3 Chapter Solid-State Diodes and Diode Circuits
Microelectronic Circuit Design
Richard C. Jaeger Travis N. Blalock
Chap 3 -1
Chapter Goals
Understand diode structure and basic layout Develop electrostatics of the pn junction Explore various diode models including the mathematical model, the ideal diode model, and the constant voltage drop model Understand the SPICE representation and model parameters for the diode Define regions of operation of the diode (forward bias, reverse bias, and reverse breakdown) Apply the various types of models in circuit analysis Explore different types of diodes Discuss the dynamic switching behavior of the pn junction diode Explore diode applications Practice simulating diode circuits using SPICE
Chap 3 -2
Diode Introduction
A diode is formed by joining an n-type semiconductor with a p-type semiconductor. A pn junction is the interface between n and p regions.
Chap 3 -3
Diode symbol
pn Junction Electrostatics
Donor and acceptor concentration on either side of the junction. Concentration gradients give rise to diffusion currents.
Chap 3 -4
Drift Currents
Diffusion currents lead to localized charge density variations near the pn junction. Gauss law predicts an electric field due to the charge distribution:
c E = s
Assuming constant permittivity,
E(x) =
1
s
(x)dx
Resulting electric field gives rise to a drift current. With no external circuit connections, drift and diffusion currents cancel. There is no actual current, since this would imply power dissipation, rather the electric field cancels the diffusion current tendency.
Chap 3 -5
Space-Charge Region Formation at the pn Junction
Chap 3 -6
Potential Across the Junction
Charge Density
Electric Field
Potential
N NA D kT = E(x)dx = T ln 2 VT = V , j q ni
Chap 3 -7
Width of Depletion Region
Combining the previous expressions, we can form an expression for the width of the space-charge region, or depletion region. It is called the depletion region since the excess holes and electrons are depleted from the dopant atoms on either side of the junction.
2s 1 1 w d 0 =(x n +x p ) = + j q N A N D
Chap 3 -8
Width of Depletion Region (Example)
Problem: Find built-in potential and depletion-region width for given diode Given data:On p-type side: NA = 1017/cm3 on n-type side: ND = 1020/cm3 Assumptions: Room-temperature operation with VT = 0.025 V Analysis:
N AND (1017 /cm 3 )(10 20 /cm 3 ) j = VT ln 2 = (0.025 V)ln = 0.979V 20 6 n (10 /cm ) i
2s 1 1 = 0.113m w = q + N j d0 N D A
Chap 3 -9
Diode Electric Field (Example)
Problem: Find the electric field and size of the individual depletion layers on either side of a pn junction for a given diode Given data: On the p-type side: NA = 1017/cm3 on the n-type side: ND = 1020/cm3 from earlier example, w= 0.113 m = 0.979V
j
d0
Assumptions: Room-temperature operation Analysis: ND NA =x wd0 =xn +x p =xn 1+ p1+ NA ND w d0 w d0 xn = =1.13104 m x p = 0.113m N N D 1+ 1+ A N N A D 2j 2(0.979V ) EMAX = w = =173kV/cm 0.113m d0
Chap 3 - 10
Internal Diode Currents
Mathematically, for a diode with no external connections, the total current expressions developed in Chapter 2 are equal to zero. The equations only dictate that the total currents are zero. However, as mentioned earlier, since there is no power dissipation, we must assume that the field and diffusion current tendencies cancel and the actual currents are zero.
n =0 x p T j p =qp pE qD p =0 x
T j n =qn nE +qDn
When external bias voltage is applied to the diode, the above equations are no longer equal to zero.
Chap 3 -11
Diode Junction Potential for Different Applied Voltages
Chap 3 -12
Diode i-v Characteristics
The turn-on voltage marks the point of significant current flow. Is is called the reverse saturation current.
