Chapter four The Equilibrium pn Junction The Electric field will create a force that will stop the diffusion of carriers reaches thermal equilibrium condition Potential difference across the depletion region is called the built-in potential barrier, or built-in voltage: ( ) ( ) V T = kt/e k = Boltzmann s constant T = absolute temperature e = the magnitude of the electronic charge = 1 ev N a = the net acceptor concentration in the p-region N d = the net donor concentration in the n-region V T = thermal voltage, [V T = kt / e] it is approximately 0.026 V at temp, T = 300 K 1 1
The Equilibrium When no external source is connected to the pn junction, diffusion and drift balance each other out for both the holes and electrons Space Charge Region: Also called the depletion region. This region includes the net positively and negatively charged regions. The space charge region does not have any free carriers. The width of the space charge region is denoted by W in pn junction formula s. Metallurgical Junction: The interface where the p- and n-type materials meet. Na & Nd: Represent the amount of negative and positive doping in number of carriers per centimeter cubed. Usually in the range of 10 15 to 10 20. n p junction at equilibrium Thermal equilibrium; no applied field; no net current flow Drift current is due to electric field at the junction; minority carriers. Diffusion current is due the to concentration gradient; majority carriers. ( ) ( ) 2 2
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. Under Equilibrium the net current is zero i.e. the drift current equals the diffusion current For a system in equilibrium the average energy must be constant (obvious). This also means the Fermi level must be constant Example Calculate the built-in potential barrier of a pn junction. Consider a silicon pn junction at T = 300 K, doped N a = 10 16 cm -3 in the p-region, N d = 10 17 cm -3 in the n-region and n i = 1.5 x 10 10 cm -3. ( ) ( ( ) ) Example 2 Consider a silicon pn junction at T = 400K, doped with concentrations of N d = 10 18 cm -3 in n-region and N a = 10 19 cm -3 in p-region. Calculate the built-in voltage V bi of the pn junction, given B and Eg for silicon are 5.23 x 10 15 cm -3 K -3/2 and 1.1 ev respectively Calculation of V T = kt / e = 86 x 10-6 ( 400 ) / 1eV = 0.0344 V Calculation of n i = BT 3/2 exp ( -Eg / 2kT ) = 5.23 x 10 15 ( 400 ) 3/2 exp -1.1 / 2 (86 x 10-6 ) (400) = 4.76 x 10 12 cm 3 Calculation of V bi = V T ln ( N a N d / n i 2 ) = 0.0344 ln 10 18 (10 19 ) / (4.76 x 10 12 ) 2 = 0.922V 3 3
pn Junction Under Bias Forward bias is a voltage applied to the pn junction that reduced the electric field at the barrier, Reverse bias increase the electric field at the junction When bias is applied the balance between drift and diffusion current is destroyed net current flow In forward bias, drift current decreases very slightly (can assume it stays the same) but diffusion current increases Net current flow In reverse bias opposite occurs with diffusion current decreasing and drift remaining same Net current flow (this one is very small) Voltage-Current Characteristic of a Diode Diode: is a semiconductor device, which conduct the current in one direction only. 4 4
Diode current equation The Shockley Equation For forward and reverse bias region: * ( ) + I s saturation current. For small signal diodes at 300K: Is ~10-14 A.(sometime called scale factor because it is proportional to the cross-section area A of the diode doubting the junction area results in a diode with double ) I s is strongly dependent on the temperature( double every 5 for silicon. η emission coefficient; n = 1.. 2 for small-signal diodes. η------- semiconductor constant η =1 for Ge, η =2 for Si. thermal voltage: T absolute temperature in K; k = 1.38 10-23 J/K the Boltzmann s constant; q = 1.60 10-19 C the charge of the electron; this equation is not applicable when Three operating regions : 5 5
1. Forward bias region Plot the result of measurement in Figure below, you get the V-I characteristic curve for a forward bias diode Increase to the right increase upward dynamic resistance decreases as you move up the curve It can be noted from the graph the current remains zero till the diode voltage attains the barrier potential. For silicon diode, the barrier potential is 0.7 V and for Germanium diode, it is 0.3 V. The barrier potential is also called as knee voltage or cur-in voltage. 6 6
