2
\$\begingroup\$

Consider a pn junction diode is forward biased. And assume the voltage applied to the diode is enough such that the depletion region is disappeared.

In this circumstance if we zoom inside the diode, can we say the following for the electric current through the diode?:

The current in p part of the diode is carried by the holes; and the current in the n part is carried by the electrons?

If so, does that mean the speed of the current flow is different at p and n parts?

edit: Is the resistance at p and n parts of the diode different in forward mode?

\$\endgroup\$
  • 1
    \$\begingroup\$ The current in p part of the diode is carried by the holes; and the current in the n part is carried by the electrons? Not really, the holes are also just electrons moving. In forward mode the number of electrons in the P-region is so high that there aren't many holes. Nearly all holes recombine due to the flood of electrons coming from the N-side. Also, speed of current flow, what does that mean ? Current flow means x electrons per second. The speed of the electrons is irrelevant. \$\endgroup\$ – Bimpelrekkie Aug 13 '17 at 20:01
  • \$\begingroup\$ I know that holes are also just electrons moving but its a bit different. ´The absences of electrons moving and electrons do not move like as in n type. But I got your point. If "nearly all holes recombine due to the flood of electrons coming from the N-side" so what you mean is the current flow is actually the electron flow in the forward mode? And there are almost no holes? \$\endgroup\$ – atmnt Aug 13 '17 at 20:07
  • \$\begingroup\$ @user134429 Electrons flow through the n-region towards the junction; holes flow through the p-region towards the junction; with these two kinds of carriers recombining in the vicinity of the junction (over the diffusion length.) The electrons and holes travel in opposite directions. But having opposite charges the overall current seen from the outside is all in the same direction. \$\endgroup\$ – jonk Aug 13 '17 at 20:31
  • \$\begingroup\$ @user134429 Because holes have lower mobility, the P material in a simple diode is usually doped to a higher concentration than the N material. In this way, hole and electron flow vs \$x\$ is relatively matched. However, the mean velocity of any specific charge is a function of its mobility and the electric field intensity. Since the electric field intensity is the same for any charge (whether +1 or -1) at any particular point through the diode, the mean free path and therefore the mean velocity will be different. Hence, the need for different doping concentrations. \$\endgroup\$ – jonk Aug 13 '17 at 20:33
  • \$\begingroup\$ In a p-n junction diode if you forward biased the junction enough to have no more depletion region you would have an immense amount of current. Diodes are never biased this far. \$\endgroup\$ – Matt Aug 14 '17 at 1:05
2
\$\begingroup\$

Short version:

Yes, the current in a diode is carried out by both electrons and holes. Mainly electrons on the n side and mainly holes on the p side

Long version:

When an electron moves from the n side of the junction to the p side it will continue to travel on its way until it recombines with a hole. The recombination rate is a function of varying factoring including acceptor concentration and trap density. Likewise for a hole moving from the p side to the n side.

Most diodes have fairly high dopant concentrations so these recombination rates will be quite high, so the electrons or holes won't make it too far into the p or n sides, respectively.

You can look at the "drift velocity" (\$\nu\$) of the electrons and holes in the semiconductor. Thats probably the closest thing to the "speed of the current flow". Drift velocity looks at the average speed of charge carriers, but there is a very wide range of velocities that make up this average.

The drift velocity is a function of carrier mobility \$\mu\$ and electric field \$E\$:

\$\nu = \mu E\$

While carrier mobility is a function mainly of temperature, total impurity concentration, and the semiconductor material in use. In most semiconductors the hole mobility is generally much lower than the electron mobility given the same impurity concentration. Therefore, it is generally safe to say that drift velocity in the p region is lower than that in the n region. Unless there is a very strong difference in dopant concentrations in the two regions.

The resistivity (\$\rho\$) of a semiconductor is a function of mobility as well:

\$\rho = \frac{1}{q(\mu_nn + \mu_pp)}\$

These values don't change based on the bias of the diode. However, it isn't really useful to look at resistance of the p and n regions in a vacuum since the main thing that controls the current is the barrier height.

\$\endgroup\$
0
\$\begingroup\$

A hole is simply the absence of an electron, if a hole moves, it just means an electron from a near-by (say) silicon atom came in the place where the hole was, thereby, creating a hole where the electron itself initially was. The flow of current is actually the flow of electrons, and since in the P-doped region holes are moving slower (ie electrons are moving slower because there are not many holes due to recombination) it means that the P-doped region has a higher resistance to current flow than the n-doped region, now even if one goes to a macroscopic scale, you can connect a 10ohm resistance in series with a 100k ohm resistance, though the electrons will feel a higher resistance in the 100k ohm resistor but since the resistors are in series the same current will flow through the 10 ohm resistor which flows through the 100k ohm one. Same goes for the PN junction, one can say that the N junction has a smaller resistance and the P type has a higher one, but at the end of the day since they are connected in series, they will have the same amount of current flow through them.

\$\endgroup\$
  • \$\begingroup\$ Bimpelrekkie says the holes in p part disappear in forward mode. Do you agree? It is against what you say. \$\endgroup\$ – atmnt Aug 13 '17 at 20:10
  • \$\begingroup\$ not all holes disappear in forward mode, some of them do go in the recombination effect but it is pretty small, small enough to allow electrons to make it to the positive terminal of the battery, here, this will help (physics.stackexchange.com/q/57237) \$\endgroup\$ – Syed Mohammad Asjad Aug 13 '17 at 20:18
  • \$\begingroup\$ A hole is not "simply the absence of an electron" that's a vastly oversimplified way to look at it. This simplification leads to another mistake: "The flow of current is actually the flow of electrons, and since in the P-doped region holes are moving slower (ie electrons are moving slower because there are not many holes due to recombination) it means that the P-doped region has a higher resistance to current flow than the n-doped region" Given the same doping a p and an n region will have their resistances determined by electron and hole mobilities. These mobilities arise as a quantum effect. \$\endgroup\$ – Matt Aug 14 '17 at 1:12
  • \$\begingroup\$ No, a hole is not an "absence of an electron," and a flow of holes is not a flow of negative charges. (Think: a vacuum is "an absence of electrons," but a vacuum has no positive charge!) Instead, a hole is a positive semiconductor ion. Each hole carries a genuine positive charge; a charge produced by a proton. In other words, a hole is an atom which had been neutral in the past, but then loses one electron, and therefore a proton in its nucleus becomes "exposed" or "un-cancelled." Holes are mobile positive charges, and any explanations which contradict this basic fact are simply wrong. \$\endgroup\$ – wbeaty Aug 14 '17 at 2:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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