Chap 3 -13
Diode Equation
v qv D iD = S I exp I exp 1= S D nkT nVT
where IS
1
vD q k T n VT
= = = = = = =
reverse saturation current (A) voltage applied to diode (V) electronic charge (1.60 x 10-19 C) Boltzmanns constant (1.38 x 10-23 J/K) absolute temperature nonideality factor (dimensionless) kT/q = thermal voltage (V) (25 mV at room temp.)
IS is typically between 10-18 and 10-9 A, and is strongly temperature dependent due to its dependence on ni2. The nonideality factor is typically close to 1, but approaches 2 for devices with high current densities. It is assumed to be 1 in this text. Chap 3 -14
Diode Voltage and Current Calculations (Example)
Problem: Find diode voltage for diode with given specifications Given data: IS = 0.1 fA, ID = 300 A Assumptions: Room-temperature dc operation with VT = 0.025 V Analysis: With IS = 0.1 fA ID 3 -4 VD = T ln 1+ = nV 1(0.0025V )ln(1+ 10 A )= 0.718V IS -16 A 10
With IS = 10 fA
VD = 0.603V VD = 0.748V
Chap 3 -15
With ID = 1 mA, IS = 0.1 fA
Diode Current for Reverse, Zero, and Forward Bias
Reverse bias:
v iD = S D S [ ] S I exp 1 I 0 1 I nVT v iD = S D S [ 1] 0 I exp 1 I 1 nVT
Zero bias:
Forward
v vD D iD = S I exp 1I S exp nVT nVT bias:
Chap 3 -16
Semi-log Plot of Forward Diode Current and Current for Three Different Values of IS
I S [ A] = S [B] = S [C ] 10I 100I
Chap 3 -17
Diode Temperature Coefficient
Diode voltage under forward bias: D kT D kT D i i i v D = T ln += ln + ln V 1 1 I I S q S q IS Taking the derivative with respect to temperature yields
dv D k D kT 1 dI S v D i 1 dI S v D GO T V 3V = ln = T V = dT q S q I S dT I T I S dT T
V/K
Assuming iD >> IS, IS ni2, and VGO is the silicon bandgap energy at 0K. For a typical silicon diode
0.65 1.12 0.075) V dv D ( = = -1.82 mV/K - 1.8 mV/ C dT 300K
Chap 3 -18
Reverse Bias
External reverse bias adds to the built-in potential of the pn junction. The shaded regions below illustrate the increase in the characteristics of the space charge region due to an externally applied reverse bias, vD.
Chap 3 -19
Reverse Bias (cont.)
External reverse bias also increases the width of the depletion region since the larger electric field must be supported by additional charge. 2 1 1 s w d =(x n +x p ) = + j +v R q N A N D
(
)
w d =w d 0 1+
j
vR
2 1 1 s where w d 0 =(x n +x p ) = + j q N A N D
Chap 3 -20
Reverse Bias Saturation Current
We earlier assumed that the reverse saturation current was constant. Since it results from thermal generation of electron-hole pairs in the depletion region, it is dependent on the volume of the space charge region. It can be shown that the reverse saturation gradually increases with increased reverse bias.
v I S =I S 0 1 + R
j
IS is approximately constant at IS0 under forward bias.
Chap 3 -21
Reverse Breakdown
Increased reverse bias eventually results in the diode entering the breakdown region, resulting in a sharp increase in the diode current. The voltage at which this occurs is the breakdown voltage, VZ. 2 V < VZ < 2000 V
Chap 3 -22
Reverse Breakdown Mechanisms
Avalanche Breakdown Si diodes with VZ greater than about 5.6 volts breakdown according to an avalanche mechanism. As the electric field increases, accelerated carriers begin to collide with fixed atoms. As the reverse bias increases, the energy of the accelerated carriers increases, eventually leading to ionization of the impacted ions. The new carriers also accelerate and ionize other atoms. This process feeds on itself and leads to avalanche breakdown.
