Week 2: PN Junction Diode

ELE 2110A Electronic Circuits

Week 2: PN Junction Diode

ELE2110A © 2008

Lecture 02 - 1

Topics to cover…

z

Physical operation of pn junction diode Terminal I-V characteristics Breakdown and junction capacitances Diode circuits analysis

z

Reading Assignment: Chap 3.1-3.7, 3.10-3.13 of Jaeger & Blalock

z z z

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Lecture 02 - 2

Why diodes? z z

R, C, L are linear circuit elements Many signal processing functions need nonlinear elements, like signal rectification (to convert negative signal to be positive).

Nonlinear signal processing

An AC-DC converter ELE2110A © 2008

Lecture 02 - 3

Why diodes? z

Basic function of diode: to allow current to flow only in one direction. Diode symbol

z

Applications: – Rectifiers – Waveform clipping and clamping circuits – DC-DC converters

Waveform clipping ELE2110A © 2008

Waveform clamping Lecture 02 - 4

Physical Structure of pn Junction Diode

Diode symbol

Metal Contact p-type

n-type

Note: both p and n regions are charge neutral.

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Lecture 02 - 5

Diffusion of Electrons and Holes z z z

z

Electrons tend to diffuse from n side to p side Holes tend to diffuse from p side to n side The resulted current is diffuse current. It is due to the movement of majority carriers. Q: What are the directions of electron and hole diffusion currents?

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Lecture 02 - 6

Diffusion of Electrons and Holes z

z

If diffusion processes were to continue unabated, there would eventually be a uniform distribution of electrons and holes. But there is a counter force…

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Lecture 02 - 7

Formation of Depletion Region

Holes and electrons are depleted from the dopant atoms

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Lecture 02 - 8

Built-in Electric Field Charge Density

Electric Field

ρ ∇⋅ E = εs (Gauss’ Law)

E(x) =

1

εs

∫ ρ(x)dx

Potential

E(x)=-∇φ(x)

built-in potential

φ j = − ∫ E ( x)dx

(1 dimension)

Solid-state physics tells us that:

kT ⎛ N A N D ⎞ ⎟⎟ φj = ln⎜⎜ 2 q ⎝ ni ⎠

The resulting potential gives rise to drift currents. ELE2110A © 2008

Lecture 02 - 9

Built-in Electric Field Charge Density

Electric Field

Potential

built-in potential

Direction of drift current: Drift current is due to the movement of minorities. For a diode with no external connections, the total current through it must be zero. Therefore, under thermal equilibrium (no external voltage), drift and diffusion currents have the same magnitude but opposite direction. ELE2110A © 2008

Lecture 02 - 10

Width of Depletion Region Charge Density

E(x) =

Electric Field

1

εs

∫ ρ(x)dx

φ j = − ∫ E ( x)dx

Depletion width is an important device parameter. It can be shown that:

Potential

built-in potential

φj =

kT ⎛ N A N D ⎞ ⎟⎟ ln⎜⎜ 2 q ⎝ ni ⎠

2εs ⎛ 1 1 ⎞ wd 0 = (x n + x p ) = + ⎜ ⎟φ j q ⎝ NA ND ⎠ ELE2110A © 2008

Lecture 02 - 11

Forward-Biased PN Junction

Forward biasing means applying an external voltage vD > 0: z Built-in potential drops (barrier lowers) z Majority electrons and holes easily cross the junction z Balance between diffusion and drift breaks z Current appears at the terminals: iD = Idiffusion – Idrift > 0 In fact, Idiffusion increases exponentially with VD. ELE2110A © 2008

Lecture 02 - 12

Reverse-Biased PN Junction

Reverse biasing means applying an external voltage vD < 0: z Built-in potential increases/barrier increases; z Majority electrons and holes can hardly cross the junction – Diffusion decreases z

Drift current almost unchanged as there are little supply of carriers (minorities). ∴ iD = Idiffusion – Idrift < 0 and tends to –Idrift.

