Course Code: Course Name: Electronic Devices

Course Code: 802311 Course Name: Electronic Devices
By Prof. Iqbal Ahmad Khan Department of Electrical Engineering Faculty of Engineering & Islamic Architecture Umm Al Qura University, Makka Al Mukarrama Kingdom of Saudi Arabia Prof. Iqbal A. Khan, EED, UQU

Prof. Iqbal A. Khan, EED, UQU
Course Number: 802311 Units: (Lec., Lab., Tot.): (3, 1 , 4) Course Name: ELECTRONIC DEVICES Prerequisite: & Contact Hours: 6 Course Topics: Semiconductor Theory PN Junction Diode Other Diodes and Devices Transistors DC and AC Analysis of Transistors Field-Effect Transistors Prof. Iqbal A. Khan, EED, UQU

Text Book: Thomas l. Floyd, “ELECTRONIC DEVICES”, Nineth Edition, Pearson Education International, 2012. References : R. Boylestad and L. Nashelsky, “Electronic Devices and Circuit Theory”, 10th Edition, Pearson Education International, 2010. A. S. Sedra and K. C. Smith, “Microelectronics Circuits”, Oxford University Press, 5th Edition, 2008. Prof. Iqbal A. Khan, EED, UQU

Classification of Materials
Electrically Materials can be classified into three categories: 1. Insulators, 2. Conductors, 3. Semiconductors. 1. Insulators: The materials in which all electrons are tightly bounded to atoms are Insulators. Examples: Glass, Ceramics, Plastic, Rubber. 2. Conductors: The materials in which the outermost atomic electrons are free to move around are Conductors. Conductors typically have ~1 “free electron” per atom. Examples: Gold, Silver, Copper, Aluminum. 3. Semiconductors. The materials in which electrons are not tightly bound and can be easily “promoted” to a free state are Semiconductors. Examples: Silicon, Germanium, Gallium Arsenide. Prof. Iqbal A. Khan, EED, UQU

Conductors, Semiconductors and Insulators
Gold Silicon Glass Silver Germanium Plastic Copper Gallium Arsenide Ceramics Aluminum Rubber Prof. Iqbal A. Khan, EED, UQU

Semi-conductors: The most commonly used semiconductor material is Silicon. It has four valence electrons in its outer most shell which it shares with its adjacent atoms in forming covalent bonds. The structure of the bond between two silicon atoms is such that each atom shares one electron with its neighbour making the bond very stable. The diagram above shows the structure and lattice of a 'normal' pure crystal of Silicon Prof. Iqbal A. Khan, EED, UQU

Donors and Acceptors in the Periodic Table: I II III IV V VI VII ZERO H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Zn Ga Ge As Se Br Kr Rb Cd In Sn Sb Te Xe Acceptors Impurity Donors Prof. Iqbal A. Khan, EED, UQU

N-type Semiconductor In order for our silicon crystal to conduct electricity, we need to introduce an impurity atom such as Arsenic, Antimony or Phosphorus into the Si crystalline structure. These atoms have five outer electrons in their outermost co-valent bond to share with other atoms and are commonly called "Pentavalent" impurities. The resulting semiconductor material has an excess of current-carrying electrons, each with a negative charge, and is therefore referred to as "N-type" material with the electrons called "Majority Carriers" The diagram above shows the structure and lattice of the donor impurity atom Antimony. Prof. Iqbal A. Khan, EED, UQU

P-Type Semiconductor If a "Trivalent" (3-electron) impurity is introduced into the Si crystal structure, such as Aluminum, Boron, Gallium or Indium, only three valence electrons are available in the outermost covalent bond meaning that the fourth bond cannot be formed. The vacancy of an electron in the bond is known as a hole. Therefore, a complete connection is not possible, giving the semiconductor material an abundance of positively charged carriers known as "holes" in the structure of the Si crystal. Addition of Boron causes conduction to consist mainly of positive charge carriers results in a "P-type" material and the positive holes are called "Majority Carriers" while the free electrons are called "Minority Carriers". The diagram above shows the structure and lattice of the acceptor impurity atom Boron. Prof. Iqbal A. Khan, EED, UQU