Chap 3 -23
Reverse Breakdown Mechanisms (cont.) Zener Breakdown
Zener breakdown occurs in heavily doped diodes. The heavy doping results in a very narrow depletion region at the diode junction. Reverse bias leads to carriers with sufficient energy to tunnel directly between conduction and valence bands moving across the junction. Once the tunneling threshold is reached, additional reverse bias leads to a rapidly increasing reverse current.
Breakdown Voltage Temperature Coefficient Temperature coefficient is a quick way to distinguish breakdown mechanisms. Avalanche breakdown voltage increases with temperature, whereas Zener breakdown decreases with temperature. For silicon diodes, zero temperature coefficient is achieved at approximately 5.6 V.
Chap 3 -24
Breakdown Region Diode Model
In breakdown, the diode is modeled with a voltage source, VZ, and a series resistance, RZ. RZ models the slope of the i-v characteristic. Diodes designed to operate in reverse breakdown are called Zener diodes and use the indicated symbol.
Chap 3 -25
Reverse Bias Capacitance
Changes in voltage lead to changes in depletion width and charge. This leads to a capacitance that we can calculate from the chargevoltage dependence.
N N Qn = D x n A = A D d A qN q w N N A+ D
C j0A dQn Cj = = dv R v 1+ R
Coulombs
F/cm2 where C j 0 =
s
wd 0
j
Cj0 is the zero bias junction capacitance per unit area.
Chap 3 -26
Reverse Bias Capacitance (cont.)
Diodes can be designed with hyper-abrupt doping profiles that optimize the reverse-biased diode as a voltage controlled capacitor.
Circuit symbol for the variable capacitance diode (Varactor)
Chap 3 -27
Forward Bias Capacitance
In forward bias operation, additional charge is stored in neutral region near edges of space charge region.
QD = iD T Coulombs
T is called diode transit time and depends on size and type of diode.
Additional diffusion capacitance, associated with forward region of operation is proportional to current and becomes quite large at high currents.
iD I dQD ( + S )T iD Cj = = T F dv D VT VT
Chap 3 -28
Schottky Barrier Diode
One semiconductor region of the pn junction diode can be replaced by a non-ohmic rectifying metal contact.A Schottky contact is easily formed on n-type silicon. The metal region becomes the anode. An n+ region is added to ensure that the cathode contact is ohmic.
Schottky diodes turn on at a lower voltage than pn junction diodes and have significantly reduced internal charge storage under forward bias.
Chap 3 -29
Diode Spice Model
Rs represents the inevitable series resistance of a real device structure. The current controlled current source models the ideal exponential behavior of the diode. Capacitor C includes depletion-layer capacitance for the reverse-bias region as well as diffusion capacitance associated with the junction under forward bias. Typical default values: Saturation current IS = 10 fA, Rs = 0 , transit time TT = 0 seconds, N = 1
Chap 3 -30
PN-JUNCTION DIODE PARAMETER EXTRACTION
Determination of the Saturation Current IS
A straight-line t of the forward-bias ID (VPN ) characteristics on a log-linear plot for values of VPN > 0.1 V intercepts the y-axis at IS .
Determination of the Ideality Factor n
Select two values of the bias voltage VPN on the straight-line part of the current-voltage characteris-tics, e.g, VP N,1 and VP N,2. Determine the values of the corresponding currents ID(VPN,1) and ID(VP N,2).
Determination of the Ideality Factor n
Determination of the Series Resistance RS
At bias voltages larger than 0.8 to 0.9 V, the eect of the diode series resistance starts aecting the ID (VPN ) characteristics as shown. Select a value of the diode current ID,1 in this bias range, and determine the dierence between the straight-line t and the diode characteristics. This dierence is ID,1RS . Knowing ID,1, determine RS . Repeat at dierent values of the current and calculate an average value of RS .