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Lecture 02 - 13

Topics to cover… z z z z


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Lecture 02 - 14

Diode Terminal I-V Characteristics ⎡ ⎛ vD i D = I S ⎢exp ⎜⎜ ⎢⎣ ⎝ VT

⎞ ⎤ ⎟⎟ − 1⎥ ⎠ ⎥⎦

where IS = reverse saturation current [A]

vD = voltage applied to diode [V] VT = kT/q = thermal voltage [V] (25 mV at room temp.) q = electronic charge [1.60 x 10-19 C] k = Boltzmann’s constant [1.38 x 10-23 J/K] T = absolute temperature [K] ELE2110A © 2008

Lecture 02 - 15

Diode Current for Reverse, Zero, and Forward Bias z

Reverse bias:

z

⎡ ⎛ vD iD = I S ⎢exp ⎜⎜ ⎣ ⎝ VT Zero bias:

⎞ ⎤ ⎟⎟ − 1⎥ ≈ I S [0 − 1] ≈ − I S for vD < −4VT ⎠ ⎦

⎡ ⎛ vD i D = I S ⎢ exp ⎜⎜ ⎝ VT ⎣ z

⎞ ⎤ ⎟⎟ − 1⎥ = I S [1 − 1] = 0 ⎠ ⎦

Forward bias:

⎡ ⎛ vD iD = I S ⎢exp ⎜⎜ ⎝ VT ⎣ ELE2110A © 2008

⎞ ⎤ ⎛ vD ⎞ ⎟⎟ − 1⎥ ≈ I S exp ⎜⎜ ⎟⎟ for vD > 4VT ⎝ VT ⎠ ⎠ ⎦ Lecture 02 - 16

Example 1 Problem: Find diode voltage for diode with given specifications Given data: IS=0.1 fA, ID= 300 µA / 1mA Assumptions: Room-temperature dc operation with VT =25mV Analysis: With IS=0.1 fA, ID= 300 µA

I ⎞⎟ -4 A 3 × 10 D ⎟ = (0.025V)ln(1+ V = V ln 1+ ) = 0.718V D T I ⎟⎟ 16 10 A S⎠ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝

With ID= 1 mA, IS=0.1 fA

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V = 0.748V D

Lecture 02 - 17

Diode i-v Characteristics

Turn-on voltage marks point of significant current flow. Is : reverse saturation current ELE2110A © 2008

Lecture 02 - 18

Semi-log Plot of Diode Current ⎛ vD iD = I S exp ⎜⎜ ⎝ VT

⎞ ⎟⎟ ⎠

⎛ v D1 − v D 2 i D1 = exp ⎜⎜ Let iD 2 ⎝ VT

⎞ ⎟⎟ = 10 ⎠

vD1 − vD 2 = VT ln 10 = 2.3VT ≈ 60 mV

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Lecture 02 - 19

Current for Three Different Values of IS I SA = 10 I SB = 100 I SC ⎛ vD iD = I S exp ⎜⎜ ⎝ VT

⎞ ⎟⎟ ⎠

Let iDA=iDB ⎛ v DB − v DA iSA = exp ⎜⎜ iSB ⎝ VT

⎞ ⎟⎟ = 10 ⎠

vDB − vDA = VT ln 10 ≈ 60 mV

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Lecture 02 - 20



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Lecture 02 - 21

Reverse Breakdown Breakdown region: Rapid increase in ID when reverse bias voltage exceeds a break down voltage VZ Typical value of Vz: 2 V < VZ < 2000 V

Temp. Coeff.

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Lecture 02 - 22

Breakdown Mechanism 1: Avalanche Breakdown

z

z

z

z

Carriers in the SCR are accelerated by the internal electric field and collide with fixed atoms As the reverse bias increases, the energy of the accelerated carriers increases, eventually breaking covalent bonds and generating electron-hole pairs The newly created carriers also accelerate, collide with atoms, and generate other electron-hole pairs. This process feeds on itself and leads to avalanche breakdown Si diodes with VZ > 5.6 V enter breakdown through this mechanism

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Lecture 02 - 23

Breakdown Mechanism 2: Zener Breakdown z

Zener breakdown: occurs when the electric field in the depletion region increases to the point where it can break covalent bonds and generate electron-hole pairs.

z

VZ < 5.6 V for diodes entering breakdown through this mechanism

z

In avalanche breakdown, VZ increases with temperature In Zener breakdown, VZ decreases with temperature

z

z

Breakdown is not a destructive process provided that the maximum specified power dissipation is not exceeded

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Lecture 02 - 24

pn Junction Capacitance: Reverse Bias z

External reverse bias leads to: – Increases of the built-in potential – Increases of the SCR, and thus fixed space charges

z

This behaviors like a capacitor, and is referred to as depletion capacitance.

