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Why is BOTH the electron & hole drift velocity positive?

by Oz Alikhan
Tags: drift, electrons, holes, mobility
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Oz Alikhan
#1
Oct15-12, 02:05 AM
P: 12
Good Morning,

The equations under investigation:

Electrons: J = enμE + eD(dn/dx)
Holes: J = epμE - eD(dp/dx)

n or p = concentration of electrons or holes respectively
D = diffusion constant
μ = mobility


The question in mind is as follows:
If holes are traveling in the direction on conventional current, then shouldn't the drift velocity for holes be fully negative, i.e. -epμE - eD(dp/dx)?
I understand by definition that, μ = mobility, is the absolute value of drift velocity (speed) over the electric field hence the direction is lost in the process (or is the same for both a hole and an electron). Therefore, shouldn't the expression for holes written above also technically factor in a negative sign corresponding to the hole drift 'velocity' part of the equation?

Please dispel my confusion lest more holes puncture brain.
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DrDu
#2
Oct15-12, 03:23 AM
Sci Advisor
P: 3,593
No, mobility is proportional to the inverse of the effective mass. However, at the top of the valence band, the effective mass of an electron is negative and they have a negative mobility. Hence the holes have a positive mobility.
Ratch
#3
Oct15-12, 08:44 AM
P: 315
Oz Alikhan,

The question in mind is as follows:
If holes are traveling in the direction on conventional current, then shouldn't the drift velocity for holes be fully negative, i.e. -epμE - eD(dp/dx)?
Why would you think that?

I understand by definition that, μ = mobility, is the absolute value of drift velocity (speed) over the electric field hence the direction is lost in the process (or is the same for both a hole and an electron).
You are correct in that mobility is defined to be always positive. Direction is set by the sign of the term with respect to the electric field. A positive sign means the J direction is the same as the electric field, whereas, a negative sign means the J direction is opposite with respect the electric field.

Therefore, shouldn't the expression for holes written above also technically factor in a negative sign corresponding to the hole drift 'velocity' part of the equation?
No, the first term is the drift current, which by definition, is the charged-particle motion in response to an applied electric field. Hole drift current is in the same direction as the electric field. The second term is the diffusion current, which is caused by thermal motion of the charged-particles. This causes the higher concentration of particles to seek a position where the concentration is lower. If there is a positive concentration gradient, i.e., the particles increase in concentration as the distance along the x-axis increases, then it can be expected that the current will be in a minus x-axis direction. That is why the minus sign is present for the hole diffusion term. By the way, velocity does not enter into the "J" equation.

Ratch

Enthalpy
#4
Oct16-12, 09:31 AM
P: 661
Why is BOTH the electron & hole drift velocity positive?

It's because holes have a positive charge, so the same direction of electric field acts in the opposite direction than on electrons.
Ratch
#5
Oct16-12, 10:53 AM
P: 315
Enthalpy,

To whom are you addressing your comments, and what question are you answering?

Ratch
Oz Alikhan
#6
Oct16-12, 11:46 AM
P: 12
Quote Quote by DrDu View Post
No, mobility is proportional to the inverse of the effective mass. However, at the top of the valence band, the effective mass of an electron is negative and they have a negative mobility. Hence the holes have a positive mobility.
With that piece of information, it makes a lot of sense. The only thing that bothers me then though is the other definition of mobility which is, μ = |Drift Velocity| |E|, hence that would imply that enμE , epμE > 0 thus positive. However, aren't holes 'traveling' the opposite direction of electrons hence opposite signs for enμE & epμE?
Oz Alikhan
#7
Oct16-12, 11:51 AM
P: 12
Quote Quote by Ratch View Post
Oz Alikhan,

The question in mind is as follows:
If holes are traveling in the direction on conventional current, then shouldn't the drift velocity for holes be fully negative, i.e. -epμE - eD(dp/dx)?


Why would you think that?


Ratch
Well, the idea in mind that I have is that holes travel in the opposite direction of electrons therefore they follow the path of conventional current. I know I am assuming and treating holes as an existing positive charged particle; is that the reason why the hole current density expression (epμE - eD(dp/dx)) does not have a negative drift velocity (-epμE)? Or have I gone on a tangent to the moon?
Ratch
#8
Oct16-12, 01:18 PM
P: 315
Oz Alikhan,

Well, the idea in mind that I have is that holes travel in the opposite direction of electrons therefore they follow the path of conventional current.
Yes, that is correct. Conventional current asserts that the positive charges determine the direction of the current.

I know I am assuming and treating holes as an existing positive charged particle;
That's fine. At the quantum level, holes have the same status as a positively charged particle.

is that the reason why the hole current density expression (epμE - eD(dp/dx)) does not have a negative drift velocity (-epμE)? Or have I gone on a tangent to the moon?
I think you are still confused. Let's assume that the electric field E is from left-to-right (LR). It will push or force holes to move in a LR direction. It will attract or force electrons in a RL direction. But a negative particle moving is one direction (RL) is the same as a positive particle moving in the opposite direction (LR). Therefore, whether the electric field is moving negative or positive charges, the conventional current is always in accordance with the direction of the electric field. So the current density due to the drift term always has a positive sign regardless of the sign of the charge.

However, aren't holes 'traveling' the opposite direction of electrons hence opposite signs for enμE & epμE?
Holes traveling in one direction are equivalent to electrons traveling in the opposite direction. Therefore, no sign change is necessary.

Ratch
Enthalpy
#9
Oct18-12, 02:53 PM
P: 661
Quote Quote by Ratch View Post
Enthalpy,

To whom are you addressing your comments, and what question are you answering?

Ratch
I was answering the first post.


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