* ( ) + ( ) 2. Reverse bias region Breakdown voltage not a normal operation of pn junction devices the value can be vary for typical Si Reverse current= Ideally, the reverse current is independent of reverse bias In reality, reverse current is larger than and also increases somewhat with the reverse bias Temperature dependence: reverse current doubles for every 10 C 0 rise in temperature 7 7
( ) 3. Breakdown region The breakdown region is entered when the magnitude of the reverse voltage exceed a threshold called breakdown voltage or the zener knee voltage. In the breakdown region the current increase rapidly with a very small increase in the associated voltage. There is an avalanche of electrons flowing across the junction with the result that the diode overheat. Provided that the power dissipated in the diode is limited by external circuitry to a safe level (typically specified in the data sheets) breakdown won t be destructive. Complete V-I Characteristic curve Terminal Characteristic of Junction Diodes The Forward-Bias Region, determined by The Reverse-Bias Region, determined by The Breakdown Region, determined by 8 8
Mathematic Model: ( ) For positive values of the first term of the equation above will grow very quickly and overpower the effect of the second term. The result is that for positive values of, will be positive and grow as the function appearing in Fig. becomes ( ) as appearing in Fig. 1.19. For negative values of the first term will quickly drop off below, resulting in, which is simply the horizontal line of Fig. 1.19. The break in the characteristics at is simply due to the dramatic change in scale from to. Example: Consider a silicon diode with n=1.5. Find the change in voltage if current changes from 0.1 ma to 10 ma. Sol: 9 9
i I1 I2 v nv T I S e n 1.5 V T 0.025 I S e v1 nv T v2 I1 0.00 01 I2 0.01 I1 I2 e ( v1 v2) nv T nv T I S e v1_v2 nv T ln I1 v1_v2 0.173 I2 Example :A diode for which the forward voltage drop is 0.7 V at 1 ma and for which n=1 is operated at 0.5 V. What is the value of the current? V i I S e v nv T 0.001 I S e 0.7 10.025 I I I S e 0.5 10.025 0.2 0.025 0.0 01e 10 6 I 0.3 35 ua (Temperature effect on the diode V-I Characteristic When the temperature is increasing the knee voltage V knee decreases by about 2mV/K For a silicon pn junction, the ideal reverse-saturation current density will 11 10
increase by approximately a factor of 4 for every 10 C increase in temperature. As temperature increases, less forward-bias voltage is required to obtain the same diode current. If the voltage is held constant, the diode current will increase as t emperature increases. 11 11
Forward biased dioed T, : for a given value of I F For a given I, V F F Barrier potential decrease as T increase Reverse current breakdown small & can be neglected DC or Static Resistance The application of a dc voltage to a circuit containing a semiconductor diode will result in an operating point on the characteristic curve that will not change with time. The resistance of the diode at the operating point can be found simply by finding the corresponding levels of V D and I D as shown in Fig. below and applying the following equation: The dc resistance levels at the knee and below will be greater than the resistance levels obtained for the vertical rise section of the characteristics. The resistance levels in the reverse-bias region will naturally be quite high. 12 12
Example Determine the dc resistance levels for the diode of Fig. above at (a) I D = 2 ma (b) I D =20 ma (c) V D = 10 V (a) At I D = 2 ma, V D = 0.5 V (from the curve) and (b) At I D = 20 ma, V D = 0.8 V (from the curve) and (c) At V D= -10 V, I D =-Is=-1 (from the curve) and AC or Dynamic Resistance The varying input will move the instantaneous operating point up and down a region of the characteristics and thus defines a specific change in current and voltage as shown in Fig. below. With no applied varying signal, the point of operation would be the Q-point appearing on Fig. below determined by the applied dc levels. The designation Q-point is derived from the word quiescent, which means still or unvarying. 13 13
A straight line drawn tangent to the curve through the Q-point as shown in Fig. above will define a particular change in voltage and current that can be used to determine the ac or dynamic resistance for this region of the diode characteristics. The steeper the slope, the less the value of or the same change in and the less the resistance The ac resistance in the vertical-rise region of the characteristic is therefore quite small, while the ac resistance is much higher at low current levels. For the characteristics of Fig. below: (a) Determine the ac resistance at I D =2 ma. (b) Determine the ac resistance at I D = 25 ma. (c) Compare the results of parts (a) and (b) to the dc resistances at each current level. 14 14
