TF EIM Chapt4 - Revision history

Foretony: /* The P-N junction Diode/rectifier */

2022-03-29T00:11:56Z

The P-N junction Diode/rectifier

Foretony: /* The P-N junction Diode/rectifier */

2022-03-29T00:08:47Z

The P-N junction Diode/rectifier

Foretony: /* Electron Energy Distribution */

2015-02-19T18:29:40Z

Electron Energy Distribution

Foretony: /* Electron Energy Distribution */

2015-02-19T18:25:45Z

Electron Energy Distribution

Foretony: /* Crystal Lattice */

2015-02-19T18:20:24Z

Crystal Lattice

Foretony: /* The P-N junction Diode/rectifier */

2015-02-19T17:56:54Z

The P-N junction Diode/rectifier

Foretony: /* The P-N junction Diode/rectifier */

2015-02-19T17:56:44Z

The P-N junction Diode/rectifier

Foretony: /* The P-N junction Diode/rectifier */

2015-02-19T17:54:41Z

The P-N junction Diode/rectifier

Foretony: /* Electron Energy Distribution */

2015-02-19T17:46:32Z

Electron Energy Distribution

Foretony: /* Fermi Function (F[E]) */

2015-02-17T19:13:40Z

Fermi Function (F[E])

@@ Line 175: / Line 175: @@
 Note:  The material still has a net charge of zero.  The semi-conductor has excess FREE electrons on the n-side and excess HOLES on the p-side of the junction.  When a junction is made excess electrons move over to the p-side and any positive "holes" move to the n-side.  Notice this looks like a charged up capacitor with a voltage difference to be in order for charge to move/conduct.
-AN electric field in the depletion region<math> (\vec{E}_d)</math> points from the n-type (donor) side to the p-type (acceptor) side.  Electrons near the junction are pulled toward the n-type material and holes are pulled to the p-type material.
+AN electric field in the depletion region<math> (\vec{E}_d)</math> points from the n-type (donor) side to the p-type (acceptor) side.  An electron pushed near the junction would be pulled toward the n-type material and holes are pulled to the p-type material.
 This electric field <math>E_d</math> results in an effective voltage <math>V_{contact} = E_d \times d</math>, where <math>d</math> is the thickness of the depletion region.

@@ Line 173: / Line 173: @@
 [[File:TF_EIM_SiliconLattice_np_junction.jpg| 400 px]][[File:TF_EIM_p-n_junction.jpg| 400 px]]
-Note:  The material still has a net charge of zero.  The semi-conductor has excess FREE electrons on the n-side and excess HOLES on the p-side of the junction.  Notice this looks like a charged up capacitor with a voltage difference to be in order for charge to move/conduct.
+Note:  The material still has a net charge of zero.  The semi-conductor has excess FREE electrons on the n-side and excess HOLES on the p-side of the junction.  When a junction is made excess electrons move over to the p-side and any positive "holes" move to the n-side.  Notice this looks like a charged up capacitor with a voltage difference to be in order for charge to move/conduct.
 AN electric field in the depletion region<math> (\vec{E}_d)</math> points from the n-type (donor) side to the p-type (acceptor) side.  Electrons near the junction are pulled toward the n-type material and holes are pulled to the p-type material.

@@ Line 135: / Line 135: @@
 :<math>N_{cb} = \int_{E_{gap}}^{\infty}N(E) dE = \int_{E_{gap}}^{\infty} \frac{2^{7/2}m^{3/2} \pi}{h^3}(E-E_{gap})^{1/2} \times e^{-\left( E-E_f\right)/kT}dE</math>
-::<math>= \frac{2^{5/2}(m\pi kT)^{3/2} }{h^3}e^{-E_{gap}/2kT}</math>
+::<math>= \frac{2^{5/2}(m\pi kT)^{3/2} }{h^3}e^{-E_{gap}/2kT}</math> : <math>E_F= E_{gap}/2</math>
 :: <math>= 4.6 \times 10^{15} \frac{\mbox{electrons}}{\mbox{cm}^3 \mbox{degree K}^{3/2}} T^{3/2}e^{-E_{gap}/2kT}</math>
 :: <math>\propto T^{3/2}e^{-E_{gap}/2kT}</math>

← Older revision		Revision as of 18:25, 19 February 2015
Line 118:		Line 118:

	:<math>\rho(E) = \frac{2^{7/2}m^{3/2} \pi}{h^3}E^{1/2}=</math> density of states for electrons in a periodic crystal lattice		:<math>\rho(E) = \frac{2^{7/2}m^{3/2} \pi}{h^3}E^{1/2}=</math> density of states for electrons in a periodic crystal lattice
−	::<math>= \frac{2^{7/2}m^{3/2} \pi}{h^3}(E-E_{gap})^{1/2}</math>	+	::<math>= \frac{2^{7/2}m^{3/2} \pi}{h^3}(E-E_{gap})^{1/2}</math> = density of states for electrons in the conduction band where E=0 at the valence band

