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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来自 乔治亚理工学院 的课程

电子学基础

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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.

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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 are going to determine BJT parameters from curve tracer measurements.

In our previous lesson, we examined BJT parameters, and our objectives for

this lesson are to introduce the curve tracer and to solve for

parameters from measured data.

So here I'm showing the beloved Tectronics 370B curve tracer.

It is a piece of test instrumentation that allow us to

obtain the characteristic curves for various devices.

So somewhere here in this socket, right there,

I have inserted a two-end 441 transistor, an NPN BJT transistor.

And I have set the controls to produce, in this case,

a set of output characteristic curves.

By changing the controls, I can generate a transfer characteristic curve.

And then I can save this data either as a picture or as numerical data.

And, what I want to do here in the following, is use the data

obtained in the curve tracer to determine the parameters of this transistor.

So if we have this transistor with unknown parameters,

we can use the curve tracer to make a measurements.

Then solve for those parameters and

then we know how this transistor will behave when operated in its active region.

So here I'm showing a set of measured output characteristic curves from

the curve tracer.

And what I want to do is use this data to determine the two transistor parameters,

the early voltage and beta naught, using the equations that we derived earlier.

Now remember the output characteristic curves.

Indicate the relationship between the collector current and

the collector to emitter voltage for the transistor.

So I want to pick two points on a single IB curve on the output characteristics.

And I'm going to choose this point here and

this point here, no, let's choose this point here.

So I want to find the IC, VCE pairs for each of these two points.

Now the scale for the for the curves is indicated over here on the right.

Per each horizontal division we have the two volt change, and for

every vertical division we have 1 milli-amp of change.

So, this distance here vertically is 1 milli-amp and

this distance here horizontally is 2 volts.

So this point that I picked here would be a VCE of 2 volts zero, 2 volts,

and an IC of 1, 2, 3, 4, 5, 6, 7, 8, 9.2 milliamps.

So we have one pair at 2 volts and

9.2 milliamps, and

we have a second pair here at 2, 4, 6, 8 volts and 9.5 milliamps.

So 8 volts and 9.5 milliamps.

Now I can use these two points to calculate the slope of this line here.

And then I can use the slope of that line and

one of the points the IC, VCE pair to determine the early voltage.

So the slope would be equal to the rise over the run,

the rise is the change in currents, so

I would have 9.5 minus 9.2 divided by 8 minus 2,

the change in VCE which should be equal to

0.3 divided by 6 milliamps per volt

is equal to 0.05 milliamps per volt.

Now I can use that slope of this line in this equation.

To calculate VA.

So I can write the VA.

And I can do this with either one of these two points, but

I'm going to use this pair.

VA is equal to 9.5 divided by

0.05 minus 8, which is equal

to 182 volts for the early voltage.

Now, we can also use the output characteristics to determine beta naught,

another transistor parameter.

And to do this we need to know IB and

one IC VCE pair, along with the VA we just calculated.

So, we can see that per step on this set of output characteristic curves we

have five micro amps per step in base current.

So this would be zero micro amps.

Five, ten, 15, 20, 25, 30, 35, 40, 45, 50.

So the top characteristic is the IB equals 50 microamps characteristic.

So I can write that beta naught is equal to IC and

I can again use this pair 9.5 milliamps.

I’d better put the units.

Milliamps divided by IB, which is 50

micro amps over 1 plus VCE which would be,

for this case, 8 volts, 8 volts

Divided by the early voltage 182 which gives

us a beta naught of surprisingly enough 182.

Now you can see that for transistors with large early voltages this, this expression

is approximately one, because the early voltage is typically much larger than VCE.

So a good approximation for

beta is just the ratio of the collector current to the base current.

So, from the output characteristic curves, we get two of the transistor parameters.

The early voltage, and beta.

So here are the measured transfer characteristics for

that same BJT transistor.

And remember transfer characteristics relate the collector current

to the change in base emitter voltage.

And I need to tell you that VCE, for this case is equal to 12 volts.

So again, like before, we have our scale.

1 milliamp, again, per vertical division, but

each horizontal division is equal to 100 millivolts.

So 500, 600, 700.

So about what we'd expect, right?

When the transistors on we approximate VBE as about 0.7 volts.

Now what I want to do is use this expression for

ISO that we found previously to calculate the value.

So what we need is a single point on this curve, and

I'm going to pick the 0.5, 0.6, 0.7.

Right there.

So, at a VBE of 0.7 volts, we have a collector current of 6 milli-amps.

So, we have this pair 0.7 volts with an IC of 6 milli-amps.

Then I can plug this in, for this IC VBE pair.

We know the thermal voltage 0.0259 volts.

We know VCE and we know VA from our previous calculation.

So ISO is equal to 6 milli-amps, e to the minus

0.7 volts, divided by a 25.9 milli-volts.

Divided by 1 plus 12 over 182,

which gives us, a IS zero of 1.03

times 10 to the minus 14 amps.

[SOUND] And again this quantity is approximately one because the early

voltage, in this case, is much, much larger than VCE.

So if you didn't exactly know what VCE is, you would still get

a good numerical result if you just assume this denominator was 1.

So, from this data, from the curve tracer we

were able to determine the three dominant parameters for the BJT.

Beta naught, the early voltage, and IS0.

So in summary, during this lesson,

we determined BJT parameters from measured data.

And in our next lesson,

we will look at a particular application of the bipolar junction transistor.

The use of the transistor as an electronic switch.

So thank you, and until next time.