Basic Physics of Semiconductors

Transcription

1 Basic Physics of Semiconductors Semiconductor materials and their properties PN-junction diodes Reverse Breakdown EEM 205 Electronics I Dicle University, EEE Dr. Mehmet Siraç ÖZERDEM Semiconductor Physics Semiconductor devices serve as heart of microelectronics. PN junction is the most fundamental semiconductor device. 1

2 Charge Carriers in Semiconductor To understand PN junction s I/V characteristics, it is important to understand charge carriers behavior in solids, how to modify carrier densities, and different mechanisms of charge flow. Periodic Table This abridged table contains elements with three to five valence electrons, with Si being the most important. 2

3 Silicon Si has four valence electrons. Therefore, it can form covalent bonds with four of its neighbors. When temperature goes up, electrons in the covalent bond can become free. Electron-Hole Pair Interaction With free electrons breaking off covalent bonds, holes are generated. Holes can be filled by absorbing other free electrons, so effectively there is a flow of charge carriers. 3

4 Free Electron Density at a Given Temperature Eg, or bandgap energy determines how much effort is needed to break off an electron from its covalent bond. There exists an exponential relationship between the free electron density and bandgap energy. For silicon, Eg=1.12eV = 1.792x10-19 J Modification of Carrier Densities Doping (N type) Pure Si can be doped with other elements to change its electrical properties. For example, if Si is doped with P (phosphorous), then it has more electrons, or becomes type N (electron). 4

5 Modification of Carrier Densities Doping (P type) If Si is doped with B (boron), then it has more holes, or becomes type P. Modification of Carrier Densities The pure type of silicon or intrinsic semiconductor has a very high resistance. It is possible to modify the resistivity of silicon by replacing some of the atoms in the crystal with atoms of another material. The unit ev (electron volt) represents the energy necessary to move one electron across a potential difference of 1 V. (1 ev=1.6x10-19 J) In an intrinsic semiconductor, the electron density, n (=n i ), is equal to the hole density, p. np=n i 2 But, how about these densities in a doped material? Even in this case, the equation works. 5

6 Modification of Carrier Densities How can np remain constant while we add more donor atoms and increase n? np=n i 2 reveals that p must fall below its intrinsic level as more n-type dopants are added to the crystal. This occurs because many of the new electrons donated by the dopant recombine with the holes that were created in the intrinsic material. Example A piece of crystalline silicon is doped uniformly with phosphorus (P) atoms. The doping density is atom/cm 3 Determine the electron and hole densities in this material at the room temperature. Solution The addition of P atoms introduces the same number of free electrons per cubic centimeter. n = electrons/cm 3 p = (n i 2 / n) p = 1.17x10 4 holes/cm 3 6

7 Summary of Charge Carriers Electron and Hole Densities For N type For P type The product of electron and hole densities is ALWAYS equal to the square of intrinsic electron density regardless of doping levels. 7

8 First Charge Transportation Mechanism: Drift The process in which charge particles move because of an electric field is called drift. Charge particles will move at a velocity that is proportional to the electric field. For silicon, the mobility of electrons, μ n =1350cm 2 /(V s), and that of holes, μ p =480cm 2 /(V s). Electrons move in a direction opposite to the electric field Example A uniform piece of n type of silicon that is 1μm long senses a voltage of 1 V. Determine the velocity of the electrons Solution L=1 μm, E=V/L then E= 1V/1μm = V/cm 8

9 Current Flow: General Case Bar Volume L (length) t=1second v = L/t=L Electric current is calculated as the amount of charge in v-meters that passes thru a cross-section if the charge travel with a velocity of v m/s. Current Flow: Drift I = - v W h n q J n = I / (W h) = - v n q v = -μ n E Current density : the current passing through a unit cross section area (A/cm 2 ) The total current density consists of both electrons and holes. 9

10 Example It is desired to obtain equal electron and hole drift currents. How should the carrier densities be chosen? Solution We must impose Note that np=n i 2 For example, in silicon, μ n /μ p =1350/480=2.81 p=1.68n i and n=0.596n i Velocity Saturation In reality, velocity does not increase linearly with electric field. It will eventually saturate to a critical value. This is because the carriers collide with the lattice so frequently and the time between the collisions is so short that they cannot accelerate much. μ o : Low-field mobility μ : effective mobility 10