Electron and Hole Mobility In solid-state physics, the electron mobility characterizes how quickly an electron can move through a metal or semiconductor, when pulled by an electric field. In semiconductors, there is an analogous quantity for holes, called hole mobility. The term carrier mobility refers in general to both electron and hole mobility in semiconductors. Electron and hole mobility are special cases of electrical mobility of charged particles in a fluid under an applied electric field. When an electric field E is applied across a piece of material, the electrons respond by moving with an average velocity called the drift velocity ( vd ). Then the electron mobility μ is defined as Electron mobility is almost always specified in units of cm2/(V·s). This is different from the SI unit of mobility, m2/(V·s). They are related by 1m2/(V·s) = 104cm2/(V·s). The hole mobility is smaller than that of the electron. μN = 580 Cm2/ V.Sec and μP = 230 Cm2 / V.Sec Prof. Iqbal A. Khan, EED, UQU

The PN-junction When the P-type and N-type materials are joined (or fuse) together then a P-N Junction is formed. the resulting device that has been made is called a PN-junction Diode or Rectifier Diode. Symbol of the PN-Junction Diode Prof. Iqbal A. Khan, EED, UQU

The Basic Diode Symbol and Static I-V Characteristics. ID = IS(eVD /VT – 1) Prof. Iqbal A. Khan, EED, UQU

Diode Charateristic Equation: The diode V-I relationship is characterized by the following equation: ID = IS(eVD/VT – 1) VD = Bias Voltage ID = Current through Diode. ID is Negative for Reverse Bias and Positive for Forward Bias IS = Reverse Saturation Current  is the emission coefficient for the diode. For a silicon diode  is around 2 for low currents and goes down to about 1 at higher currents VT is the thermal equivalent voltage and is approximately 26 mV at room temperature. The equation to find VT at various temperatures is: k = 1.38 x J/K, T = temperature in Kelvin, q = 1.6 x C Prof. Iqbal A. Khan, EED, UQU

Forward Biased Diode With forward biased the depletion layer is reduced and after the threshold level of voltage the majority charge carriers cross the depletion layer and the diode conducts or ON. The diode in forward mode after the threshold has low resistance. Forward Characteristics Curve for a Diode. Prof. Iqbal A. Khan, EED, UQU

A Reverse Biased Diode With the reversed biased junction the depletion layer is increased and the Majority charge carriers can not cross the depletion region. This condition represents the high resistance and the diode is said to be OFF. Reverse Characteristics Curve for a Diode. Prof. Iqbal A. Khan, EED, UQU

Forward and Reversed Biased Diode Then we can say that an ideal small signal diode conducts current in one direction (forward-conducting) and blocks current in the other direction (reverse-blocking). Signal Diodes are used in a wide variety of applications such as a switch in rectifiers, limiters, snubbers or in wave-shaping circuits. Anode is positive with respect to Cathode Cathode is positive with respect to Anode Prof. Iqbal A. Khan, EED, UQU

Half-wave Rectifier Circuit During each "positive" half cycle of the AC sinewave, the diode is Forward Biased and current flows through it. The voltage across the load is then Vout = Vs. During each "negative" half cycle of the AC sinewave, the diode is Reverse Biased and No current flows through it. Therefore, in the negative half cycle of the supply, The output voltage Vout = 0. =VmSinθ 2π π 2π π The current on the DC side of the circuit flows in one direction only making the circuit Unidirectional and the value of the DC voltage VDC (i,.e., the average value) across the load resistor is calculated as follows. Prof. Iqbal A. Khan, EED, UQU

Half-wave Rectifier with Smoothing Capacitor When rectification is used to provide a direct voltage power supply from an alternating source, the amount of ripple can be reduced by using larger value capacitors as shown in the Figure. After the peak voltage the capacitor discharges through the load resistor at the slower rate and thus increases the average value or the DC value. Half-wave Rectifier with Smoothing Capacitor Prof. Iqbal A. Khan, EED, UQU

Example-1: Calculate the current (IDC) flowing through a 100Ω resistor connected to a 240v single phase half-wave rectifier as shown above, and also the power consumed by the load. Solution: Prof. Iqbal A. Khan, EED, UQU

Full-wave Rectifier Circuit-Full Wave Rectifier In a full-wave rectifier circuit two diodes are now used, together with a transformer whose secondary winding is split equally into two and has a common center tapped connection, (C). Now each diode conducts in turn when its Anode terminal is positive with respect to the center point C as shown in Figure. The output voltage(Vdc) can be analysed as follows. +ve Half Cycle -ve Half Cycle π π 2π Prof. Iqbal A. Khan, EED, UQU

