# 3.7 Physical Operation of Diodes

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1 10/4/2005 3_7 Physical Operation of Diodes blank.doc 1/2 3.7 Physical Operation of Diodes Reading Assignment: pp , A. Semiconductor Materials Q: So, what exactly is a junction diode made of? A: HO: Intrinsic Silicon Q: We call Silicon a semiconductor. Can current flow in a semiconductor? A: HO: Drift Current HO: Diffusion Current Q: So, is a junction diode just a single hunk of intrinsic Silicon?

2 10/4/2005 3_7 Physical Operation of Diodes blank.doc 2/2 A: HO: Doped Silicon B. pn Junction Diode Operation Q: So, exactly how is a junction diode formed? A: HO: The pn Junction Diode Q: How does this simple device result in the complex diode iv characteristic that we studied earlier? HO: The pn Junction Diode in Forward Bias A: HO: The pn Junction Diode in Reverse Bias HO: The pn Junction Diode in Breakdown

3 10/4/2005 Intrinsic Silicon.doc 1/2 Intrinsic Silicon Silicon has 4 electrons in an outer valence shell that requires 8 electrons! Each Si atom therefore forms a covalent bond with 4 other Si atoms they complete their outer valence shell by sharing electrons. A Silicon crystal lattice is created! = electron BUT, thermal agitation breaks covalent bonds. Electrons break free from lattice!! A hole is left where the bound electron used to be.

4 10/4/2005 Intrinsic Silicon.doc 2/2 We now have both free electrons and holes existing in Silicon. = hole * The warmer the Silicon, the more free electrons (and thus holes) are produced. We can therefore define a particle density, defined as either holes/unit volume, or free electrons/unit volume. * In pure Silicon, the number of holes equals the number of free electrons. * Note however, silicon is electrically neutral. In other words, the net charge density within the material is zero, as the number of electrons equals the number of protons (in the nucleus).

5 10/4/2005 Drift Current.doc 1/4 Drift Current Say an electric field is applied to pure Silicon. Q: An electric field E ( r )! Tell me, what will happen? A: From EECS 220 we know what happens! Since electrons and protons have electric charge, the electric field applies a force on them; a force that is proportional to the magnitude of the charge. Q: A force! So, the electrons and protons move, right??? A: Not necessarily! The protons, as well as (generally speaking) the bound electrons, are held in place by the lattice.

6 10/4/2005 Drift Current.doc 2/4 The electric field pulls at them, but atomic forces hold in place the charged particles within the lattice. Q: Within the lattice? The free electrons do not reside within the lattice. Do they move?? A: YES! Free electrons are free to move they are not bound by atomic forces to the lattice. Moving Charge! Moving charge is current! * Charge that moves in response to an applied electric field is known as Drift Current. * Drift current in Silicon has two components current due to moving free electrons, and current due to moving holes. Q: Moving holes!? How can moving holes create current? A hole is nothing!

7 10/4/2005 Drift Current.doc 3/4 A: Let s examine a Crystal Lattice as a function of time, while an electric field is applied to it: t=1 t=2 t=3 t=4 * Note over time, the free electrons move from left to right, in response to the electric field. * But note also that some of the bound electrons also move from left to right, provided that there is a hole in the lattice for them to move into.

8 10/4/2005 Drift Current.doc 4/4 * As a result, the hole appears to moving from right to left! * A positive charge would move right to left in the applied electric field the hole appears to have positive charge! * In fact, holes behave as if they are positively charged particles, with a charge equal to that of an electron (only positive!). So, we have negative charge (free electrons) moving left to right, and positive charge (holes) moving right to left. Both result in (drift) current moving from right to left. Current i is either: or

9 10/4/2005 Diffusion Current.doc 1/4 Diffusion Current Q: So, it seems to me that, if there is no electric field present, there can be no current flow in Silicon. Right?? A: NO! There can still be current flow in a Silicon lattice, even if there is no electric field applied to it! This kind of current is different from drift current we call this current diffusion current. Q: But, how can a charged particle move if there is no force applied to it?!? A: Electrons, whether free or bound, have both a charge and a mass. There is a source of energy within the Silicon lattice that can move particles, independent of their charge. This energy is heat! Q: HEAT! Seems like heat would result in the particles moving randomly in the lattice, as opposed to moving in a specific direction. The average current would be zero right??