The PN-Junction Diode Capacitance-Voltage Characteristics
The PN-Junction Diode Capacitance-Voltage Characteristics
The PN-Junction Diode Capacitance-Voltage Characteristics
Diode Layout
Chap 3 -39
Diode Circuit Analysis: Basics
The loop equation for the diode circuit is:
V = I D R + VD
This is also called the load line for the diode. The solution to this equation can be found by: Graphical analysis using the load-line method. Analysis with the diodes mathematical model. Simplified analysis with the ideal diode model. Simplified analysis using the constant voltage drop (CVD) model.
Chap 3 -40
V and R may represent the Thvenin equivalent of a more complex 2terminal network. The objective of diode circuit analysis is to find the quiescent operating point for the diode. Q-Point = (ID, VD)
Load-Line Analysis (Example)
Problem: Find diode Q-point Given data: V = 10 V, R = 10k. Analysis: 10 = I D 104 + VD To define the load line we use,
ForVD = D =10V 10k= 0,I ( ) 1mA ForVD = 5V,ID =5V 10k= ( ) 0.5mA
These points and the resulting load line are plotted.Q-point is given by intersection of load line and diode characteristic: Q-point = (0.95 mA, 0.6 V)
Chap 3 -41
Analysis using Mathematical Model for Diode
Problem: Find the Q-point for a given diode characteristic. Given data: IS =10-13 A, n=1, VT =0.0025 V Analysis: Make initial guess VD0 . Evaluate f and its derivative f for this value of VD. Calculate new guess for VD using 0 f VD 1 0 VD = VD 0 f ' (VD )
()
V I D = I S exp D 1 = 1013 [ exp( 40VD ) 1] nVT Repeat steps 2 and 3 till convergence. 10 = 1041013 [ exp( 40VD ) 1] + VD Using a spreadsheet we get : Q-point = ( 0.9426 mA, 0.5742 V) The solution is given by a transcendental Since, usually we dont have accurate equation. A numerical answer can be found saturation current values and significant by using Newtons iterative method. tolerances exist for sources and passive
f = 10 1041013 [ exp( 40VD ) 1] we VD
components, need answers precise to only 2or 3 significant digits.
Chap 3 -42
Analysis using Ideal Model for Diode
If an ideal diode is forward-biased, the voltage across the diode is zero. If an ideal diode is reverse-biased, the current through the diode is zero. vD =0 for iD >0 and vD =0 for vD < 0 Thus, the diode is assumed to be either on or off. Analysis is conducted in following steps: Select a diode model. Identify anode and cathode of the diode and label vD and iD. Guess diodes region of operation from circuit. Analyze circuit using diode model appropriate for assumed region of operation. Check results to check consistency with assumptions.
Chap 3 -43
Analysis using Ideal Model for Diode: Example
Since source appears to force positive current through diode, assume diode is on. (10 0)V ID = = 1mA | ID 0 10k Our assumption is correct, and the Q-Point = (1 mA, 0V)
Since source is forcing current backward through diode assume diode is off. Hence ID = 0 . Loop equation is: 10 + D + 4 ID = V 10 0
VD = 10V | VD < 0
Our assumption is correct and the Q-Point = (0, -10 V)
Chap 3 -44
Analysis using Constant Voltage Drop Model for Diode
Analysis:
Since the 10-V source appears to force positive current through the diode, assume diode is on. (10 Von )V 10k (10 0.6)V = =0.940 mA 10k ID =
Chap 3 -45
vD = Von for iD >0 and vD = 0 for vD < Von.
Two-Diode Circuit Analysis
Analysis: The ideal diode model is chosen. Since the 15-V source appears to force positive current through D1 and D2, and the -10-V source is forcing positive current through D2, assume both diodes are on. Since the voltage at node D is zero due to the short circuit of ideal diode D1, (15 0)V I1 = =1.50mA 10k 0 ( 10V ) ID 2 = =2.00mA 5k I1 =ID1 +ID 2 | ID1 =1.50 2.00 = 0.500mA The Q-points are (-0.5 mA, 0 V) and (2.0 mA, 0 V) But, ID1 < 0 is not allowed by the diode, so try again.
