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.

Â This is Dr. Robinson.

Â In this lesson,

Â we are going to determine BJT parameters from curve tracer measurements.

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

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

Â parameters from measured data.

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

Â It is a piece of test instrumentation that allow us to

Â obtain the characteristic curves for various devices.

Â So somewhere here in this socket, right there,

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

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

Â a set of output characteristic curves.

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

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

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

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

Â So if we have this transistor with unknown parameters,

Â we can use the curve tracer to make a measurements.

Â Then solve for those parameters and

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

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

Â the curve tracer.

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

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

Â Now remember the output characteristic curves.

Â Indicate the relationship between the collector current and

Â the collector to emitter voltage for the transistor.

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

Â And I'm going to choose this point here and

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

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

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

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

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

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

Â this distance here horizontally is 2 volts.

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

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

Â So we have one pair at 2 volts and

Â 9.2 milliamps, and

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

Â So 8 volts and 9.5 milliamps.

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

Â And then I can use the slope of that line and

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

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

Â the rise is the change in currents, so

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

Â the change in VCE which should be equal to

Â 0.3 divided by 6 milliamps per volt

Â is equal to 0.05 milliamps per volt.

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

Â To calculate VA.

Â So I can write the VA.

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

Â I'm going to use this pair.

Â VA is equal to 9.5 divided by

Â 0.05 minus 8, which is equal

Â to 182 volts for the early voltage.

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

Â another transistor parameter.

Â And to do this we need to know IB and

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

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

Â have five micro amps per step in base current.

Â So this would be zero micro amps.

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

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

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

Â I can again use this pair 9.5 milliamps.

Â Iâ€™d better put the units.

Â Milliamps divided by IB, which is 50

Â micro amps over 1 plus VCE which would be,

Â for this case, 8 volts, 8 volts

Â Divided by the early voltage 182 which gives

Â us a beta naught of surprisingly enough 182.

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

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

Â So a good approximation for

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

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

Â The early voltage, and beta.

Â So here are the measured transfer characteristics for

Â that same BJT transistor.

Â And remember transfer characteristics relate the collector current

Â to the change in base emitter voltage.

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

Â So again, like before, we have our scale.

Â 1 milliamp, again, per vertical division, but

Â each horizontal division is equal to 100 millivolts.

Â So 500, 600, 700.

Â So about what we'd expect, right?

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

Â Now what I want to do is use this expression for

Â ISO that we found previously to calculate the value.

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

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

Â Right there.

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

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

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

Â We know the thermal voltage 0.0259 volts.

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

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

Â 0.7 volts, divided by a 25.9 milli-volts.

Â Divided by 1 plus 12 over 182,

Â which gives us, a IS zero of 1.03

Â times 10 to the minus 14 amps.

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

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

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

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

Â So, from this data, from the curve tracer we

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

Â Beta naught, the early voltage, and IS0.

Â So in summary, during this lesson,

Â we determined BJT parameters from measured data.

Â And in our next lesson,

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

Â The use of the transistor as an electronic switch.

Â So thank you, and until next time.

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