10 10/4/2005 Diffusion Current.doc 2/4 A: It is true that if the particles are uniformly distributed within the lattice, then the particles will remain uniformly distributed the average current is zero. However, if (for some reason) the particles are concentrated in one region of the lattice, entropy will ensure that they move from the region of high concentration into regions of low concentration. Moving particles mean moving charge in other words current. For example, consider the situation below, where a collection of particles (e.g., holes or free electrons), are uniformly distributed throughout the volume: However, since these particles posses thermal energy (i.e., heat) they are moving randomly. As a result, at some later time, the particles have all moved, but are still uniformly distributed throughout the volume:

11 10/4/2005 Diffusion Current.doc 3/4 In contrast, consider the situation where the particles are not uniformly distributed, but instead are concentrated in one region of the volume: If there is no other force holding them in place, then thermal energy will cause these particles to randomly redistribute themselves over time, again uniformly across the volume:

12 10/4/2005 Diffusion Current.doc 4/4 Said another way, the thermal energy will maximize the entropy (i.e., randomness) of the volume! But, notice what has happened. The charged particles have moved from one region of the volume into another. This movement, although having nothing to due with electromagnetics, is current! Diffusion Current Current generated by this mechanism is referred to as Diffusion Current. Recapping: 1) Drift current is a result of electrons having charge. They move due to energy supplied by an electric field. 2) Diffusion current is a result of electrons having mass. They move due to energy supplied by heat.

13 10/4/2005 Doped Silicon.doc 1/6 Doped Silicon We can add impurities to Silicon to change the lattice characteristics. Specifically, we can alter the particle densities (i.e., either hole or f.e. densities) of the lattice, such that there are more holes than free electrons, or more free electrons than holes. For example, we can add Phosphorus (P) to Silicon. Phosphorus has 5 valence electrons (one more than Silicon). Problem!!! There is no room for this extra electron in the lattice! As a result, a Silicon lattice that has been doped with Phosphorus has an abundance of free electrons! P P P P

14 10/4/2005 Doped Silicon.doc 2/6 Or, we can dope the Silicon with Boron (B). Boron has 3 valence electrons (one less than Silicon). As a result, there are holes left in the lattice. An abundance of holes is the result!!! B B B B Silicon doped with Phosphorus, such that there is an abundance of free electrons, is called ntype Silicon. Likewise, Silicon doped with Boron is called ptype Silicon. But note that due to thermal agitation, there are still holes in ntype Silicon, and free electrons in ptype Silicon. 1) For ntype Silicon we call free electrons the majority carrier, and holes the minority carrier. 2) Conversely, holes are the majority carrier in ptype Silicon, and free electrons the minority carrier.

15 10/4/2005 Doped Silicon.doc 3/6 Therefore, unlike intrinsic (i.e., pure) Silicon, the particle density (i.e., concentration) of free electrons does not equal the particle density of holes! Q: We learned that holes have positive charge, and of course free electrons have negative. Since in doped Silicon the concentrations of each are unequal, isn t the charge density of doped Silicon nonzero?? A: NO! Remember, a Phosphorus atom has one more electron that a Silicon atom, but it also has one more proton! Likewise, a Boron atom has the same number of electrons as protons. In other words, the lattice remains electrically neutral no ions are present! So, generally speaking, in doped Silicon the charge densities (electrons and protons) are in balance, but the particle densities (holes and free electrons) are out of balance. Now, lets consider the case where the particle concentrations in Silicon become rebalanced! Q: Rebalanced! How could this possibly occur?

16 10/4/2005 Doped Silicon.doc 4/6 A: The majority carriers can move out of a region of the lattice. Recall moving charge is current. We now know that holes and free electrons can flow in the lattice due to either drift current or diffusion current. So, for example, ntype Silicon might look like this: P P P P Note the free electrons are gone. We say that this region of the lattice has been depleted. This looks a lot like intrinsic Silicon, in that the particle densities are now equal there are as many holes as free electrons. But, think about what has happened. The free electrons associated with the Phosphorus atoms have left, but no protons left with them positive Phosphorus ions are created!!!

