This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

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From the course by Georgia Institute of Technology

Introduction to Electronics

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This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

From the lesson

Bipolar Junction Transistors

Learning Objectives: 1. Develop an understanding of the NPN BJT and its applications. 2. Develop an ability to analyze BJT circuits.

- Dr. Bonnie H. FerriProfessor

Electrical and Computer Engineering - Dr. Robert Allen Robinson, Jr.Academic Professional

School of Electrical and Computer Engineering

Welcome back to electronics this is Dr. Robinson.

Â In this lesson, we're going to look at the BJT used as a common emitter amplifier.

Â In the previous lesson, we examined the BJT switch and our objectives for

Â today's lesson are to introduce the BJT common emitter or CE amplifier.

Â And to examine biasing of the CE amplifier.

Â So here I've drawn a schematic of the common emitter amplifier.

Â I'm showing the active device here a BJT, along with the external

Â circuitry necessary to bias the transistor in the active region of operation, and to

Â set various parameters of this amplifier, such as its bandwidth and its gain.

Â This is the input to the amplifier, Via a AC or time-varying signal.

Â The circuit amplifies the signal, and at the output, here,

Â across the resister RL, produces a larger signal than was input.

Â Hence the circuit acts as an amplifier.

Â So once these components are chosen, we can consider the circuit to be.

Â We draw a box around this part.

Â This box, once the values are chosen to be considered to be a gain stage with some

Â gain A, so in other words we have an input voltage VN into this amplifier module.

Â And on the output of the amplifier we have a load which in this case is resistor RL.

Â We have some gain A.

Â Now this is an idealized block, we can add make it more realistic by

Â showing that this block actually requires a DC power supply to operate, so

Â we can indicated that power supply is required.

Â In this case 15 volts.

Â So if we apply a 15 volt dc voltage to this box then apply a time

Â varying input here, we should at the output get a larger output,

Â which is equal to the input multiplied by the gain A.

Â Now, these external components set the gain A and they also

Â control other factors of the amplifier like, for example, its bandwidth.

Â So we have frequency here.

Â The ideal amplifier would operate with a gain A for any frequency, but in reality

Â this sort of circuit has a bandwidth, which is typically a band pass filter.

Â But if we operate in the mid band region of the band pass filter, it has a gain A.

Â Now remember, these capacitors have an impedance,

Â z sub c is equal to one over j omega c.

Â So the magnitude of the impedance of the capacitor is equal to one over omega c.

Â At the regions of interest of this amplifier the frequencies at

Â which we operate this amplifier.

Â The impedance of the capacitor can be considered to be very small or

Â a short circuit.

Â However to the DC voltages and

Â currents that set the operating point or the biasing of this transistor.

Â These capacitors appear to be open circuits.

Â So in other words no DC current can flow through into this branch,

Â this branch or this branch.

Â Because the impedance is large at that very low frequency at DC.

Â But at higher frequencies these appear to be short circuits, so

Â we can apply an AC voltage here which can be amplified and appear at the output.

Â So here I'm showing a table that indicates the possible regions of

Â operation for a BJT.

Â And remember, to operate the BJT as an amplifier,

Â we want it to be biased in its active region.

Â On the output characteristic curves, remember the active region is here.

Â So we want to pick the external circuitry I showed you in the previous schematic so

Â that the collector current and VCE for

Â the BJT are such that the bias point lands in this region.

Â So I wanted to work out for you how I obtained some of

Â the component values in the schematic we've been looking at.

Â So I've redrawn the schematic with all the capacitors made open circuits.

Â So we're left with only the portion of the circuit that controls the operating point

Â or the biasing of the transistors.

Â These components RB1, RB2, RC and RE1.

Â Now to solve for these values, I started out with some assumptions.

Â So we're going to assume that beta for the transistor is equal to 99.

Â The VBE when the transistor is in it's active region is equal to as we expect,

Â approximately .7 volts.

Â I want the emitter current IE to be equal to 2.5 milliamps.

Â I'm also going to assume that the current, IB1, is equal to 19 times IB,

Â where IB1 is the current through RB1.

Â So, IB1.

Â This is the phase current IB.

Â Here is the emitter current IE.

Â And here is the collector current IC.

Â Now a typical rule of thumb for

Â biasing a BJT is to assume that one third of the power supply voltage in

Â this case 15 is distributed across the emitter resistor.

Â One third is distributed across the BJT and the remaining on third is across RC.

Â So in other words we're going to assume that VE

Â the voltage that the emitter is equal to five volts and

Â we're going to assume the collector voltage VC is equal to 10 volts.

Â Now we know that when the transistor is biased in it's

Â active region, that we have IC is equal to beta IB.

Â So I can write.

Â What I want to do is related IE, our known current, to the base current.

Â So I can write, if you remember, that this is also equal to alpha IE.

Â So IB is equal to alpha over beta.

Â IE is equal to remember alpha is equal to beta over beta plus one.

Â So I have beta over beta plus one, times one over beta IE.

Â Which implies that the base current IB is equal to the emitter

Â current IE the value we know, divided by beta plus one.