11 Example A uniform piece of semiconductor L=0.2 μm long sustains a voltage of 1 V. If the low-field mobility is equal to μ o =1350 cm 2 /(V s) and the saturation velocity of the carriers v sat =10 7 cm/s, determine the effective mobility (μ). Second Charge Transportation Mechanism: Diffusion Charge particles move from a region of high concentration to a region of low concentration. It is analogous to an every day example of an ink droplet in water. 11

12 Current Flow: Diffusion Diffusion current is proportional to the gradient of charge (dn/dx) along the direction of current flow. n : The carrier concentration If each carrier has a charge equal to q, and the semiconductor has a cross section area of A Thus Dn : Diffusion constant (cm 2 /s), In intrinsic silicon, Dn=34cm 2 /s for electrons Dn=12cm 2 /s for holes Current Flow: Diffusion Its total current density consists of both electrons and holes. 12

13 Example Suppose the electron concentration is equal to N at x=0 and falls linearly to zero at x=l. Determine the diffusion current. Solution The current is constant along the x-axis; i.e., all of the electrons entering the material at x=0 successfully reach the point at x=l. Example Repeat the above example but assume an exponential gradient. L d : constant Solution The current is not constant along the x-axis. That is, some electrons vanish while traveling from x=0 to the right. 13

14 Einstein's Relation While the underlying physics behind drift and diffusion currents are totally different, Einstein s relation provides a mysterious link between the two. PN Junction (Diode) p side : anode n side : cathode When N-type and P-type dopants are introduced side-byside in a semiconductor, a PN junction or a diode is formed. 14

15 Diode s Three Operation Regions In order to understand the operation of a diode, it is necessary to study its three operation regions: equilibrium, reverse bias, and forward bias. PN Junction in Equilibrium The PN junction with no external connections, i.e., the terminals are open and no voltage is applied across the device. We say the junction is in equilibrium. 15

16 PN Junction in Equilibrium Current Flow Across Junction: Diffusion Because each side of the junction contains an excess of holes or electrons compared to the other side, there exists a large concentration gradient. Therefore, a diffusion current flows across the junction from each side. The diffusion currents transport a great deal of charge from each side to the other, but they must eventually decay to zero. Example A pn junction employs the following doping levels: N A =10 16 cm -3 and N D =5x10 15 cm -3. Determine the hole and electron concentrations on the two sides. Solution p side n side 16

17 Depletion Region For every electron that departs from the n side, a positive ion is left behind. As free electrons and holes diffuse across the junction, a region of fixed ions is left behind. This region is known as the depletion region. PN Junction in Equilibrium Current Flow Across Junction: Drift The fixed ions in depletion region create an electric field that results in a drift current. PN Junction in Equilibrium 17

18 Current Flow Across Junction t0 electrons move from left to right holes move from right to left Diffusion Current After a net (nonzero) charge creates an electric field t1 electrons move from right to left holes move from left to right Drift Current The junction reaches equilibrium once the electric field is strong enough to completely stop the diffusion currents PN Junction in Equilibrium Current Flow Across Junction: Equilibrium At equilibrium, the drift current flowing in one direction cancels out the diffusion current flowing in the opposite direction, creating a net current of zero. PN Junction in Equilibrium 18

19 Built-in Potential & Where p n and p p are the hole concentrations at x1 and x2, respectively. From Einstein s relation PN Junction in Equilibrium Built-in Potential (Cont.) the potential barrier PN Junction in Equilibrium 19

20 Exercise Repeat the same peocedure for electron and derive an equation for Vo in terms of n n and n p. Example A silicon pn junction employs N A =2x10 16 cm -3 and N D =4x10 16 cm -3. Determine the built-in potential at room temperature (T=300K) Exercise: By what factor should N D be changed to lower Vo by 20 mv? PN Junction in Equilibrium 20

21 Diode in Reverse Bias When the N-type region of a diode is connected to a higher potential than the P-type region, the diode is under reverse bias, which results in wider depletion region and larger built-in electric field across the junction. Reverse Biased Diode s Application: Voltage-Dependent Capacitor The PN junction can be viewed as a capacitor. By varying V R, the depletion width changes, changing its capacitance value; therefore, the PN junction is actually a voltage dependent capacitor. 21

22 Reverse Biased Diode s Application: Voltage-Dependent Capacitor C1 C2 L1 L2 L1 > L2 C2<C1 Voltage-Dependent Capacitance The capacitance of the junction per unit area represents the dielectric constant of silicon and is equal to 11.7x8.85x10-14 F/cm The equations that describe the voltage-dependent capacitance are shown above 22

