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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Del curso dictado por 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.

De la lección

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.

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

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

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

And to examine biasing of the CE amplifier.

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

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

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

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

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

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

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

Hence the circuit acts as an amplifier.

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

We draw a box around this part.

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

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

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

We have some gain A.

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

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

we can indicated that power supply is required.

In this case 15 volts.

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

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

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

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

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

So we have frequency here.

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

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

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

Now remember, these capacitors have an impedance,

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

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

At the regions of interest of this amplifier the frequencies at

which we operate this amplifier.

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

a short circuit.

However to the DC voltages and

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

These capacitors appear to be open circuits.

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

this branch or this branch.

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

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

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

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

operation for a BJT.

And remember, to operate the BJT as an amplifier,

we want it to be biased in its active region.

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

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

that the collector current and VCE for

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

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

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

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

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

or the biasing of the transistors.

These components RB1, RB2, RC and RE1.

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

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

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

approximately .7 volts.

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

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

where IB1 is the current through RB1.

So, IB1.

This is the phase current IB.

Here is the emitter current IE.

And here is the collector current IC.

Now a typical rule of thumb for

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

this case 15 is distributed across the emitter resistor.

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

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

the voltage that the emitter is equal to five volts and

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

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

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

So I can write.

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

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

So IB is equal to alpha over beta.

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

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

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

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

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

Is equal to

0.025 milliamps.

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

We know that the emitter voltage is five volts and

we know that the VB drop is .7 volts.

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

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

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

volts, is equal to 5.7 volts.

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

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

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

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

Or, I can write it like this, VB,

VB minus the voltage here which is

zero divided by the current which

is 19 times IB is equal to RB1.

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

Gives up 12 KMs.

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

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

we know the current through RB2.

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

Now how do we know what IB2 is?

Because we know the currents here must sum to zero.

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

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

So we can write that IB2 is equal

to 19 plus one IB is equal to 20 IB.

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

VCC minus VB divided by the current

which is 20 IB is equal to

RB2 is equal to 18.6 K Ohms.

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

We know the voltage here is five volts and

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

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

So RE1 is equal to VE over IE is

equal to five volts divided by

2.5 milliamps is equal to 2Kamps.

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

IE and IC.

We know that IC is equal to alpha.

IE is equal to 0.99

times 2.5 milliamps.

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

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

So we have that VCC minus VC,

divided by the current IC is equal to RC,

is equal to 15 minus ten,

divided by 0.99 times 2.5 milliamps.

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

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

And it turns out to be equal to 2.2K.

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

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

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

circuit to flow through the two resistors.

So that any variation in the base current here,

does not change the voltage here by very much.

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

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

circuit using the values that we calculated.

And in this table I am showing the bias values.

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

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

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

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

The negative sign just indicates that the direction of

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

So these, these component values,

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

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

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

So overall, these values of components give us

good agreement with our desired specifications.

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

operating point would land.

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

and a collector to emitter voltage of,

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

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

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

which would put us somewhere in here.

So we have an operating point of approximately right there,

which is in the active region.

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

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

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

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

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

at the emitter by about this is about point,

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

And the collector base junction has be reverse biased.

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

So the transistor is definitely operating in the active region,

so can be used as an amplifier.

So in summary during this lesson you were introduced to

the BJT common emitter amplifier.

And we solved a common emitter biasing example.

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

Thank you and till next time

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