(a) For I D = 2 ma; the tangent line at I D =2 ma was drawn as shown in the figure and a swing of 2 ma above and below the specified diode current was chosen. At I D = 4 ma, V D =0.76 V, and at I D =0 ma, V D =0.65 V. The resulting changes in current and voltage are and the ac resistance: (b) For I D =25 ma, the tangent line at I D =25 ma was drawn as shown on the fig-ure and a swing of 5 ma above and below the specified diode current was chosen. At I D =30 ma, V D =0.8 V, and at I D 20 ma, V D =0.78 V. The resulting changes in current and voltage are and the ac resistance: (c)= for and which far exceeds the r d of 27.5. For I D =25 ma, V D =0.79 V and which far exceeds the r d of 2. 15 15
Dynamic Resistance The dynamic resistance of the diode is mathematically determined as the inverse of the slope of the transconductance curve. Therefore, the equation for dynamic resistance is: The dynamic resistance is used in determining the voltage drop across the diode in the situation where a voltage source is supplying a sinusoidal signal with a dc offset. The ac component of the diode voltage is found using the following equation: The voltage drop through the diode is a combination of the ac and dc components and is equal to: Example: Use the same circuit used for the Q point example but change the voltage source so it is an ac source with a dc offset. The source voltage is now, v in = 6 + sin(wt) Volts. It is a silicon diode so the barrier potential voltage is still 0.7 volts. The DC component of the circuit is the same as the previous example and therefore I D = 6V 0.7 V = 5.2 ma = 1 is a good approximation if the dc current is greater than 1 ma as it is in this example. Therefore (the voltage drop across the diode) 16 16
Diode Circuit Models The Ideal Diode Model The diode is designed to allow current to flow in only one direction. The perfect diode would be a perfect conductor in one direction (forward bias) and a perfect insulator in the other direction (reverse bias). In many situations, using the ideal diode approximation is acceptable. Ideal diode characteristics An diode is a two terminal device: Anode: the positive terminal Cathode: the negative terminal Forward biased turned on short perfect conductor zero voltage drop Reverse biased turned off open Example: Assume the diode in the circuit below is ideal. Determine the value of I D if a) V A = 5 volts (forward bias) and b) V A = -5 volts (reverse bias) 17 17
a) With V A > 0 the diode is in forward bias and is acting like a perfect conductor so: I D = V A /R S = 5 V / 50 = 100 ma b) With V A < 0 the diode is in reverse bias and is acting like a perfect insulator, therefore no current can flow and I D = 0. Practical diode The Ideal Diode with Barrier Potential This model is more accurate than the simple ideal diode model because it includes the approximate barrier potential voltage. Remember the barrier potential voltage is the voltage at which appreciable current starts to flow. Example: To be more accurate than just using the ideal diode model include the barrier potential. Assume V = 0.3 volts (typical for a germanium diode) Determine the value of I D if V A = 5 volts (forward bias). 18 18
With V A > 0 the diode is in forward bias and is acting like a perfect conductor so write a KVL equation to find I D : Complete diode The Ideal Diode with Barrier Potential and Linear Forward Resistance This model is the most accurate of the three. It includes a linear forward resistance that is calculated from the slope of the linear portion of the transconductance curve. However, this is usually not necessary since the R F (forward resistance) value is pretty constant. For low-power germanium and silicon diodes the R F value is usually in the 2 to 5 ohms range, while higher power diodes have a R F value closer to 1 ohm. 19 19
Example: Assume the diode is a low-power diode with a forward resistance value of 5 ohms. The barrier potential voltage is still: V = 0.3 volts (typical for a germanium diode) Determine the value of I D if V A = 5 volts. Once again write a KVL Example: Determine the forward voltage and forward current [forward bias] for each of the diode model also find the voltage across the limiting resistor in each cases. Assumed at the determined value of forward current. 21 20