Â So we have that IB is equal to IE which is two point five milliamps divided by 100.

Â Is equal to

Â 0.025 milliamps.

Â Now next thing I want to do is calculate the voltage VB here at the base.

Â We know that the emitter voltage is five volts and

Â we know that the VB drop is .7 volts.

Â So if I start at five volts here, and go up by one VB drop,

Â I have the base voltage, which is five VB is equal to five volts.

Â The emitter voltage, plus a VBE drop, which is 0.7

Â volts, is equal to 5.7 volts.

Â Now, I can use that voltage to calculate the value for

Â RB1, because I know that IB1 is equal to 19 IB, and we've calculated IB.

Â So I can use Ohm's law across this resistor,

Â the voltage here is equal to the current times the resistance.

Â Or, I can write it like this, VB,

Â VB minus the voltage here which is

Â zero divided by the current which

Â is 19 times IB is equal to RB1.

Â Which is equal to for this case VB is 5.7, zero and 19 times IB.

Â Gives up 12 KMs.

Â Now I can use the same technique to solve for RB2.

Â Because we know now this voltage and we know this voltage and

Â we know the current through RB2.

Â Let me label it, the current here as, IB2.

Â Now how do we know what IB2 is?

Â Because we know the currents here must sum to zero.

Â So we have IB in this direction, and 19 IB in this direction,

Â which means that IB2 must be equal to 19 plus one, or 20 IB.

Â So we can write that IB2 is equal

Â to 19 plus one IB is equal to 20 IB.

Â So using Ohms law here I can calculate RB2 as let me write it like this.

Â VCC minus VB divided by the current

Â which is 20 IB is equal to

Â RB2 is equal to 18.6 K Ohms.

Â Now to find a value for RE1 it is simply by inspection.

Â We know the voltage here is five volts and

Â we know the current through the resister is IE which is 2.5 milliamps.

Â So the ration that voltage to that current must be equal to RE1.

Â So RE1 is equal to VE over IE is

Â equal to five volts divided by

Â 2.5 milliamps is equal to 2Kamps.

Â And finally to solve for RC we need to find the relationship between IB or

Â IE and IC.

Â We know that IC is equal to alpha.

Â IE is equal to 0.99

Â times 2.5 milliamps.

Â So, at the collector here, we can again use Ohm's law,

Â this voltage minus this voltage, divided by this current must be equal to RC.

Â So we have that VCC minus VC,

Â divided by the current IC is equal to RC,

Â is equal to 15 minus ten,

Â divided by 0.99 times 2.5 milliamps.

Â So here we have five over 2.5 milliamps, which we know is 2K, so

Â the value would be something a little bit bigger than 2K.

Â And it turns out to be equal to 2.2K.

Â Now in the base bias circuit here, you might want to know why we picked

Â 19 IB here, and a total of 20 IB through, or into the base node.

Â And the reason for that is, you want most of the current that biases the base

Â circuit to flow through the two resistors.

Â So that any variation in the base current here,

Â does not change the voltage here by very much.

Â So here, I am showing you the results of the simulation of that circuit.

Â So, I have used a circuit analysis program to simulate the behavior of that

Â circuit using the values that we calculated.

Â And in this table I am showing the bias values.

Â This is the collector current, the base current, the emitter current,

Â the collector voltage, the base voltage and the emitter voltage.

Â So we designed the circuit to have an emitter current of about 2.5 milliamps.

Â And you can see that we are very close to that value, 2.47.

Â The negative sign just indicates that the direction of

Â the current is taken opposite to what we think it would be.

Â So these, these component values,

Â using our analysis get us close to the IE value that we wanted.

Â Remember we designed for a collector voltage of about ten volts.

Â We had 9.66, we wanted an emitter voltage of about five volts, we have 4.93 volts.

Â So overall, these values of components give us

Â good agreement with our desired specifications.

Â Now, let's see on the output characteristic curves where our

Â operating point would land.

Â We know that we have a collector current of about 2.5,

Â and a collector to emitter voltage of,

Â from our table, VCE is equal to 9.66 minus 4.93.

Â Or about 4.7 or so for VCE, which would be somewhere here.

Â And for that VCE we have a collector current of 2.43,

Â which would put us somewhere in here.

Â So we have an operating point of approximately right there,

Â which is in the active region.

Â Now, another way to be sure that we're in the active region is to check that both

Â the collector-emitter junction, or to check that the base-emitter

Â junction is forward-biased and the collector-base junction is reverse-biased.

Â So you can see that the base-emitter junction is forward-biased.

Â The voltage at the base is greater than the voltage at the,

Â at the emitter by about this is about point,

Â 0.65 volts, which is close to the 0.7 we, we assumed.

Â And the collector base junction has be reverse biased.

Â And it is, because the collector voltage is greater than the base voltage.

Â So the transistor is definitely operating in the active region,

Â so can be used as an amplifier.

Â So in summary during this lesson you were introduced to

Â the BJT common emitter amplifier.

Â And we solved a common emitter biasing example.

Â Our next lesson we'll look more at the BJT CE amplifier.

Â Thank you and till next time

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