17 10/4/2005 Doped Silicon.doc 5/6 Now the particle densities are balanced, but the charge density is not the charge density is now positive! When a free electron is removed from a region of ntype Silicon, we say that a positive ion has been uncovered. Now, let s consider what a depleted region of ptype Silicon would look like: B B B B Note what has happened here is that the holes have left, leaving the concentration of holes and free electrons equal. But, the holes left when electrons took their places in the lattice. Each Boron atom therefore has an extra electron. Negative ions have been uncovered! What s more, the charge density within the lattice is now negative!

18 10/4/2005 Doped Silicon.doc 6/6 Recapping: If we dope Silicon with impurities, then we create an imbalance in the number of holes and number of free electrons within the lattice the particle densities are unequal. However, the charge density within the lattice is still zero. If the majority carries are depleted, then ions are uncovered. The particle densities of holes and free electrons are now equal, while the charge density is now nonzero.

19 10/4/2005 The pn Junction Diode.doc 1/13 The pn Junction Diode (Open Circuit) We create a pn junction diode simply by sticking together a hunk of ptype Silicon and a hunk of ntype Silicon! anode ptype ntype cathode Now, let s think about what happens here: 1) The concentration of holes in the anode is much greater than that of the cathode. 2) The concentration of free electrons in the cathode is much greater than that of the anode.

20 10/4/2005 The pn Junction Diode.doc 2/13 Diffusion is the result! 1) Holes begin to migrate (diffuse) across the junction from the anode to the cathode. 2) Free electrons begin to migrate (diffuse) across the junction from the cathode to the anode. Q: Oh, I see! This is entropy at work. This diffusion will occur until the concentration of holes and free electrons become uniform throughout the diode, right? A: Not so fast! There are more phenomena at work here than just diffusion! For instance, think about what happens when holes leave the ptype Silicon of the anode, and the free electrons leave the ntype Silicon of the cathode: They uncover ions!!! As a result, the charge density of the anode along the junction becomes negative, and the charge density of the cathode along the junction becomes positive.

21 10/4/2005 The pn Junction Diode.doc 3/13 This region of uncovered ions along the junction is known as the depletion region. Depletion Region anode ptype ntype cathode Diffusion Current = I Df Now, something really interesting occurs! The uncovered ions of opposite polarity generate an electric field across the junction. E ( r )

22 10/4/2005 The pn Junction Diode.doc 4/13 Recall that an electric field exerts a force on charge particles charged particles like holes and free electrons! Let s see what this force is on both holes and free electrons: For holes: Using the Lorentz force equation, we find that the force vector F on a hole (with charge Q = e ) located at position r is: F = Q E r ( ) Note that since the electric field vector in the depletion region is pointing from right to left (i.e., from the ntype Si cathode to the ptype Si cathode), and since the charge Q of a hole is positive, the force vector likewise extends from right to left: anode ptype Hole Diffusion ntype cathode F Look what happens! The electric field in the depletion region applies a force on the holes that is opposite of the direction of hole diffusion!

23 10/4/2005 The pn Junction Diode.doc 5/13 In other words, the electric field begins to hold back the tide of holes attempting to diffuse into the ntype cathode region. For free electrons: Now, let s see what effect this electric field has on free electrons. Using the Lorentz force equation, we find that the force vector F on a free electron (with chargeq = e ) located at position r is: F = Q E r ( ) Note that since the electric field vector in the depletion region is pointing from right to left (i.e., from the ntype Si to the ptype Si), and since the charge Q of a free electron is negative, the force vector extents in the opposite direction E r from left to right: of ( ) anode ptype ntype Free Electron Diffusion cathode Look what happens! The electric field in the depletion region likewise applies a force on the free electrons that is opposite of the direction of free electron diffusion! F