← Older revision		Revision as of 18:20, 19 February 2015
Line 34:		Line 34:
	:When you add the atoms to form the lattice, the potential barrier is lowered. This allows the electrons in the higher (more valence) states to become shared among adjacent atoms.( Delocalized)		:When you add the atoms to form the lattice, the potential barrier is lowered. This allows the electrons in the higher (more valence) states to become shared among adjacent atoms.( Delocalized)
	:Another effect is that the energy bands become wider. As the electrons in the higher energy states are shared between atoms forming a bond, the time any one electron spends bound to any one atom becomes smaller. SInce<math> \Delta E \Delta t \approx \hbar</math> one finds that <math>\Delta E</math> gets bigger for the higher state electrons (less bound) since their time \Delta t in a given bound state becomes less. The electron energy states become band like structures which get wider as you move to higher energy (less bound) electron states.		:Another effect is that the energy bands become wider. As the electrons in the higher energy states are shared between atoms forming a bond, the time any one electron spends bound to any one atom becomes smaller. SInce<math> \Delta E \Delta t \approx \hbar</math> one finds that <math>\Delta E</math> gets bigger for the higher state electrons (less bound) since their time \Delta t in a given bound state becomes less. The electron energy states become band like structures which get wider as you move to higher energy (less bound) electron states.
		+
		+
		+	;also
		+	:The amount of energy needed for an electron to move from the valence region to the conduction region is known as <math>E_{gap}</math>. This gap energy is on the order of an election Volt (eV). The available thermal energy (kT) at room temperature is about 0.025 eV (1/40). At room temperature you would not expect to see an electron move from the valence region (band) to the conduction region (band). A battery though could be used to accomplish this by pulling valence electrons out of their bound states and putting them into the conduction region. Or you could heat up the material to 50 times room temperature.

← Older revision		Revision as of 17:56, 19 February 2015
Line 167:		Line 167:


−	[[File:TF_EIM_SiliconLattice_np_junction.jpg\| ~~200~~ px]][[File:TF_EIM_p-n_junction.jpg\| 400 px]]	+	[[File:TF_EIM_SiliconLattice_np_junction.jpg\| 400 px]][[File:TF_EIM_p-n_junction.jpg\| 400 px]]

	Note: The material still has a net charge of zero. The semi-conductor has excess FREE electrons on the n-side and excess HOLES on the p-side of the junction. Notice this looks like a charged up capacitor with a voltage difference to be in order for charge to move/conduct.		Note: The material still has a net charge of zero. The semi-conductor has excess FREE electrons on the n-side and excess HOLES on the p-side of the junction. Notice this looks like a charged up capacitor with a voltage difference to be in order for charge to move/conduct.

@@ Line 167: / Line 167: @@
-[[File:TF_EIM_SiliconLattice_np_junction.jpg| 200 px]][[File:TF_EIM_p-n_junction.jpg| 200 px]]
+[[File:TF_EIM_SiliconLattice_np_junction.jpg| 200 px]][[File:TF_EIM_p-n_junction.jpg| 400 px]]
 Note:  The material still has a net charge of zero.  The semi-conductor has excess FREE electrons on the n-side and excess HOLES on the p-side of the junction.  Notice this looks like a charged up capacitor with a voltage difference to be in order for charge to move/conduct.
@@ Line 177: / Line 177: @@
 The "depletion" region is located at the junction and is a place which has been "depleted" of mobile charge carriers.  The ionized acceptor and donor impurity atoms are fixed in the crystal lattice at the junction.
-[[File:TF_EIM_PotBarrier_p-n_junct.jpg | 200 px]]
+[[File:TF_EIM_PotBarrier_p-n_junct.jpg | 400 px]]

← Older revision		Revision as of 17:54, 19 February 2015
Line 180:		Line 180:


−	A forward biasing of the above p-n junction means that you oppose the ~~direct~~ of the junction field. The positive side of the battery is put on the "p" side of the junction and the negative side on the "n" side of the junction. This reduces the junction barrier (squeezes the depletion region) making it easier for current to flow through the junction. The bias electric field <math>E_B</math> is directed opposite to ~~teh~~ depletion electric field <math>E_d</math>. When you reverse bias the p-n junction you are increasing the junction barrier making it harder for current to flow. At some point though, the junction will break down and you will force current to flow.	+	A forward biasing of the above p-n junction means that you oppose the direction of the junction field. The positive side of the battery is put on the "p" side of the junction and the negative side on the "n" side of the junction. This reduces the junction barrier (squeezes the depletion region) making it easier for current to flow through the junction. The bias electric field <math>E_B</math> is directed opposite to the depletion electric field <math>E_d</math>. When you reverse bias the p-n junction you are increasing the junction barrier making it harder for current to flow. At some point though, the junction will break down and you will force current to flow.

@@ Line 99: / Line 99: @@
 ;Note: At room temperature: kT = 1/40 eV
-[[File:TF_EIM_FermiDistrib.jpg| 200 px]]
+[[File:TF_EIM_FermiDistrib.jpg| 400 px]]
-[[File:TF_EIM_E-vs-FermiFunct.png| 200 px]]
+[[File:TF_EIM_E-vs-FermiFunct.png| 400 px]]
 :<math>E_f = E_{gap}/2 =</math> (half of the energy needed to move from the valence to the conduction band