The Diode Bridge Rectifier Another type of circuit that produces full-wave rectification is that of the Bridge Rectifier. This type of single phase rectifier uses 4 individual diodes connected in a "bridged" configuration to produce the desired output but does not require a special center tapped transformer, thereby reducing its size and cost. The single secondary winding is connected to one side of the diode bridge network and the load to the other side as shown below. The Negative Half-cycle The Positive Half-cycle Prof. Iqbal A. Khan, EED, UQU

Full-wave Rectifier with Smoothing Capacitor The full-wave bridge rectifier however, gives us a greater mean DC value (0.636Vmax) with less superimposed ripple while the output wveform is twice that of the frequency of the input supply frequency. We can therefore increase its average DC output level even higher by connecting a suitable smoothing capacitor across the output of the bridge circuit. Prof. Iqbal A. Khan, EED, UQU

Zener Diode I-V Charcateristics Zener Diodes are used in the "REVERSE" bias mode. We can see that the zener diode has a region in its reverse bias characteristics of almost a constant voltage regardless of the current flowing through the diode. This voltage across the diode (it's Zener Voltage, Vz) remains nearly constant even with large changes in current through the diode caused by variations in the supply voltage or load. This ability to control itself can be used to great effect to regulate or stabilise a voltage source against supply or load variations. The diode will continue to regulate until the diode current falls below the minimum Iz value in the reverse breakdown region. Prof. Iqbal A. Khan, EED, UQU

The Zener Regulator Zener Diodes can be used to produce a stabilized voltage output by passing a small current through it from a voltage source via a suitable current limiting resistor, (RS). The DC output voltage from the half or full-wave rectifiers contains ripple. By connecting a simple zener stabilizer circuit as shown below across the output of the rectifier a more stable dc output voltage can be produced. =(VS - VZ ) / IZ Zener Diode Stabiliser Prof. Iqbal A. Khan, EED, UQU

Example -1. A 5.0v stabilized power supply is required from a 12V d.c. input source. The maximum power rating of the Zener diode is 2W. Using the circuit above calculate: a) The maximum current flowing in the Zener Diode. b) The value of the series resistor, RS c) The load current IL if a load resistor of 1kΩ is connected across the Zener diode. d) The total supply current IS Prof. Iqbal A. Khan, EED, UQU

Zener Diodes with different voltages and power ratings BZX55 Zener Diode Power Rating 500mW 2.4V 2.7V 3.0V 3.3V 3.6V 3.9V 4.3V 4.7V 5.1V 5.6V 6.2V 6.8V 7.5V 8.2V 9.1V 10V 11V 12V 13V 15V 16V 18V 20V 22V 24V 27V 30V 33V 36V 39V 43V 47V BZX85 Zener Diode Power Rating 1.3W 5.6 51V 56V 62V Prof. Iqbal A. Khan, EED, UQU

rms and average relationship: Form Factor = Vrms = 1.11 Vav Prof. Iqbal A. Khan, EED, UQU

Diode Clippers The diode in a series clipper “clips” any voltage that does not forward bias it: A reverse-biasing polarity A forward-biasing polarity less than 0.7 V (for a silicon diode) Prof. Iqbal A. Khan, EED, UQU

Parallel Clippers The diode in a parallel clipper circuit “clips” any voltage that forward bias it. DC biasing can be added in series with the diode to change the clipping level. Prof. Iqbal A. Khan, EED, UQU

Summary of Clipper Circuits
Prof. Iqbal A. Khan, EED, UQU

Clampers A diode and capacitor can be combined to “clamp” an AC signal to a specific DC level. Prof. Iqbal A. Khan, EED, UQU

Biased Clamper Circuits The input signal can be any type of waveform such as sine, square, and triangle waves. The DC source lets you adjust the DC clamping level. Prof. Iqbal A. Khan, EED, UQU

Summary of Clamper Circuits
Prof. Iqbal A. Khan, EED, UQU 33

Voltage Doubler: Positive Half-Cycle D1 conducts D2 is switched off Capacitor C1 charges to Vm Negative Half-Cycle D1 is switched off D2 conducts Capacitor C2 charges to 2Vm Vout = VC2 = 2Vm Prof. Iqbal A. Khan, EED, UQU