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Lecture 02 - 25

pn Junction Capacitance: Forward Bias

Forward bias minority distribution in neutral regions ELE2110A © 2008

Lecture 02 - 26

pn Junction Capacitance: Forward Bias Excess charge stored in neutral region near edges of space charge region is

QD = iDτ T Coulombs tT is called diode transit time and depends on size and type of diode. The Q depends on iD and in turn on vD. This capacitance behavior is referred to as diffusion capacitance and given by:

dQ D dQ D di D i Dτ T = ≅ [F] Cj = dv D di D dv D VT ⎡ ⎛ vD iD = I S ⎢exp ⎜⎜ ⎝ VT ⎣

⎞ ⎤ ⎛ vD ⎞ ⎟⎟ − 1⎥ ≈ I S exp ⎜⎜ ⎟⎟ ⎠ ⎦ ⎝ VT ⎠

Diffusion capacitance is proportional to current and becomes quite large at high currents. ELE2110A © 2008

Lecture 02 - 27

Diode Circuit Analysis: Basics Objective: to find the DC current (ID) and voltage (VD), or quiescent operating point (Q-point) of the diode.

Full set of equations: 1. Diode equation:

⎡ ⎛ VD I D = I S ⎢exp ⎜⎜ ⎣ ⎝ VT 2. KVL along the loop: V and R may represent Thevenin equivalent of a more complex 2-terminal network.

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V = I D R + VD

⎞ ⎤ ⎟⎟ − 1⎥ ⎠ ⎦ nonlinear

This is also called the load line equation of the diode, as if the V-source and R were the load of the diode. Lecture 02 - 28



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Lecture 02 - 29

Diode Circuit Analysis Approaches z

Graphical analysis (load-line method)

z

Numerical Analysis with diode’s mathematical model

z

Simplified analysis with ideal diode model.

z

Simplified analysis using constant voltage drop model.

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Lecture 02 - 30

Load-Line Analysis (Example 2) Problem: Find Q-point Given data: V=10 V, R=10kΩ. Diode I-V curve. Analysis: 10 = I D 10 4 + VD To define the load line we use,

VD = 0, I D = (10V / 10kΩ) = 1 mA VD = 5, I D = 0.5 mA 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)

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Lecture 02 - 31

Numerical Analysis (Example 3) Problem: Find Q-point for given diode characteristics. Given data: IS =10-13A,VT =25 mV, V = 10 V, R = 10kΩ. Analysis:

⎧I D = I S [exp(VD / VT ) − 1] = 10−13 [exp(40VD ) − 1] ⎨ 4 10 = I 10 + VD D ⎩ ⇒ 10 = 10410−13 [exp(40VD ) − 1] + VD

This is a transcendental equation and does not have analytical solutions. A numerical answer can be found by using Newton’s iterative method. Let:

f = 10 − 10 −9 [exp (40VD ) − 1] − VD

We desire to find VD for which f=0.

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Lecture 02 - 32

Newton’s Iteration Method for f (vD) = 0 3.

Make initial guess VD0 Evaluate f and its derivative f ’ for this value of VD Calculate new guess for VD using 1 0

4.

Repeat steps 2 and 3 till convergence

1. 2.