1. Ideal Model: V F 0 VBIAS 10V I F 10mA R 1000 3 3 VR I F RLIMIT (1010 A)(1 10 ) 10V LIMIT 2. Practical Model V I F F 0.7V ( V R BIAS V LIMIT F ) 10V 0.7V 9.3mA 1000 3 3 VR I F RLIMIT (9.310 A)(110 ) 9. 3V LIMIT 3. Complete model I V F F V R BIAS LIMIT 0.7V ' r 0.7V I d r ' F d 10V 0.7V 1k 10 9.21mA 0.7V (9.21mA)(10) 792mV VR I F RLIMIT (9.21mA)(1k ) 9. 21V LIMIT Transition and diffusion capacitances Diffusion Capacitance (CD) or storage capacitance. Diffusion Capacitance is the capacitance due to transport of charge carriers between two terminals of a device, for example, the diffusion of carriers from anode to cathode in forward bias mode of a diode or from emitter to base forward-biased junction for a transistor. The variation of minority carrier charges stored in the neutral regions of the pn junction under forward bias contributes a small-signal capacitance known as the diffusion capacitance, C d, Charge stored in bulk region changes with the change of voltage across pn junction gives rise to capacitive effect. If the applied voltage changes to a different value and the current changes to a different value, a different amount of charge will be in transit in the new circumstances. The 21 21
change in the amount of transiting charge divided by the change in the voltage causing it is the diffusion capacitance. According to the definition C d dq dv The charge stored in bulk region is obtained from below equations: Q Q Q p n Aq [ pn( x) pno xn Aq [ pn( xn) pno] I p I n p n ] dx L p The expression for diffusion capacitance C d d [ T I dv T ( ) IQ V T T ( ) I VT 0 Q s e V V T ] Forward-bias, linear relationship Reverse-bias, almost inexistence Transition Capacitance or Depletion Capacitance or Junction Capacitance Transition capacitance : A reverse biased PN-junction has a region of high resistivity (depletion layer) sandwiched in between two regions of relatively low resistivity. The P- N regions act as the plates of a capacitor and the depletion layer acts as the dielectric This is known as the transition capacitance or depletion capacitance. Charge stored in depletion layer changes with the change of voltage across pn junction gives rise to capacitive effect. Recall that the basic equation for the capacitance of a parallel-plate capacitor is defined by, where ε is the permittivity of the dielectric (insulator) between the plates of area A separated by a distance. 22 22
In the reverse-bias region there is a depletion region (free of carriers) that behaves essentially like an insulator between the layers of opposite charge. Since the depletion width ( ) will increase with increased reverse-bias potential, the resulting transition capacitance will decrease, as shown in Fig. According to the definition C j dq dv R V R V Q Actually this capacitance is similar to parallel plate capacitance. C j A W dep = C j0 V (1 R 2 1 [ ( q N V o ) A A 1 N B )( V 0 v R ) W dep 2 1 1 [ ( )( V0 vr ) q N N A B A more general formula for depletion capacitance is : C j C (1 V R j0 V 0 ) m Where m is called grading coefficient. If the concentration changes sharply, Forward-bias condition, C j 2C j 0 Reverse-bias condition, C C j d 1 1 m ~ 3 2 1 m 2 23 23
Diode Switching Times Switching Time of a diode is the time it takes to switch the diode between two states (ON and OFF states) Switching diodes Switching diodes, sometimes also called small signal diodes, are single diodes in a discrete package. A switching diode provides essentially the same function as a switch. Below the specified applied voltage it has high resistance similar to an open switch, while above that voltage it suddenly changes to the low resistance of a closed switch. 24 24
R src V D V 1 V src I D V 2 t = 0 t = T Excess charge Space charge V D ON OFF ON Time Small -Signal Model and Load Line small -signal model Small signal approximation ( ) ( ) ( ) ( ) Under small signal condition : ( ) ( ) ( ) I D associates with V D i d associates with v d dc operating point Q small signal response The diode exhibits linear I V characteristics under small signal conditions (v d 10mV) Diode small signal resistance and conductance at operating point 25 25
Symbol convention: i D (t) Lowercase symbol, uppercase subscript stands for total instantaneous qualities. I D Uppercase symbol, uppercase subscript stands for dc component. i d (t) Lowercase symbol, lowercase subscript stands for ac component or incremental signal qualities. I d (t) Uppercase symbol, lowercase subscript stands for the rms(root-mean-square) of ac. The diode small signal model Choose proper dc analysis technique or model to obtain the operation point Q The small signal model is determined once Q is provided The small signal model is used for circuit analysis when the diode is operating around Q 26 26