Voltage Tripler and Quadrupler
Prof. Iqbal A. Khan, EED, UQU

Light-Emitting Diodes: Light-emitting diodes are designed with a very large bandgap so movement of carriers across their depletion region emits photons of light energy. Lower bandgap LEDs (Light-Emitting Diodes) emit infrared radiation, while LEDs with higher bandgap energy emit visible light. Many stop lights are now starting to use LEDs because they are extremely bright and last longer than regular bulbs for a relatively low cost. The arrows in the LED representation indicate emitted light. A K Schematic Symbol for a Light-Emitting Diode Prof. Iqbal A. Khan, EED, UQU

Photodiodes: Photodiodes are sensitive to received light. They are constructed so their PN junction can be exposed to the outside through a clear window or lens. In Photoconductive mode the saturation current increases in proportion to the intensity of the received light. This type of diode is used in CD players. In Photovoltaic mode, when the PN junction is exposed to a certain wavelength of light, the diode generates voltage and can be used as an energy source. This type of diode is used in the production of solar power. Schematic Symbols for Photodiodes A K  Prof. Iqbal A. Khan, EED, UQU

Recombination produces light!!
LIGHT EMITTING DIODE-LED: LED are semiconductor p-n junctions that under forward bias conditions can emit radiation by electroluminescence in the UV, visible or infrared regions of the electromagnetic spectrum. The qaunta of light energy released is approximately proportional to the band gap of the semiconductor. P-n junction Electrical Contacts Schematic Symbols for LED A K Recombination produces light!! Junction is biased to produce even more e-h and to inject electrons from n to p for recombination to happen Prof. Iqbal A. Khan, EED, UQU

The BJT – Bipolar Junction Transistor
The Two Types of BJT Transistors: npn pnp n p n p n p E C E C Cross Section Cross Section C B E B B C B E Schematic Symbol Schematic Symbol Collector doping is usually ~ 106 Base doping is slightly higher ~ 107 – 108 Emitter doping is much higher ~ 1015 Prof. Iqbal A. Khan, EED, UQU

BJT Structure In this process, all steps are performed from the surface of the wafer Prof. Iqbal A. Khan, EED, UQU

BJT Relationships - Equations
IB - + C E IE IC B VBE VBC VCE IB IE IC - + VEB VCB B C E VEC npn IE = IB + IC VCE = -VBC + VBE pnp IE = IB + IC VEC = VEB - VCB Note: The equations seen above are for the transistor, not the circuit. Prof. Iqbal A. Khan, EED, UQU

Transistor Configurations: Input = VBE & IB Output = VCE & IC Input = VEB & IE Output = VCB & IC Input = VBC & IB Output = VEC & IE Prof. Iqbal A. Khan, EED, UQU

DC  and DC   = Common-base current gain  = Common-emitter current gain The relationships between the two parameters are: Note:  and  are sometimes referred to as dc and dc because the relationships being dealt with in the BJT are DC. Prof. Iqbal A. Khan, EED, UQU

Example1: For a Common-Base NPN Circuit Configuration, Given: IB = 50  A , IC = 1 mA, Find: IE ,  , and . Solution: IE = IB + IC = 0.05 mA + 1 mA = 1.05 mA = IC / IB = 1 mA / 0.05 mA = 20  = IC / IE = 1 mA / 1.05 mA =  could also be calculated using the value of  with the formula from the previous slide. Prof. Iqbal A. Khan, EED, UQU

Cutoff Region IB = 0, IC = 0 (Transistor is OFF)
Output Characteristics of Common Emitter Configuration There are three regions of operation: Active Region: The region where current curves are practically flat. In this region the JEB is forward bias and the JCB is reversed bias. In this region transistor acts as an amplifier. IB2 VCE IC Active Region Saturation Region (Transistor is ON) VSAT = 0.2V Cutoff Region IB = 0, IC = 0 (Transistor is OFF) IB4 IB3 IB1 2. Saturation Region: In this region both the junctions JEB and JCB are forward bias, as a result the barrier potential of the junctions cancel each other out causing a virtual short between collector and emitter terminals i. e. the transistor is ON. Output Characteristics of CE-Configuration 3. Cutoff Region: In this region both the junctions JEB and JCB are reverse bias, and thus the currents reduced to zero. In this region transistor behaves like an open switch, i. e. the transistor is OFF. Prof. Iqbal A. Khan, EED, UQU

Output Characteristics of Common-Base Configuration: Although the Common-Base configuration is not the most common biasing type, it is often helpful in the understanding of how the BJT works. IE Active Region Saturation Region IC VCB Cutoff IE = 0 Prof. Iqbal A. Khan, EED, UQU