VD = VD

( ) 0

f VD − 0 f ' (VD )

( ) f ' (V ) 0

D

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0

f VD − 0 = 0 1 VD − VD

Lecture 02 - 33

Solution from a Spreadsheet Iteration #

V_D [V]

f

f'

I_D [A]

0

0.8000

-7.895E+04

-3.159E+06

7.896E+00

1

0.7750

-2.904E+04

-1.162E+06

2.905E+00

2

0.7500

-1.068E+04

-4.276E+05

1.069E+00

3

0.7250

-3.927E+03

-1.575E+05

3.936E-01

4

0.7001

-1.442E+03

-5.806E+04

1.452E-01

5

0.6753

-5.281E+02

-2.150E+04

5.374E-02

6

0.6507

-1.918E+02

-8.048E+03

2.012E-02

7

0.6269

-6.817E+01

-3.103E+03

7.754E-03

8

0.6049

-2.281E+01

-1.289E+03

3.220E-03

9

0.5872

-6.455E+00

-6.357E+02

1.587E-03

10

0.5770

-1.148E+00

-4.238E+02

1.057E-03

11

0.5743

-5.989E-02

-3.804E+02

9.486E-04

12

0.5742

-1.876E-04

-3.780E+02

9.426E-04

13

0.5742

-1.858E-09

-3.780E+02

9.426E-04

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Q-point Lecture 02 - 34

Analysis using Ideal Diode Model Ideal diode model: Forward-biased: voltage = zero, iD >0 Reverse-biased: current = zero, vD 0, our assumption is correct. Q-point is (1 mA, 0V). ELE2110A © 2008

Lecture 02 - 36

Example 5

Since source is forcing current backward through diode, assume diode is off. Hence ID =0.

10 + VD + 10 4 I D = 0 ∴VD = −10 V Since VD < 0, our assumption is right. Q-point is (0, -10 V) ELE2110A © 2008

Lecture 02 - 37

Constant Voltage Drop Model

Forward-bias: vD = Von, iD >0 Reverse-bias: iD = 0, vD < Von

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Lecture 02 - 38

Example 6

Analysis:

Since 10V source is forcing positive current through diode, we assume diode is on.

(10 − Von ) V ID = 10 kΩ (10 − 0.6) V = 0.94 mA = 10 kΩ Since ID > 0, our assumption is correct. Q-point is (0.94 mA, 0.6V). ELE2110A © 2008

Lecture 02 - 39

Two-Diode Circuit Analysis (Example 7) Analysis: Choose ideal diode model. Assume both diodes are on. Since VD1=0,

I1 =

I

D2

I

D1

(15 − 0) V = 1.50 mA 10 kΩ

=

0 − (−10) V = 2.00 mA 5 kΩ

= I − I =1.5 − 2 = −0.5 mA 1 D2

Q-points are (-0.5mA, 0V) and (2.0mA, 0V) But, ID1 0 Z D ELE2110A © 2008

Since IZ >0 (ID 0 . If IZ < 0, Zener diode no longer controls voltage across load resistor and the regulator is said to have “dropped out of regulation”. ⎛ ⎞ V ⎜1 ⎟ 1 S ∴I = −V ⎜ + ⎟ > 0 Z R Z⎜R R ⎟ L⎠ ⎝

R >⎛ R ⎞=R L ⎜V min ⎟ S −1⎟ ⎜ ⎟ ⎜V ⎠ ⎝ Z Lecture 02 - 46

Example 10 Zener diode model Analysis: KCL at output node: VL − 20 VL − 5 V + + L =0 5000 100 5000

Problem: Find VL and IZ using the piece-wise linear model. Given data: VS=20 V, R=5 kΩ, RZ= 0.1 kΩ, VZ=5 V

∴ V = 5.19V L

V − 5 5.19 − 5 I = L = =1.9mA > 0 Z 100 100 Diode is actually operating in breakdown region as assumed.

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Lecture 02 - 47

Line and Load Regulation

Load is modeled as an I-source

By superposition:

R Rz VL = VZ + V S − ( R z // R ) I L R + Rz R + Rz

Rz ∂V L Line regulation ≡ = ∂V S R + Rz

(characterizes how sensitive output voltage is to input voltage changes)

∂V L Load regulation ≡ = − ( R z // R ) [ Ω ] (characterizes how sensitive output ∂I L voltage is to load current changes)

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Lecture 02 - 48