Chap 3 -46
Two-Diode Circuit Analysis (contd.)
Since the current in D1 is zero, ID2 = I1, 1510,000I1 5,000ID2 (10)=0 25V I1 = =1.67mA 15,000 VD1 =1510,000I1 =1516.7 =1.67V Q-Points are D1 : (0 mA, -1.67 V):off D2 : (1.67 mA, 0 V) :on Now, the results are consistent with the assumptions.
Chap 3 -47
Analysis: Since current in D2 is valid, but that in D1 is not, the second guess is D1 off and D2 on.
Analysis of Diodes in Reverse Breakdown Operation (-5 V, -3 Choose 2 points (0V, -4 mA) and
mA) to draw the load line. It intersects the i-v characteristic at the Q-point: (-2.9 mA, -5.2 V). Using the piecewise linear model: IZ = I D > 0
Using load-line analysis: =D + 20 V 5000ID
205100IZ 5=0 (205)V IZ = =2.94mA 5100
Since IZ > 0 (ID < 0), the solution is consistent with Zener breakdown.
Chap 3 -48
Voltage Regulator Using the Zener Diode
V V (205)V IS = S Z = =3mA R 5k V 5V IL = Z = =1mA| IZ =IS IL =2mA RL 5k For proper regulation, Zener current must be positive. If the Zener current < 0, the Zener diode no longer controls the voltage across the load resistor and the voltage regulator is said to have dropped out of regulation. VS R 1 + 1 >0 | R > IZ = VZ =R L R RL min R V S 1 V Z
Chap 3 -49
The Zener diode keeps the voltage across load resistor RL constant. For Zener breakdown operation, IZ > 0.
Voltage Regulator Using a Zener Diode: Example Including Zener Resistance
VZ 20V VL 5V V + + L =0 5000 100 5000 VL =5.19V Problem: Find the output voltage and Zener diode current for a Zener diode regulator. Given data: VS = 20 V, R = 5 k, RZ = 0.1 k, VZ = 5 V Analysis: The output voltage is a function of the current through the Zener diode. IZ = VL 5V 5.19V5V = =1.9mA >0 100 100
Chap 3 -50
Line and Load Regulation
Line regulation characterizes how sensitive the output voltage is to input voltage changes. dVL LineRegulation= mV/V dVS RZ Forafixedloadcurrent,LineRegulation = R+ RZ Load regulation characterizes how sensitive the output voltage is to changes in load current withdrawn from regulator. dV LoadRegulation= L dI L
Forchangesinloadcurrent,LoadRegulation=RZ R
Load regulation is the Thvenin equivalent resistance looking back into the regulator from the load terminals.
Chap 3 -51
Rectifier Circuits
A basic rectifier converts an ac voltage to a pulsating dc voltage.
A filter then eliminates ac components of the waveform to produce a nearly constant dc voltage output. Rectifier circuits are used in virtually all electronic devices to convert the 120-V 60-Hz ac power line source to the dc voltages required for operation of electronic devices. In rectifier circuits, the diode state changes with time and a given piecewise linear model is valid only for a certain time interval.
Chap 3 -52
Half-Wave Rectifier Circuit with Resistive Load
For the positive half-cycle of the input, the source forces positive current through the diode, the diode is on, and vO = vS. During the negative half cycle, negative current cant exist in the diode. The diode is off, current in resistor is zero, and vO =0 .
Chap 3 -53
Half-Wave Rectifier Circuit with Resistive Load (cont.)
Using the CVD model, during the on-state of the diode vO = (VP sint)- Von. The output voltage is zero when the diode is off. Often a step-up or step-down transformer is used to convert the 120-V, 60-Hz voltage available from the power line to the desired ac voltage level as shown.