Output Characteristics of Common Collector Configuration: The Common-Collector configuration circuit is basically equivalent to the common-emitter biased circuit except instead of looking at IC as a function of VCE and IB we are looking at IE. Also, since  ~ 1, and  = IC/IE that means IC~IE VCE IE Active Region IB Saturation Region Cutoff Region IB = 0 Active: Small base current controls a large collector current Saturation: VCE(sat) ~ 0.2V, VCE increases with IC Cutoff: Achieved by reducing IB to 0, Ideally, IC will also equal 0. Prof. Iqbal A. Khan, EED, UQU

Common-Base Configuration:
_ + IC IE IB VCB VBE E C B VCE Circuit Diagram: NPN Transistor The Table Below lists assumptions that can be made for the attributes of the common-base configuration circuit in the different regions of operation. Given for a Silicon NPN transistor. Region of Operation IC VCE VBE E-B Bias C-B Bias Active IB =VBE+VCB ~0.7V Fwd. Rev. Saturation Max ~0V Cutoff ~0 Prof. Iqbal A. Khan, EED, UQU

Self Bias Circuit for Active Region: + VBE - IB IC IE Prof. Iqbal A. Khan, EED, UQU

Example: In the self bias circuit the R1 = 30K, R2 = 10K, RC = 4.3K, RE = 1.3K, VCC =12V and β = 100. Find IB , IC , IE and VB , VC , VE , VCE. Solution: The self bias circuit and its Thevenin’s equivalent is given as follows. RC VC VE VB Prof. Iqbal A. Khan, EED, UQU

VBB = IERE + VBE + IBRBB and IE = (1+β)IB Therefore, 3V = (1+100)IB x1.3 x V + IB x 7.5 x 1000 (101x )x1000xIB = 3 – 0.7 138.8x1000xIB = 2.3 IB = 2.3 / (138.8x1000) A IB = 2.3 / (138.8) mA IB = 2.3x1000 / (138.8) µA IB = µA Hence, IC = βIB =100x16.57 µA =1.657 mA IE = (1+β)IB = (1+100)x16.5 µA = mA Prof. Iqbal A. Khan, EED, UQU

VB = VBB – IB RB = 3V – 16.57x10-6x7.5x10+3 = 2.87V VB = 2.87V VE = IERE = mA)(1.3K) = 2.17 V VE = 2.17V VBE =VB – VE = 2.87V – 2.17V = 0.7V VC = 12 – ICRC = 12 - (1.657 mA)x(4.3K) = 4.87 V VC = 4.87V Thus VC > VB ; The Collector Junction JC is reversed bias. VB > VE ; The Emitter Junction JE is forward bias. Therefore the transistor is in Active Region of operation. VCE = VC – VE = = 2.7V Prof. Iqbal A. Khan, EED, UQU

Variation of IC and VCE as a result of variation in IB : DC Load Line VCC VCC /RC DC Load line equation Prof. Iqbal A. Khan, EED, UQU

RC Coupled Amplifier Circuit: RC coupled amplifier is an inverting voltage amplifier. CC is coupling capacitor used to block the DC CE is the bypass capacitor used to bypass the AC signals through it to avoid the gain loss due to RE . For AC signal operation the capacitors CC and CE are replaced by effective short circuit because their values are selected so that XC (=1/ωC) ≈ 0 at the signal frequency. AC Ground-The DC source is replaced by a ground. Assuming that the DC source has the zero internal resistance so that no AC voltage is developed across the DC source. Therefore, the VCC terminal is at zero volt AC and is called AC ground. Prof. Iqbal A. Khan, EED, UQU

Simple r-parameter transistor model: IC = αIe = βIb Vb = Iere’ = Icre’ Where, Prof. Iqbal A. Khan, EED, UQU

RC Coupled Amplifier- AC Equivalent: Vb Vc Voltage Gain Av with CE: Voltage Gain Av without CE: Prof. Iqbal A. Khan, EED, UQU

Rin(Total) = R1 ǁ R2 ǁ Rin(base) Overall Voltage Gain Overall Power Gain (Ap) =Voltage gain x Current gain = Av x Ai Prof. Iqbal A. Khan, EED, UQU

BJT Hybrid Model: Since hre is negligible and Rout = 1/hoe =1MΩ is very high and can be taken as open circuit. Thus the reduced BJT hybrid-model is as: h-parameters Typical Values hie 1KΩ hre 0.5x10-4 hfe 100 hoe 1x Ʊ Prof. Iqbal A. Khan, EED, UQU

