Diodes - Electronics I - Lecture Slides, Slides of Electronics

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Kristin Ackerson, Virginia Tech EE

Table of Contents

• What are diodes made out of?____________________slide Kristin Ackerson, Virginia Tech EE
• N-type material_________________________________slide
• P-type material_________________________________slide
• The pn junction_________________________________slides 6-
• The biased pn junction___________________________slides 8-
• Properties of diodes_____________________________slides 10-
• Diode Circuit Models ____________________________slides 12-
• The Q Point____________________________________slides 17-
• Dynamic Resistance_____________________________slides 19-
• Types of diodes and their uses ___________________ slides 21-
• Sources_______________________________________slide

N-Type Material

Kristin Ackerson, Virginia Tech EE

N-Type Material: When extra valence electrons are introduced

into a material such as silicon an n-type material is produced. The extra valence electrons are introduced by putting impurities or dopants into the silicon. The dopants used to create an n-type material are Group V elements. The most commonly used dopants from Group V are arsenic, antimony and phosphorus. The 2D diagram to the left shows the extra electron that will be present when a Group V dopant is introduced to a material such as silicon. This extra electron is very mobile. +4 +

+4 +4 + +4 +

P-Type Material

Kristin Ackerson, Virginia Tech EE

P-Type Material: P-type material is produced when the dopant

that is introduced is from Group III. Group III elements have only 3 valence electrons and therefore there is an electron missing. This creates a hole (h+), or a positive charge that can move around in the material. Commonly used Group III dopants are aluminum, boron, and gallium. The 2D diagram to the left shows the hole that will be present when a Group III dopant is introduced to a material such as silicon. This hole is quite mobile in the same way the extra electron is mobile in a n-type material. +4 +

+4 +4 + +4 +

The PN Junction

Steady State Kristin Ackerson, Virginia Tech EE P (^) n

---

---

---

---

+ + + + + + + + + + + + + + + + + + + + 

Na Nd Metallurgical Junction Space Charge ionized Region acceptors ionized donors E-Field

_ _ h+ drift == h+ diffusion e- diffusion == e-^ drift 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 .

The Biased PN Junction

Kristin Ackerson, Virginia Tech EE

P n

_

Applied Electric Field Metal Contact “Ohmic Contact” (Rs~0)

_

V

applied

I

The pn junction is considered biased when an external voltage is applied. There are two types of biasing: Forward bias and Reverse bias. These are described on then next slide.

Properties of Diodes

Kristin Ackerson, Virginia Tech EE Figure 1.10 – The Diode Transconductance Curve 2

• VD = Bias Voltage
• ID = Current through Diode. ID is Negative for Reverse Bias and Positive for Forward Bias
• IS = Saturation Current
• VBR = Breakdown Voltage
• V = Barrier Potential Voltage

V

D

I

D (mA) (nA)

V

BR

~V



I

S

Properties of Diodes

The Shockley Equation

Kristin Ackerson, Virginia Tech EE

• The transconductance curve on the previous slide is characterized by the following equation:

I

D

= I

S

(e

V

D

/V

T – 1)

• As described in the last slide, ID is the current through the diode, IS is the saturation current and VD is the applied biasing voltage.
• VT is the thermal equivalent voltage and is approximately 26 mV at room temperature. The equation to find VT at various temperatures is: VT = kT q k = 1.38 x 10-^23 J/K T = temperature in Kelvin q = 1.6 x 10-^19 C
•  is the emission coefficient for the diode. It is determined by the way the diode is constructed. It somewhat varies with diode current. For a silicon diode  is around 2 for low currents and goes down to about 1 at higher currents

Diode Circuit Models

Kristin Ackerson, Virginia Tech EE

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. Example: Assume the diode in the circuit below is ideal. Determine the value of ID if a) VA = 5 volts (forward bias) and b) VA = - 5 volts (reverse bias)

_

VA

ID

RS = 50 

a) With VA > 0 the diode is in forward bias and is acting like a perfect conductor so: ID = VA/RS = 5 V / 50  = 100 mA b) With VA < 0 the diode is in reverse bias and is acting like a perfect insulator, therefore no current can flow and ID = 0.

Diode Circuit Models

Kristin Ackerson, Virginia Tech EE

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 ID if VA = 5 volts (forward bias).

_

VA

ID

RS = 50 

With VA > 0 the diode is in forward bias and is acting like a perfect conductor so write a KVL equation to find ID: 0 = VA – IDRS - V ID = VA - V = 4.7 V = 94 mA RS 50  V

V

Diode Circuit Models

The Ideal Diode

with Barrier

Potential and

Linear Forward

Resistance

Kristin Ackerson, Virginia Tech EE 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 ID if VA = 5 volts.

_

VA

ID

RS = 50 

V

RF Once again, write a KVL equation for the circuit: 0 = VA – IDRS - V - IDRF ID = VA - V = 5 – 0.3 = 85.5 mA RS + RF 50 + 5

Diode Circuit Models

Kristin Ackerson, Virginia Tech EE Values of ID for the Three Different Diode Circuit Models Ideal Diode Model Ideal Diode Model with Barrier Potential Voltage Ideal Diode Model with Barrier Potential and Linear Forward Resistance

ID 100 mA^ 94 mA^ 85.5 mA

These are the values found in the examples on previous slides where the applied voltage was 5 volts, the barrier potential was 0.3 volts and the linear forward resistance value was assumed to be 5 ohms.

The Q Point

I

D (mA)

V

D (Volts) 2 4 6 8 10 12 0.2 0.4 0.6 0.8 1.0 1.2 1. The transconductance curve below is for a Silicon diode. The Q point in this example is located at 0.7 V and 5.3 mA.

Kristin Ackerson, Virginia Tech EE

Q Point: The intersection of the load line and the transconductance curve.

Capacitance and Voltage of PN

Junctions

Diode Operation – Animation Webpage Link