charge, $\tau$ is the overall “carrier lifetime” or “carrier relaxation time,” and $m_{\rm eff}$ is the effective mass of the carrier. The individual contributions from qualitatively different scattering processes can be included using Matthiessen’s rule:$$\frac{1}{\tau} = \frac{1}{\tau_{\rm phonon}} + \frac{1}{\tau_{\rm impurity}} + \frac{1}{\tau_{\rm defects}} + \ldots$$where (for example) $\tau_{\rm phonon}$ would be the time interval between successive collision of a carrier with a phonon.

Now, in a ballistic device, the carrier travels through the entire device without scattering even once. This is accomplished by manipulating the device geometry, material parameters, temperature, etc., which in turn affect the contribution of the different scattering sources. However, as I pointed out earlier, we are cheating. That’s because the mobility, being solely defined in the terms of scattering rate, is undefined in a ballistic device. In other words, if I am telling you that a device has mobility $\mu$, then it is implied that the device is not ballistic. Hence you cannot have a fixed well-defined mobility and still have the current be independent of mobility. This is pretty much the same thing @JonCuster pointed out.

If you are interested in how one would obtain ballistic transport, then that’s a separate question, which I would be glad to answer if necessary. In this post, I simply pointed out that mobility alone cannot completely describe semiconductor transport in all parameter regimes. If you are interested, you can also find several examples of ballistic devices in the transport literature.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.