Other BJT Models: Prof. Iqbal A. Khan, EED, UQU

Voltage Gain of a CE Amplifier: With CE : Without CE : Prof. Iqbal A. Khan, EED, UQU

Example1: For the given RC-coupled amplifier Vcc = 12V R1 = 22KΩ, R2=6.8KΩ, RC=1KΩ, RE= 560Ω, CC=1μF. , CE=10μF and βDC =150 and βAC =160. Calculate the over all voltage gain Av‘ with and without CE. Solution: Voltage gain with CE: Rin(Total) = R1 ǁ R2 ǁ Rin(base) Prof. Iqbal A. Khan, EED, UQU

VB – IBxRB – VBE- (1+β)IBxRE = 0 2.83V - IBx 5.19K – 0.7V – 151x0.56KxIB = 0 IB ( )K = = 2.13V IB = 2.13V/89.75K = mA = 23.7μA Therefore IE = (1+β)IB = 151x mA = 3.58mA Hence re’ = VT/ IE = 26mV / 3.58mA = 7.26Ω The gain with CE and RS = 0: The gain without CE and RS = 0: Prof. Iqbal A. Khan, EED, UQU

With CE and RS = 500Ω: Rin(Base) = βre’ = 160x7.26Ω = 1.16KΩ Rin(Total) = RB||Rin(Base) = 5.19Kx1.16K/(5.19K+1.16K) = 5.19Kx1.16K/6.35K Rin(Total) = 6.35KΩ Without CE and RS = 500Ω : Rin(Base) = β(re’+RE) = 160x((7.26Ω + 560) = 90.76KΩ Rin(Total) = RB||Rin(Base) = 5.19Kx90.76K/(5.19K K) = 5.19Kx90.76K/95.95K Rin(Total) = 4.9K Prof. Iqbal A. Khan, EED, UQU

Voltage Gain AVS With CE and Without CE and RS = 0 RS = 500Ω -127.6 -1.76 -1.6 Rin(Base) βACre’ =160x7.26Ω = 1.16KΩ βAC(re’+RE) = 160x((7.26Ω+ 560) = 90.76KΩ Thus it is observed that the gain is drastically increased with bypass capacitor CE. Also the source resistance RS reduces the gain. Prof. Iqbal A. Khan, EED, UQU

Amplifier Types 1. Voltage amplifier: input & output signals are voltages Voltage Gain AV = Vout / Vin 2. Current amplifier: input and output signals are currents Current Gain AI = Iout / Iin 3. Transconductance amplifier: input signal is voltage; output signal is current Transconductance Gain GT = Iout / Vin 4. Transresistance amplifier: input signal is current; output signal is voltage Transresistance Gain RT = Vout / Iin Prof. Iqbal A. Khan, EED, UQU

Junction Field Effect Transistor-JFET: D-Drain G-Gate S-Source Prof. Iqbal A. Khan, EED, UQU

ID = IS for all JFET Prof. Iqbal A. Khan, EED, UQU

Pinch off Voltage(VP): For VGS = 0, the value of VDS (= VP ) at which ID becomes essentially constant is called pinch-off voltage. For a given JFET the pinch-off voltage VP has fixed value. At VGS = 0, the constant current ID = IDS = IDSS, the drain to source current with Gate shorted is always specified by the Data Sheets. IDSS is maximum drain current that a specific JFET can produce regardless of the external circuit. The Breakdown occurs at VDS = VBD and ID increases rapidly with further increase in VDS . The break down may result in irreversible damage to the device so JFETs are always operated below VBD. Prof. Iqbal A. Khan, EED, UQU

N-Channel JFET Characteristics: ID = 0 at VGS(OFF) ; ID = IDSS at VGS = 0 ID = K(VGSOFF – VGS)2 ; IDSS = K V2GSOFF ; Prof. Iqbal A. Khan, EED, UQU

Example-1: For the JFET in following Fig., VGSOFF = - 4V and IDSS = 12mA. Determine the minimum value of VDD required to put the device in the constant current area of operation. Solution: Since VGSOFF = - 4V, VP = 4V. The minimum value of VDS for the JFET to be in its constant-current region is VDS = VP = 4V In the constant –current area with VGS = 0V, ID = IDSS =12mA Hence, VDD = IDxRD +VDS = 12mAx = 10.7V = ANS. Example-2: For a 2N5459 JFET IDSS = 9mA and VGSOFF = - 8V. Determine the drain current for VGS = 0, -1V and -4V. Solution: For VGS = 0, ID = IDSS = 9mA For VGS = -1V, For VGS = -4V, Prof. Iqbal A. Khan, EED, UQU

