CHAPTER 3 Range of power supply Voltage Regulator VRi = VPS - VZ Assume: IZmin = 0.1 IZ(max) Range of power supply Voltage Regulator The load resistor sees a constant voltage regardless of the current VL = VZ VRi = VPS - VZ II = IZ + IL Zener Diode CHAPTER 3

Vm is the peak value of the output voltage Half Wave V r = V m T P RC If the ripple is very small, we can approximate T’ = Tp where Tp is the period of the cycle VC = Vm e – t / RC Full Wave V r = V m T P 2RC Vm is the peak value of the output voltage Capacitor Discharge V r = V m T ′ RC Multiple Diode Circuit Ripple Voltage, Vr Vo = Vs - V Filter Vo = Vs - 2V PIV = 2Vspeak - V Peak and RMS V rms = V peak 2 Center-tapped Duty Cycle Bridge PIV = Vspeak - V Full Wave Vo = Vs - V CHAPTER 3 Rectifier Half Wave PIV = Vspeak

Chapter 4 Bipolar Junction Transistor

REMEMBER THIS Current flow in the opposite direction of the electrons flow; same direction as holes e e e I h h h

Transistor Structures The bipolar junction transistor (BJT) has three separately doped regions and contains two pn junctions. Bipolar transistor is a 3-terminal device. Emitter (E) Base (B) Collector (C) The basic transistor principle is that the voltage between two terminals controls the current through the third terminal. Current in the transistor is due to the flow of both electrons and holes, hence the name bipolar.

3 Regions of Operation Active Operating range of the amplifier. Base-Emitter Junction forward biased. Collector-Base Junction reverse biased Cutoff The amplifier is basically off. There is voltage but little current. Both junctions reverse biased Saturation The amplifier is full on. There is little voltage but lots of current. Both junctions forward biased

OPERATIONS - npn VBE = V FORWARD ACTIVE MODE VBE = V + VBE The base-emitter (B-E) junction is forward biased and the base-collector (C-B) junction is reverse-biased,. - Since the B-E junction is forward biased, electrons from the emitter are injected across the B-E junction into the base  IE Once in the base region, the electrons are quickly accelerated through the base due to the reverse-biased C-B region  IC iB Some electrons, in passing through the base region, recombine with majority carrier holes in the base. This produces the current IB

TO ILLUSTRATE E B C - VBE + Imagine the marbles as electrons A flat base region with gaps where the marbles may fall/trapped – recombine A sloping collector region represents high electric field in the C-B region Hence, when enough energy is given to the marbles, they will be accelerated towards to base region with enough momentum to pass the base and straight ‘fly’ to the collector

MATHEMATICAL EXPRESSIONS IB + IE VBE - IE = IS [ e VBE / VT -1 ] = IS e VBE / VT Based on KCL: IE = IC + IB No. of electrons crossing the base region and then directly into the collector region is a constant factor  of the no. of electrons exiting the base region IC =  IB No. of electrons reaching the collector region is directly proportional to the no. of electrons injected or crossing the base region. Ideally  = 1, but in reality it is between 0.9 and 0.998. IC =  IE

Based on KCL: IE = IC + IB IC =  IB IC =  IE IE =  IB + IB = IB(  + 1) IE = IB(  + 1) Now With IC =  IB  IB = IC /  Hence, IE = [ IC / ] (  + 1) IC = IE [  /  + 1 ] Comparing with IC =  IE  = [  /  + 1 ]

OPERATIONS - pnp IE = IS [ e VEB / VT -1 ] = IS e VEB / VT C E OPERATIONS - pnp IB IC FORWARD ACTIVE MODE - The emitter – base (E- B) junction is forward biased and the base-collector (B- C) junction is reverse-biased,. VEB IE + VEB = V IE = IS [ e VEB / VT -1 ] = IS e VEB / VT **Notice that it is VEB Based on KCL: IE = IC + IB

SUMMARY: Circuit Symbols and Conventions Based on KCL: IE = IC + IB npn bipolar transistor simple block diagram and circuit symbol. Arrow on the emitter terminal indicates the direction of emitter current pnp bipolar transistor simple block diagram and circuit symbol.

IC =  IB IC =  IE IE = IB(  + 1) Based on KCL: IE = IC + IB NPN PNP IE = IS [ e VBE / VT ] IE = IS [ e VEB / VT] IC =  IB IC =  IE IE = IB(  + 1) 𝜶= 𝜷 𝟏+𝜷 𝜷= 𝜶 𝟏−𝜶 Based on KCL: IE = IC + IB

EXAMPLE   Calculate the collector and emitter currents, given the base current and current gain. Assume a common-base current gain,  = 0.97 and a base current of iB = 25 µA . Also assume that the transistor is biased forward in the forward active mode. Solution: The common-emitter current gain is The collector current is And the emitter current is

Examples EXAMPLE 1 Given IB = 6.0A and IC = 510 A. Determine ,  and IE Answers:  = 85 = 0.9884 IE = 516 A EXAMPLE 2 NPN Transistor Reverse saturation current Is = 10-13A with current gain,  = 90. Based on VBE = 0.685V, determine IC , IB and IE Answers: IE = 10-13 (e 0.685/0.026) = 0.0277 A IC = (90/91)(0.0277) = 0.0274 A IB = IE – IC = 0.3 mA

BJT: Current-Voltage Characteristic IC versus VCE

Common-Emitter Configuration - npn The Emitter is common to both input (base-emitter) and output (collector-emitter). Since Emitter is grounded, VC = VCE With decreasing VC (VCE), the junction B-C will become forward biased too. The current IC quickly drops to zero because electrons are no longer collected by the collector Node B 0V

Characteristics of Common-Emitter - npn