Mechatronics Chapter 3-1 Semiconductor devices Diode

MEMS1082 Mechatronics Chapter 3-1 Semiconductor devices Diode

Semiconductor: Si

Semiconductor

N-type and P-type Semiconductors There are two types of impurities: N-type - In N-type doping, phosphorus or arsenic is added to the silicon in small quantities. Phosphorus and arsenic each have five outer electrons, so they're out of place when they get into the silicon lattice. The fifth electron has nothing to bond to, so it's free to move around. It takes only a very small quantity of the impurity to create enough free electrons to allow an electric current to flow through the silicon. N-type silicon is a good conductor. Electrons have a negative charge, hence the name N-type. P-type - In P-type doping, boron or gallium is the dopant. Boron and gallium each have only three outer electrons. When mixed into the silicon lattice, they form "holes" in the lattice where a silicon electron has nothing to bond to. The absence of an electron creates the effect of a positive charge, hence the name P-type. Holes can conduct current. A hole happily accepts an electron from a neighbor, moving the hole over a space. P-type silicon is a good conductor.

N-type and P-type Semiconductors

Semiconductor device-diode A diode is the simplest possible semiconductor device. A diode allows current to flow in one direction but not the other. You may have seen turnstiles at a stadium or a subway station that let people go through in only one direction. A diode is a oneway turnstile for electrons. When you put N-type and P-type silicon together as shown in this diagram, you get a very interesting phenomenon that gives a diode its unique properties.

Diodes

Diode Electron flow direction Current direction

Diode depletion region

pn junction PN Junction

Diode depletion region

Diode forward and reverse bias

Shockley diode equation

Diode current and voltage

Diode Characteristic

Diode Characteristic at different scale

Diode measurement Meter with a Diode check function displays the forward voltage drop of 0.548 volts instead of a low resistance

Measurement of a diode Measuring forward voltage of a diode without diode check meter function: (a) Schematic diagram. (b) Pictorial diagram

Load line of diode A circuit with a diode

Example For circuit, determine the current i

Example Circuit reduction to Thévenin equivalent circuit

Example Thévenin equivalent circuit

Example Draw load line to determine the diode voltage and current

Example Determine current i

Example Determine the current and voltage of the diode in the circuit. The diode characteristic is given in the right figure.

Example

Piecewise-linear approximation and small signal analysis Diode is nonlinear resistor

Piecewise-linear approximation and small signal analysis Diode piecewise-linear approximation

Piecewise-linear approximation and small signal analysis

Piecewise-linear approximation and small signal analysis Small signal analysis

Piecewise-linear approximation and small signal analysis If we are only interested in the portion due to v s (t), we may set E s =0, and E f =0, then Often, for practical purpose, we can assume E f =0 in small signal equivalent circuit of a diode. For typical diodes, the value of R f is quite small, between 1Ω and 100Ω. Thus R f can be neglected.

Piecewise-linear approximation and small signal analysis

The ideal diodes

The piecewise- linear model of a diode, using an ideal diode Ideal diode

Example Nonlinear resistors with a wide range of characteristics can be obtained, approximately, with circuit containing diodes, for example, a square-law device is two-terminal nonlinear resistor whose terminal voltage-current characteristic obey 2 i = kv where k is normalization constant. The ideal characteristic is shown

Example This device may be used in modulator, e.g., to attain a voice signal to high-frequency carrier wave, as is done in amplitude modulation (AM) radio transmission. Design a square-law device to approximate the ideal characteristics for 0 v 5V with a normalization constant k=0.001

Example A circuit using ideal diodes D 1 and D 2 and voltage sources E 1 and E 2 Use V=5V; E 1 < E 2 Initially 0 v E 1,the diodes are reverse biased and open, the curve will have slope 1/R 3 For E 1 v E 2,D 1 closes, and D 2 open, the input resistance will be R 3 llr 1 For E 2 v 5V,D 1 and D 2 close, the input resistance will be R 3 llr 1 llr 2 Suppose E 1 =2.0V and E 2 =3.5V 2 I = ke 4mA 1 1 = 2 I = ke 12. 25mA 2 2 = I = kv 2 = 25mA

Example R R Noting the slope of each portion, we obtain 3 1 E1 = = 500Ω I R 2 1 R 3 R V E2 = = 118Ω I I 2 1 R 2 E2 E1 = = 182Ω I I R 2 2 1 = 333Ω R 1 = 286Ω Replacing the actual diode with their piecewise-linear approximation using R f = 10 Ω, E f = 0. 5V R1 = 276Ω R 2 = 323Ω E 1 =1.5V and E 2 =3.0V R 3 = 500Ω

Ideal transformer

Rectifiers Half-Wave Rectifier The transformer isolates the load from the source

Rectifiers Half-Wave Rectifier π ω π π ω ω 2 0 0 sin = = t v t t V v L s L ( ) π ω ω π π s s L V t d t V V = = 0 sin 2 1 The average dc value of v L

Rectifiers Representing the Half-Wave Rectifier voltage by Fourier series v L = VL + a1 sinωt + a2 sin 2ωt +... + b1 cosωt + b2 cos 2ωt +... The Fourier coefficients can be determined as a n 2 = T T T ( t) sin nωt dt; b v ( t) cos nωt dt vl n = 2 T 0 0 For the Half-Wave Rectified voltage 2 T 1 π a 1 = vl ( t) sin t dt = Vs sin t sin t d( t) = T ω 0 ω ω ω π 0 a n 2 T 1 = vl T 0 π L Vs 2 π ( t) sin nωt dt = V sinωt sin nωt d( ωt) = 0 0 s

