1 P–N Junction Diode
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Lecture Notes: 2304154 Physics and Electronics Lecture 4 (2 nd Half), Year: 2007 Physics Department, Faculty of Science, Chulalongkorn University 18/10/2007 Contents 1 P–N Junction Diode 1 1.1 Currents in Diode ................................ 1 1.2 Diode in Dc Circuits ............................... 4 2 Problems 9 1 P–N Junction Diode Structure of P-N Junction Diode P–N junction diode: the electrical equipment constructed by attaching a slab of N material with a slab of P material. 2–D diode section and symbol 1.1 Currents in Diode Depletion Region + diffuse across surface - diffuse across surface 2304154 Physics and Electronics (file: lec04 2.tex) 1 Tianprateep, M.
Transcript of 1 P–N Junction Diode
Lecture Notes: 2304154 Physics and ElectronicsLecture 4 (2nd Half), Year: 2007Physics Department, Faculty of Science, Chulalongkorn University18/10/2007
Contents
1 P–N Junction Diode 11.1 Currents in Diode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Diode in Dc Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Problems 9
1 P–N Junction Diode
Structure of P-N Junction DiodeP–N junction diode: the electrical equipment constructed by attaching a slab of N
material with a slab of P material.
2–D diode section and symbol
1.1 Currents in Diode
Depletion Region
+diffuse across surface
−diffuse across surface
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+drift currents
E–field
+
diffusion currents
The P– and N–regions are separated by a depletion region (space–charge region) .
• drift currents: because of E–fields in depletion region (minority carriers)
• diffusion currents: because of the difference of no. of hole (e) in P–N regions (majoritycarriers)
Energy Level at Equilibrium
depletion region
Forward Bias
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Reverse Bias
Since barrier potential equals bias voltage, minority carriers produce reverse current.
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1.2 Diode in Dc Circuits
Volt-Ampere Characteristic
Shockley Equation (not for reverse breakdown)
iD = Is
[exp
(vD
nVT
)− 1
]Is = saturation current (depends on T)n = emission coefficient
VT =kT
q= thermal voltage
k = Boltzmann’s constant
For diode voltage
vD = nVT ln[(
iDIs
)]At reverse bias
iD ∼= −Is
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Diode Circuit Analysis
How to find iD and vD?From Kirchhoff’s voltage law:
Vss = R iD + vD
load line equationFrom Shockley equation (volt–ampere characteristic of diode)
iD = Is
[exp
(vD
nVT
)− 1
]ID and vD is the solution of these 2 equations.
Graphical Load Line Analysis
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Zener Diode
Zener diodes: the diodes that are intended to operate in the breakdown region.They are useful in applications for which a constant voltage in breakdown is desir-
able. (For example, voltage regulator.)
Simple Voltage Regulator Circuit
Load line equation for this circuit:
VSS + RiD + vD = 0
15V < VSS < 20V −10.5V < vD < −10V
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Figure 1 Piecewise-liner model for the diode.
Piecewise-Linear Diode ModelsHow to calculate in this model?
• Approximate the actual volt–ampere characteristic by straight–line segments.
• Model each section of the diode characteristic with a resistance in series with aconstant–voltage source.
v = Rai + Va
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iD =?
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Figure 2 For Problem 1
iD = 80 mA
2 Problems
1. Find the operating point for the circuit, which a resistor R is series with a diode, ifthe diode characteristic is shown in Fig. 2 and a) VSS = 2 V and R = 100 Ω; b) VSS
= 15 V and R = 1 kΩ; c) VSS = 1.0 V and R = 20 Ω.
2. The nonlinear circuit element, that is series with R, has ix = v3x/8. Also, we have Vs
= 6 V and Rs = 3 Ω. Use graphical load line techniques to solve for ix and vx.
3. Use the appropriate circuit model from Fig. 1 to solve for v0 in the circuit of Fig. 3if a) RL = 10 kΩ and b) RL = 1 kΩ.
4. Design a voltage–regulator circuit to provide a constant voltage of 5 V to a load froma variable supply voltage. The load current varies from 0 to 100 mA, and the sourcevoltage varies from 8 to 10 V. You may assume that ideal Zener diode are available.Resistors of any value may be specified. Draw the circuit diagram of your regulator,and specify the value of each component. Also, find the worst case (maximum) powerdissipated in each compnent in your regulator. Try to use good judment in yourdesign.
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Figure 3 For Problem 3
5. Repeat Problem 4 if the supply voltage ranges from 6 to 10 V.
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