The input impedance of JFET (RIN): Where, IGSS is reverse saturation gate current caused by minority charge carriers. Example-3 A certain JFET has its IGSS = -1nA and VGS = -10V, find its input impedance. Solution: Since Prof. Iqbal A. Khan, EED, UQU

JFET- Self Bias: VGS = VG –VS = 0 – ID x RS = -IDRS VGS = -IDRS VD = (VDD – IDxRD), and Vs = IDxRS Therefore VDS = VD - VS VDS = VDD – ID( RD + RS ) Example: Determine the value of RS required to self bias a n-channel JFET with IDSS = 25mA and VGSOFF = -15V, VGS = -5V. Solution: Since Also, VGS = -IDRS, Therefore Prof. Iqbal A. Khan, EED, UQU

EXAMPLE: For a self bias N-channel JFET of Fig. 2, determine the value of source resistance RS, if IDSS = 16 mA, VGSOFF = -8V, and VGS = -4V. Solution: Since Also, VGS = -IDRS, Therefore Prof. Iqbal A. Khan, EED, UQU

Voltage-Divider Bias: VG ID = IS , VGS = VG – VS VS = VG - VGS ID = VS/ RS or Example: Determine ID and VGS for the JFET with voltage divider bias with VDD =12V and VD =7V, R1 = 6.8MΩ, R2 = 1.0MΩ, RD = 3.3KΩ and RS = 2.2KΩ. Solution: ID = ( VDD-VD)/RD = ( 12-7)/3.3KΩ = 1.52mA VS = IDRS =1.52mAx 2.2KΩ=3.34V, VG= 1.0MΩ/7.8MΩ=1.54V VGS=VG-VS= V= - 1.8V Prof. Iqbal A. Khan, EED, UQU

Metal Oxide Semiconductor Field Effect Transistor-MOSFET Heavily doped (n-type or p-type) diffusions Very thin (<20Å) high-quality SiO2 insulating layer isolates gate from channel region Gate Source Drain Polysilicon wire Inter-layer SiO2 insulation Doped (p-type or n-type) silicon substrate Bulk Channel region: electric field from charges on gate locally “inverts” type of substrate to create a conducting channel between source and drain MOSFETs (metal-oxide-semiconductor field-effect transistors) are four-terminal voltage-controlled switches. Current flows between the diffusion terminals if the voltage on the gate terminal is large enough to create a conducting “channel”, otherwise the MOSFET is off and the diffusion terminals are not connected. Prof. Iqbal A. Khan, EED, UQU

Metal Oxide Semiconductor Field Effect Transistors-MOSFETs: E-Enhancement Type, D-Depletion Type Prof. Iqbal A. Khan, EED, UQU

Prof. Iqbal Ahmad Khan, EED, UQU
MOSFET CONSTRUCTION: 2002: L=130nm 2003: L=90nm 2005: L=65nm Today: L=18nm Key Feature: Transistor Length L Prof. Iqbal Ahmad Khan, EED, UQU

Enhancement and Depletion type NMOSFETs: Cut-Off VDS Cut-Off Prof. Iqbal A. Khan, EED, UQU

Enhance and Depletion type PMOSFETs: Cut-Off Cut-Off Prof. Iqbal A. Khan, EED, UQU

General Transfer Characteristics of E-MOSFET: Prof. Iqbal A. Khan, EED, UQU

General Transfer Characteristics of D-MOSFET: The transfer characteristics, that is out(ID) Vs Input(VGS) is given as follows. Prof. Iqbal A. Khan, EED, UQU

Example: For a D-MOSFET, IDSS = 10mA and VGS(OFF) = - 8V, then find (a) What type of channel is this MOSFET? (b) Calculate ID at (i) VGS = - 3V and (ii) at VGS = +3V. Solution: (a) The device has a negative VGS(OFF); therefore, it is an NMOS. (b) At VGS = 0, ID = IDSS =K(0-VGS(OFF))2 =K(-VGS(OFF))2; IDSS = K(-VGS(OFF))2 and since ID = K( VGS- VGS(OFF))2 ; Therefore (i) At VGS = - 3V (ii) At VGS = +3V Prof. Iqbal A. Khan, EED, UQU