Rectifiers b 2V s 2V s = 0; b2 =, b3 = 0; b4 = ; b5 3π 15π 1 = 0 Thus the Fourier series for the Half-Wave Rectified signal v L ( t) Vs Vs 2V s 2V s = + sinωt cos 2ωt cos 4ωt +... π 2 3π 15π

Rectifiers Filtering the Half-Wave Rectifier Capacitor has lower impedance to higher frequencies

Rectifiers Filtering the Half-Wave Rectifier Larger C can be used to increase the time constant RC

Rectifiers Effects of actual diodes

The Full-Wave Rectifiers The full-wave rectifier

The Full-Wave Rectifiers The full-wave rectifier The average dc value of v L V L 1 = π 2V s = π π 0 V s sinωt d ( ωt) Thus the Fourier series for the Full-Wave Rectified signal v L 2V π 4V 3π 4V 15π s s s ( t) = cos 2ωt cos 4ωt +...

The Full-Wave Rectifiers Effect of actual diodes

The Full-Wave Bridge Rectifier A bridge rectifier makes use of four diodes in a bridge arrangement to achieve full-wave rectification. This is a widely used configuration, both with individual diodes wired as shown and with single component bridges where the diode bridge is wired internally.

Bridge Rectifiers Various types of Bridge Rectifiers Note that some have a hole through their centre for attaching to a heat sink

The Full-Wave Bridge Rectifier Bridge Rectifier

The Full-Wave Bridge Rectifier Bridge Rectifier with RC Filter and LC filter

The Voltage Limiter Limiter using ideal diodes and batteries

The Voltage Limiter Limiter using ideal diode and batteries

The Voltage Limiter Limiter using ideal diode and batteries Load voltage is limited for source voltage R L + Rs R + R V2 < vs V RL RL L s ( t) < 1

The Voltage Limiter Limiter using ideal diode and batteries

Example For a limiter shown below, assume identical piecewiselinear diodes with R f =100Ω, E f =0.5V, V 1 =V 2 =10V, R L =100Ω, R s =100Ω, and v s (t)=50sinωt V, sketch v L (t)

Zener Diodes A Zener diode is a type of diode that permits current not only in the forward direction like a normal diode, but also in the reverse direction if the voltage is larger than the breakdown voltage known as "Zener knee voltage" or "Zener voltage". The device was named after Clarence Zener, who discovered this electrical property.

Zener Diodes Device characteristic of Zener diode Piecewise-linear characteristic

Zener Diodes Piecewise-linear model

Zener Diode Regulator In this circuit, a typical voltage reference or regulator, an input voltage, U IN, is regulated down to a stable output voltage U OUT. The intrinsic voltage drop of diode D is stable over a wide current range and holds U OUT relatively constant even though the input voltage may fluctuate over a fairly wide range. Because of the low impedance of the diode when operated like this, Resistor R is used to limit current through the circuit. I Diode = (U IN - U OUT ) / R

Zener Diode Regulator R must be small enough that the current through D keeps D in reverse breakdown. The value of this current is given in the data sheet for D. For example, the common BZX79C5V6 device, a 5.6 V 0.5 W Zener diode, has a recommended reverse current of 5 ma. If insufficient current exists through D, then U OUT will be unregulated, and less than the nominal breakdown voltage. When calculating R, allowance must be made for any current through the external load, not shown in this diagram, connected across U OUT. R must be large enough that the current through D does not destroy the device. If the current through D is I D, its breakdown voltage V B and its maximum power dissipation P MAX, then I D V B < P MAX.

Zener Diode regulator V V P s,max z max I max = = + Rs + Rmin Vz V R z L V V s,min z I min = = Rs + Rmax V R z L

Example A source voltage varies between 120V and 75V. The source resistance is zero, and the load resistance is 1kΩ. It is desired to maintain the load voltage at 60V. Determine the value of a regulator resistor R that will accomplish this and the required power rating of the zener. 1. A zener having a zener voltage of 60V is selected 2. The maximum value of regulator resistance I Vz 60 = 60mA R 1000 min = = L I P V V V V R R max 3. The power rating is determined when V s =V s,max. And zener draw the maximum current max s,max z z max = = = z R L Pmax =10. 8W Vs,min Vz = = 250Ω I 0.18A min

Light Emitting Diode

Light Emitting Diode An LED will begin to emit light when the on-voltage is exceeded. Typical on voltages are 2 3 volts

Connect Light Emitting Diode in Series Connecting LEDs in series If you wish to have several LEDs on at the same time it may be possible to connect them in series. This prolongs battery life by lighting several LEDs with the same current as just one LED. All the LEDs connected in series pass the same current so it is best if they are all the same type. The power supply must have sufficient voltage to provide about 2V for each LED (4V for blue and white) plus at least another 2V for the resistor. To work out a value for the resistor you must add up all the LED voltages and use this for VL. Example calculations: A red, a yellow and a green LED in series need a supply voltage of at least 3 2V + 2V = 8V, so a 9V battery would be ideal. VL = 2V + 2V + 2V = 6V (the three LED voltages added up). If the supply voltage VS is 9V and the current I must be 15mA = 0.015A, Resistor R = (VS - VL) / I = (9-6) / 0.015 = 3 / 0.015 = 200, so choose R = 220 (the nearest standard value which is greater).