Obtaining the operating point. Referring to Figure, the current I d, when V in V g, is This is the load line equation for this circuit. The intersection of the load line and the I-V characteristic curve ( ) for the device is the operating point for the diode. This operating point is also called the quiescent point or Q-point and it gives the value of the current through the diode and the voltage across the diode. For R = 100 Ω, and V in = 2V, the load line and resulting operating point is shown on Figure. The Q Point 27 27
The operating point or Q point of the diode is the quiescent or no-signal condition. The Q point is obtained graphically and is really only needed when the applied voltage is very close to the diode s barrier potential voltage. The example 3 below that is continued on the next slide, shows how the Q point is determined using the transconductance curve and the load line. Load line Analysis The applied load will normally have an important impact on region of operation of device. If analysis is done in graphical approach, a line can be drawn on the characteristics of the device that represents applied load. The intersection of load line with the characteristics will determine the point of operation. Such an analysis is called as load-line analysis. The intersection point is called Q point or operating point. 28 28
Example on load line analysis 29 29
example For thecircuit shown, Given :Vss 10 V, R 10 k, the I - V curve of the diode Find : the diode current and voltage at the operating point V SS Ri performload - line analysis at theoperating point V DQ D v D 0.68 V,, i.e.,10 10k i i DQ D 0.93 ma v D example For thecircuit shown, Given :Vss 10 V, R 10 k, the I - V curve of the diode Find : the diode current and voltage at the operating point introduction to hetetrojunctions and double hetetrojunctions Heterojunction: the interface between two layers or regions of dissimilar crystalline semiconductors. Heterostructure: A stack of materials based on a central heterojunction can be considered a heterostructure. homo means similar or the same (as in a homojunction of p-si/n-si) hetero means different (as in a heterojunction of CdTe/CdS) If both the p-type and the n-type regions are of the same semiconductor material, the junction is called a homojunction. If the junction layers are made of different semiconductor materials, it is a heterojunction. As a matter of convention, if the n-type doped semiconductor material has larger energy gap than the p-type doped material, it is denoted a p-n heterojunction. Semiconductor N-N Heterojunction 31 30
Once a junction is made, electrons will flow from the side with higher Fermi level (1) to the side with lower Fermi level (2) Semiconductor N-N Heterojunction: Equilibrium Electrons will flow from the side with higher Fermi level (1) to the side with lower Fermi level (2) Electron flow away from semiconductor (1) will result in a region at the interface which is depleted of electrons (depletion region). Because of positively charged donor atoms, the depletion region has net positive charge density 31 31
Electron flow into semiconductor (2) will result in a region at the interface which has an accumulation of electrons (accumulation region). The accumulation region has net negative charge density Semiconductor P-N Heterojunction Consider a junction of a n-doped semiconductor (semiconductor 1) with an p-doped semiconductor (semiconductor 2). The two semiconductors are not necessarily the same. We assume that 1 has a wider band gap than 2. The band diagrams of 1 and 2 by themselves are shown below. Once a junction is made: Electrons will flow from the side with higher Fermi level (1) to the side with lower Fermi level (2) Holes will flow from the side with lower Fermi level (2) to the side with higher Fermi level (1) Semiconductor P-N Heterojunction: Equilibrium 32 32
Electron flow away from semiconductor (1) will result in a region at the interface which is depleted of electrons (depletion region). Because of positively charged donor atoms, the depletion region has net positive charge density Hole flow away from semiconductor (2) will result in a region at the interface which is depleted of holes (depletion region). Because of negatively charged acceptor atoms, the depletion region has net negative charge density Depending upon the difference between and we could have type I, type II, or type III heterojunction interfaces, as shown below. Type-I: Straddling gap Type-II: Staggered gap 33 33
Type-III: Broken gap Double Heterojunction Homojunction Single Heterojunction Double Heterojunction 34 34