MOSFET as Switch: MOSFETs (metal-oxide-semiconductor field-effect transistors) are four-terminal voltage-controlled devices. Current flows between the diffusion terminals(Drain and Source) if the voltage on the gate terminal is large enough to create a conducting “channel”, otherwise the MOSFET is off and the diffusion terminals (D & S) are not connected. INVERSION: A sufficiently strong vertical field will attract enough electrons to the surface to create a conducting n-type channel between the source and drain. Thus the layer under gate is inverted i.e., it is changed from P-type to N-type. CONDUCTION: If a channel exists, a horizontal field will cause a drift current from the drain to the source. Inversion happens here Prof. Iqbal A. Khan, EED, UQU

ON-OFF VGS for MOSFETs: Prof. Iqbal A. Khan, EED, UQU

Linear operating region of MOSFETs: Why is this bigger here than on the other side? 0 < VDS < VDsat VS VGS > VTH Larger VDS increases drift current but also reduces vertical field component which in turn makes channel less deep. At some point, electrons are traveling as fast as possible through the channel (“velocity saturation”) and the current stops growing linearly. Larger VGS creates deeper channel which increases IDS IDS proportional to μ0(W/L) VDS IDS Increasing VGS Prof. Iqbal A. Khan, EED, UQU

NMOS Regions of Operation: Operating regions: Cut-off: VGS < VTH IDS = 0 Linear: VGS ≥ VTH (=0.5V) VDS < VDsat = VGS - VTH Saturation: VGS ≥ VTH VDS ≥ VDsat = VGS – VTH IDS VDS VGS linear saturation Cut-Off μn is the Electron mobility and COX is Gate capacitance per unit area. W is the Channel width and L is the Channel Length Prof. Iqbal A. Khan, EED, UQU

The drain current is expressed as Where, μ0 =Surface mobility of the channel in (Cm2/V-s) Cox = Capacitance per unit area of the gate oxide in (F/Cm2) W = Chanel Width, L= Channel Length Since at saturation VDS = VGS - VT Assuming Therefore the drain current reduces to Or Prof. Iqbal A. Khan, EED, UQU

Channel Resistance: Since the drain current is given as Therefore The small signal channel resistance can be expressed as For VDS << VGS Prof. Iqbal A. Khan, EED, UQU

Common-Source (CS) Amplifier
The input voltage vs causes vGS to vary with time, which in turn causes iD to vary. The changing voltage drop across RD causes an amplified (and inverted) version of the input signal to appear at the drain terminal. vs + vOUT = vDS  VDD RD vIN = vGS – VBIAS  iD Prof. Iqbal A. Khan, EED, UQU

Load-Line Analysis of CS Amplifier
The operating point of the circuit can be determined by finding the intersection of the appropriate MOSFET ID vs. VDS characteristic and the load line: Load-line equation: Prof. Iqbal A. Khan, EED, UQU

NMOSFET Small-Signal Model: Where, go = 1 / ro Prof. Iqbal A. Khan, EED, UQU

Small-Signal Equivalent Circuit of NMOS Amplifier: NMOS Equivalent Vout Vin Voltage gain Prof. Iqbal A. Khan, EED, UQU

Example1: For an E-NMOSFET ID(ON) = 500mA, at VGS = 10V and VTH = 1V. Determine the drain current for VGS = 5V. Solution: Since ID = K(VGS- VTH)2 ID1 = 500 mA = K( 10-1)2 = Kx81 ID2 = K(5-1))2 = Kx16 Therefore, ID2 = 500x16 / 81 = 98.7mA = Ans Example2: For an E-NMOSFET given in Fig., if VDD = 18V, IDSS =12mA, RD = 620Ω and RG = 10MΩ, then determine the drain to source voltage VDS. Soln: Since from the Fig. it is clear that the VGS = 0, therefore ID = IDSS Thus, VDS = VDD –IDSSxRD = 18V – 12x10-3x 620 = 10.6V = Ans. Prof. Iqbal A. Khan, EED, UQU

Example-3: For an E-NMOSFET given in Fig. 2, find the value of load resistance RD, for an output voltage VDS = 10V if VDD = 15V, IDSS = 1mA, and RG = 10MΩ. Solution: Since from the Fig. 2, it is clear that the VGS = 0, therefore ID = IDSS Thus, VDS = VDD – IDSSxRD 15V – 1mAxRD = 10V RD = 5V/1mA = 5KΩ Fig. 2 Prof. Iqbal A. Khan, EED, UQU

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