24 10/4/2005 The pn Junction Diode.doc 6/13 In other words, the electric field begins to hold back the tide of free electrons attempting to diffuse into the ptype anode region. Q: So, does this electric field stop all diffusion across the junction? Is the diffusion current I Df therefore zero? A: Typically NO! The electric field will greatly reduce the diffusion across the junction, but only in certain cases will it eliminate I Df entirely (more about that later!). The amount of diffusion that occurs for a given electric field E ( r ) is dependent on how energetic the particles (holes and freeelectrons) are! Recall that these particles will have kinetic energy due to heat. If this energy is sufficiently large, a particle can still diffuse across the pn junction! To see why, consider the amount of energy E it would take to move a charged particle through this electric field. Recall from EECS 220 that this energy is: E = Q E ( r ) d For our case, Q is the charge on a particle (hole or free electron), and contour C is a path that extends across the depletion region. C

25 10/4/2005 The pn Junction Diode.doc 7/13 Moreover, we recall that this expression can be simplified by using electric potential, i.e., V = E ( r ) d Where V is the difference in potential energy (per coulomb) between a charge at either end of contour C. This of course tells us how much work must be done (per coulomb) to move a charge from one end of the contour to the other. C Of course V has units of Volts, but its more descriptive unit is joules/coulomb energy per unit charge. Therefore, the energy required to move a charge Q along some contour C can likewise be expressed as: E = QV Now, for our particular problem, the charge Q is either the charge of a free electron (Q ) or the charge of a hole (Q ). The voltage (i.e., potential difference) across the depletion region is called the barrier voltage V B (sometimes denoted V 0 ): VB = E ( r ) d where the contour C dr describes some contour across the depletion region. C dr

26 10/4/2005 The pn Junction Diode.doc 8/13 Typically, we find that when the junction diode is open circuited (i.e., v D =0 and i D =0), this barrier voltage is approximately 0.7 V! Thus, we find that the energy required for a hole to diffuse across the depletion region is: E B = Q V While the energy required for a free electron to diffuse across the depletion region is: E B B = Q V Note that both these energies are the same (positive) value! OK, here s the important part: A. If the particle has kinetic energy greater than E B, it can diffuse across the depletion region. B anode ptype ntype cathode A hole with kinetic energy > E B will.... F...diffuse across the depletion region. B. If the particle has kinetic energy less than E B, then the electric field will push it back into either the p type anode region (for holes) or the ntype cathode region (for free electrons).

27 10/4/2005 The pn Junction Diode.doc 9/13 A hole with kinetic energy < E B will.... anode ptype ntype cathode F...be pushed back by the electric field (it will not diffuse across the depletion region)! Thus, the diffusion current I Df across the pn junction will depend on three things: 1. The majority particle concentration. The more holes or free electrons there are, the more particles will diffuse across the junction. 2. The barrier voltage V B. A lower barrier means less kinetic energy is required to diffuse across the depletion region, resulting in more. 3. The diode temperature Higher temperature means holes and electrons have more kinetic energy and thus are more likely to diffuse across the depletion region.

28 10/4/2005 The pn Junction Diode.doc 10/13 Q: Wait a minute! We ve examined the behavior of holes in the ptype region and free electrons in the ntype region. These are the majority carriers for each of those Silicon types. There are also minority carriers present in each side. What does the electric field in the depletion region do to them? A: A great question! We will find that the electric field will have a profoundly different effect on minority carriers! For holes: Recall that the electric field in the depletion region applies a force on positive charges (holes) that is directed from the n type (cathode) region into the ptype (anode) region. This force of course pushes the holes in the ptype anode (the majority carriers) back into the ptype region. However, the same force will pull holes from the ntype region (the minority carriers) into the ptype region! anode ptype Hole Drift ntype cathode F

29 10/4/2005 The pn Junction Diode.doc 11/13 Any unsuspecting minority hole that drifts into the depletion region will from the ntype side will be pulled into the ptype side! Note that this result is independent of the kinetic energy of the particle it takes no energy to fall downhill. This movement of charge is completely due to the force applied by the electric field this is drift current I s! Now, for free electrons: Recall also that the electric field in the depletion region applies a force on negative charges (free electrons) that is directed from the ptype (anode) region into the ntype (cathode) region. This force of course pushes the free electrons in the ntype region (the majority carriers) back into the ntype region. However, the same force will pull free electrons from the ptype region (the minority carriers) into the ntype region! anode ptype Free Electron Drift ntype cathode F