Time-varying components in the rectifier output are removed using a filter capacitor.
Chap 3 -54
Peak Detector Circuit
As the input voltage rises, the diode is on, and the capacitor (initially discharged) charges up to the input voltage minus the diode voltage drop. At the peak of the input voltage, diode current tries to reverse, and the diode cuts off. The capacitor has no discharge path and retains a constant voltage providing a constant output voltage: Vdc = VP - Von
Chap 3 -55
Half-Wave Rectifier Circuit with RC Load
As the input voltage rises during the first quarter cycle, the diode is on and the capacitor (initially discharged) charges up to the peak value of the input voltage. At the peak of the input, the diode current tries to reverse, the diode cuts off, and the capacitor discharges exponentially through R. Discharge continues till the input voltage exceeds the output voltage which occurs near the peak of next cycle. This process then repeats once every cycle. This circuit can be used to generate negative output voltage if the top plate of capacitor is grounded instead of bottom plate. In this case, Vdc = -(VP - Von)
Chap 3 -56
Half-Wave Rectifier Circuit with RC Load (cont.)
The output voltage is not constant as in an ideal peak detector, but has a ripple voltage Vr. The diode conducts for a short time T called the conduction interval during each cycle, and its angular equivalent is called the conduction angle c. T T (VP Von ) T Vr (VP Von ) 1 RC T R C 1 2T (VP Von ) = 1 2Vr T VP RC VP
c =T =
2Vr VP
Chap 3 -57
Half-Wave Rectifier Analysis: Example
Problem: Find the dc output voltage, output current, ripple voltage, conduction interval, and conduction angle for a half-wave rectifier. Given data: secondary voltage Vrms = 12.6 (60 Hz), R = 15 , C = 25,000 F, Von = 1 V Analysis: Vdc =VP Von =(12.6 2 1)V =16.8V V V Idc = P on =16.8V =1.12A R 15 (V V ) T Vr P on =0.747V R C Using discharge interval T=1/60 s,
c =T =
c T = =
= 0.29 =0.769ms 2f 120
c
2Vr =0.290rad =16.6o VP
Chap 3 -58
Peak Diode Current
In rectifiers, nonzero current exists in the diode for only a very small fraction of period T, yet an almost constant dc current flows out of the filter capacitor to load. The total charge lost from the filter capacitor in each cycle is replenished by the diode during a short conduction interval causing high peak diode currents. If the repetitive current pulse is modeled as a triangle of height IP and width T,
IP = dc 2T = I 48.6A T
using the values from the previous example.
Chap 3 -59
Surge Current
In addition to the peak diode currents, there is an even larger current through the diode called the surge current that occurs when power is first turned on. During first quarter cycle, current through diode is approximately d V sin = CV cos id (t)=c (t) P i C t P t dt Peak values of this initial surge current occurs at t = 0+: ISC =CVP = 168A using values from previous example. Actual values of surge current wont be as large as predicted above because of the neglected series resistances associated with the rectifier diode and transformer.
Chap 3 -60
Peak Inverse Voltage Rating
The peak inverse voltage (PIV) rating of the rectifier diode is the diode breakdown voltage. When the diode is off, the reverse-bias across the diode is Vdc - vS. When vS reaches negative peak,
PIV min = VP ) P Vdc vs VP Von ( 2V
The PIV value corresponds to the minimum value of Zener breakdown voltage required for the rectifier diode.
Chap 3 -61
Diode Power Dissipation
Average power dissipation in a diode is given by IT 1T 1T PD = v D (t)iD (t)dt = Von iD (t)dt = on P on Idc V V 2T T0 T0 The simplification is done by assuming a triangular approximation for the diode current and that the voltage across diode is constant at Vdc. Average power dissipation in the diode series resistance is given by 1T 2 12 T4 2 PD = iD (t)RS dt = IDRS = T Idc RS T 3 T T0 3 This power dissipation can be reduced by minimizing peak current through the use of a minimum size of filter capacitor or by using fullwave rectifiers.