23 Example A pn junction is doped with N A =2x10 16 cm -3 and N D =9x10 15 cm -3. Determine the capacitance of the device with (a) V R =0 and V R =1V. Solution We first obtain the built-in potential: For V R =0V, q=1.6x10-19 C For V R =0V For V R =1V C j =C j0 C j =0.172 ff / μm 2 Voltage-Controlled Oscillator A very important application of a reverse-biased PN junction is V CO, in which an LC tank is used in an oscillator. By changing V R, we can change C, which also changes the oscillation frequency. 23

24 Diode in Forward Bias When the N-type region of a diode is at a lower potential than the P-type region, the diode is in forward bias. The depletion width is shortened and the built-in electric field decreased. Minority Carrier Profile in Forward Bias V T =kt/q e : Equilibrium Under forward bias, minority carriers in each region increase due to the lowering of built-in field potential. Therefore, diffusion currents increase to supply these minority carriers. In forward bias, the potential barrier is lowered by an amount equal to the applied voltage: f : Forward 24

25 Diffusion Current in Forward Bias The minority carrier concentration : rises rapidly The majority carrier concentration : relatively constant As the junction goes from equilibrium to forward bias, n p and p n increase dramatically, leading to a proportional change in the diffusion currents The hole concentration on the n side p p,f p p,e N A The electron concentration on the p side Diffusion Current in Forward Bias The diffusion currents rise by a proportional amount of minority concentrations. The diffusion current can be also calculated by I s : Reverse saturation current A : The cross section area L x : Diffusion lengths 25

26 Example A silicon pn junction employs N A =2x10 16 cm -3 and N D =4x10 16 cm -3. Determine Is at room temperature (T=300K). A=100 μm 2, L n =20μm, L p =30μm q=1.6x10-19 C, n i =1.08x10 10 electron/cm 3 D n =34 cm 2 /s, D p =12 cm 2 /s I s =1.77x10-17 A Forward Bias Condition: Summary Minority and majority carrier currents In forward bias, there are large diffusion currents of minority carriers through the junction. However, as we go deep into the P and N regions, recombination currents from the majority carriers dominate. These two currents add up to a constant value. 26

27 I/V Characteristic of PN Junction I D : diode current V D : diode voltage The current and voltage relationship of a PN junction is exponential in forward bias region, and relatively constant in reverse bias region. Parallel PN Junctions I S A Since junction currents are proportional to the junction s crosssection area. Two PN junctions put in parallel are effectively one PN junction with twice the cross-section area, and hence twice the current. 27

28 Example Each junction in Figure employs the doping levels described in previous Example. Determine the forward bias current of the composite device for VD=300mV and 800mV at T=300K Solution I s =1.77x10-17 A (Calculated in the previous example) Similarly, for V D =800mV Example A diode operates in the forward bias region with a typical current level [ i.e. I D =I s exp(v D /V T ) ]. Suppose we wish to increase the current by a factor of 10. How much change in V D is required? Solution If I 1 =10I D Thus, the diode voltage must rise by V T ln10=60mv 28

29 Constant-Voltage Diode Model Diode operates as an open circuit if V D < V D,on and a constant voltage source of V D,on if V D tends to exceed V D,on. Example Calculate I x for V x =3V and V x =1V using a) An exponential model with Is=10-16 A b) A constant-voltage model with V D,on =800mV (a) This equation must be solved by iteration Procedure 1) Guess a value for V D 2) Compute the corresponding Ix 3) Determine the new value of V D 4) If not converge then Go to step 2 29

30 Example (Cont.) (a) Let us gess V D =750mV then Vx=3V Thus With this new value of V D Ix rapidly converges The same procedure can be applied for Vx=1V Example (Cont.) (b) A constant-voltage model readily The value of Ix incurs some error, but it is obtained with much less computational effort than that in part (a). This example shows the simplicity provided by a constant voltage model over an exponential model. For an exponential model, iterative method is needed to solve for current, whereas constant-voltage model requires only linear equations. 30

31 Reverse Breakdown When a large reverse bias voltage is applied, breakdown occurs and an enormous current flows through the diode. Zener vs. Avalanche Breakdown Zener breakdown is a result of the large electric field inside the depletion region that breaks electrons or holes off their covalent bonds. Avalanche breakdown is a result of electrons or holes colliding with the fixed ions inside the depletion region. 31

32 Reference Behzad Razavi, Fundamentals of Microelectronics, Wiley, 2006 EEM 205 Electronics I Dicle University, EEE Dr. Mehmet Siraç ÖZERDEM 32