30 10/4/2005 The pn Junction Diode.doc 12/13 Any unsuspecting minority free electron that drifts into the depletion region will from the ptype side will be pulled into the ntype side! Note that this result is independent of the kinetic energy of the particle it takes no energy to fall downhill. This movement of charge is completely due to the force applied by the electric field this is also drift current I S! There are two very important differences between drift and diffusion currents in a pn junction diode: 1. Drift and Diffusion current flow in opposite directions The Diffusion current I Df flows across the pn junction from anode to cathode, while Drift current I S flows across the pn junction from cathode to anode. Drift Current = I S anode ptype ntype cathode Diffusion Current = I Df

31 10/4/2005 The pn Junction Diode.doc 13/13 2. Diffusion current depends on the barrier voltage V B, but Drift Current does not. As the barrier voltage increases, fewer and fewer of the majority carriers will have sufficient kinetic energy to cross the depletion region the diffusion current will decrease. Conversely, minority carriers require no energy to be swept across the depletion region by the electric field, the value of the barrier voltage is irrelevant to the value of drift current I S. Now, for an opencircuited (i.e., disconnected) junction diode, the total current i D through the device must be zero (i D =0). In other words, the diffusion current I Df must be equal but opposite that of the drift current I S, such that I Df I S =0 : Drift Current = I S =I Df ptype ntype i D = 0 = I Df I s Diffusion Current = I Df = I s This is the equilibrium state of a disconnected junction diode. We find that typically this drift/diffusion current is very small, generally 10 8 to Amps!

32 10/4/2005 The pn junction in forward bias.doc 1/2 The pn Junction in Forward Bias Now consider the case where we place a small, positive voltage across a junction diode. I S i D p n I Df V D > 0 1) This voltage reduces the barrier voltage, i.e., the electric field that holds back the diffusion of holes from the anode to the cathode, as well as holds back the diffusion of free electrons from the cathode to the anode. 2) Thus, diffusion current increases as diode voltage increases. In fact, this increase is exponential with the diode voltage! : IDf = I e s vd nvt

33 10/4/2005 The pn junction in forward bias.doc 2/2 3) But, the drift current does not change if v D is increased! The reduced electric field moves charges with less force, but the number of holes and free electrons swept across the depletion region does not change. Therefore, drift current I S remains at its same small value, independent of diode voltage v D. The total current i D through the diode is therefore: i = I D Df I S = = I S VD n VT VD nv T I ( e 1) S e I S Hey! this result is very familiar!!

34 10/4/2005 The pn juncton diode in reverse bias.doc 1/2 The pn Junction Diode in Reverse Bias Say we now place a significant negative voltage across the diode. I S i D p n I Df V D << 0 1) This negative voltage increases the electric field within the depletion region. 2) The barrier voltage is now so large that it stops virtually all diffusion across the junction. 3) Therefore, the diffusion current I Df = 0: vd nvt I = I e 0 for v 0 Df s D

35 10/4/2005 The pn juncton diode in reverse bias.doc 2/2 4) As with the forward bias case, the drift current remains constant. The holes and free electrons are swept through the depletion region with greater energy, but the number these charged particles remains unchanged. Therefore, the total diode current is: i D = = = I Df 0 I S I I S S This result should likewise be very familiar!

36 10/4/2005 The pn junction diode in breakdown.doc 1/1 The pn Junction Diode in Breakdown If reverse bias too large (i.e., v D < V ZK ), the covalent bonds within the depletion region will break. Therefore, free electrons are created (i.e., conductivity σ goes from zero to very high). Large electric field and high conductivity: This means high current ( J = σ Ε )!! Attempts to decrease v D past V ZK instead just causes further breaking of covalent bonds (i.e., conductivity σ increases). Therefore i D increases while v V. D ZK There are two mechanisms for breakdown. 1) Zener Effect Covalent bonds break because of large Efield. 2) Avalanche Effect Bonds break due to kinetic energy of drift current.

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