Chap 3 -62
Full-Wave Rectifiers
Full-wave rectifiers cut capacitor discharge time in half and require half the filter capacitance to achieve a given ripple voltage. All specifications are the same as for halfwave rectifiers. Reversing polarity of the diodes gives a fullwave rectifier with negative output voltage.
Chap 3 -63
Full-Wave Bridge Rectification
The requirement for a centertapped transformer in the fullwave rectifier is eliminated through use of 2 extra diodes. All other specifications are the same as for a half-wave rectifier except PIV = VP.
Chap 3 -64
Rectifier Topology Comparison
Filter capacitors are a major factor in determining cost, size and weight
in design of rectifiers. For a given ripple voltage, a full-wave rectifier requires half the filter capacitance as that in a half-wave rectifier. Reduced peak current can reduce heat dissipation in diodes. Benefits of full-wave rectification outweigh increased expenses and circuit complexity (an extra diode and center-tapped transformer). The bridge rectifier eliminates the center-tapped transformer, and the PIV rating of the diodes is reduced. Cost of extra diodes is negligible.
Chap 3 -65
Rectifier Design Analysis
Problem: Design a rectifier with given specifications. Given data: Vdc = 15 V, Vr < 0.15 V, Idc = 2 A Analysis: Use a full-wave bridge rectifier that needs a smaller value of filter capacitance, smaller diode PIV rating, and no center-tapped transformer. V V +2Von 15+2 V = P = dc = V =12Vrms 2 2 2 T /2 1 1 s =0.111F =2A C =Idc V 120 0.15V r
T = 1 2Vr 1 2(0.15V ) = =0.352ms VP 120 17V
2 T =2A 1/60s =94.7A IP =I 2 T 0.352ms Isurge =CVP =120(0.111)(17)=711A | PIV =VP =17V dc
Chap 3 -66
Three-Terminal IC Voltage Regulators
Regulators use feedback with high-gain amplifiers to reduce ripple voltage at the output. Bypass capacitors provide low-impedance paths for highfrequency signals to ensure proper operation of the regulator. Regulators provide excellent line and load regulation, maintaining constant voltage even if the output current changes by many orders of magnitude. A main design constraint is VREG which must not fall below a minimum specified dropout voltage (a few volts).
Chap 3 -67
Dynamic Switching Behavior of Diodes
The non-linear depletion-layer capacitance of the diode prevents the diode voltage from changing instantaneously and determines turn-on and recovery times. Both forward and reverse current overshoot the final values when the diode switches on and off as shown. Storage time is given by:
S =T ln 1
IF I
R
Chap 3 -68
Photo Diodes and Photodetectors
If the depletion region of a pn junction diode is illuminated with light with sufficiently high frequency, photons can provide enough energy to cause electrons to jump the semiconductor bandgap to generate electron-hole pairs:
hc EP =h= EG
h =Plancks constant = 6.626 x 10-34 J-s = frequency of optical illumination = wavelength of optical illumination c = velocity of light = 3 x 108 m/s Photon-generated current can be used in photodetector circuits to generate an output voltage
vO =PHR i
The diode is reverse-biased to enhance depletion-region width and electric field.
Chap 3 -69
Solar Cells and Light-Emitting
Diodes
In solar cell applications, optical illumination is constant, and dc current IPH is generated. The goal is to extract power from the cell, and the i-v characteristics are plotted in terms of cell current and cell voltage. For a solar cell to supply power to an external circuit, the ICVC product must be positive, and the cell should be operated near the point of maximum output power Pmax. Light-Emitting Diodes (LEDs) use recombination processes in the forward-biased pn junction diode to produce light. When a hole and electron recombine, an energy equal to the bandgap of the semiconductor is released as a photon.
Chap 3 -70
End of Chapter 3
Chap 3 -71