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 are going to take a further look at the bipolar junction

Â transistor parameters.

Â In our previous lesson you were introduced to the bipolar junction transistor and

Â our objective for

Â this lesson is to relate the BJT parameters to the characteristic curves.

Â Here are the two equations that describe the operation of

Â the BJT in the active region.

Â And what I want to do is use these equations to solve for

Â the three dominant spice parameters.

Â Beta naught, IS0, and VA.

Â So to begin I'm going to substitute this expression for beta into this equation.

Â IC is equal to beta IB.

Â So I get IC, is equal to beta naught

Â IB times 1 plus VCE over VA.

Â So remember this gives us the linear relationship between IC and VCE.

Â These straight lines in the active region here, of the output characteristics.

Â So what I want to do is find the value for VC E when IC is equal to zero.

Â So I set IC equal to zero.

Â And we know that beta naught,

Â this transistor parameter, is a non-zero quantity.

Â And if we're in the active region, we know that I B is greater than zero.

Â So for this to be true 1 plus VCE over VA

Â is equal to 0 which implies that VCE is equal to negative VA.

Â Now this is linear equation, these are straight lines and

Â when IC is equal to zero we know that we have found the X intercept.

Â So all of these lines let me extend the X axis.

Â All of these lines if we extended these lines,

Â they represent the behavior of the BGAT in the active region.

Â They would all cross at the same point.

Â At the x intercept and at the x intercept we know that VCE is equal to minus VA.

Â So if we had this measured data,

Â these measured output characteristics curves we can solve for the x intercept.

Â To give us the parameter VA the early voltage.

Â Now lets look at this equation again.

Â I'm going to solve this equation for beta not another transistor parameter.

Â So from this equation here I can write

Â that beta not [NOISE] Is equal to IC divided

Â by IB over one plus VCE over VA.

Â So if I'm given a set of output characteristic curves and

Â I pick a single IC VCE pair.

Â Say this point, and we know that this curve is associated with some value of IB.

Â If I take that point and the associated IB, if I know the early

Â voltage NBCE I can plug into this equation to get the parameter beta not.

Â If I measure two points on this same line I can

Â determine the slope of this line and then from what we know about.

Â Straight lines and slopes.

Â I can calculate the early voltage as va is equal to

Â one point on this line or one ic value on this line.

Â Ic divided by the slope of the line minus VCE [NOISE].

Â Now if we look at this equation the IC equation, we can solve this for ISO if we

Â make the substitution of this expression into here and we can find that ISO.

Â Is equal to Ice to the minus VBE over VT,

Â IC, divided by 1 plus VCE over VA.

Â So here are the three equations for the three dominant transistor parameters.

Â We can solve for VA, then use that value of VA in here to solve for beta not and

Â that value of VA in here to solve for IS not, ISO if we know an ICVBE pair.

Â So for this equation, we need an IC VBE pair.

Â And how do we find that?

Â We measure the transfer characteristic of the BJT.

Â [SOUND] From

Â the transfer characteristic.

Â Let's take a look at how changing the parameters of

Â the transistor changes the characteristic curves of the transistor.

Â So here I have two sets of output characteristic curves where I've

Â kept beta n or beta naught and

Â IS0 the same between the sets but I've changed the early voltage.

Â In this set of curves, I've set the early voltage to ten volts, and

Â in this set of curves, I've set the early voltage to 200 volts.

Â And remember the early voltage gives us an indication of

Â the x intercept of the curves.

Â So this low early voltage, we would expect all these curves to

Â intersect at a short distance, a distance of ten to the left of the origin here.

Â While the early voltage of 200 would extend a distance of, would extend out,

Â and cross the x axis at a distance of 200 volts to the left of the origin here.

Â So, a BJT with a small early voltage has steeper curves in the active region,

Â which means that as VCE changes you can get large changes in IC.

Â But a transistor with a larger early voltage results in

Â less change in collector current with changes in BCE.

Â These two sets of output characteristic curves illustrate how

Â changing the parameter beta changes the curves.

Â On this set of curves the transistor has a beta of 150 and on this set of

Â curves it has a beta, remember the base two collector current gain of 50.

Â So in both sets of curves I've stepped the base current by 20 micro amps.

Â So 0, 20, 40, 60, 80, 100.

Â Micro.

Â And the same on this set of curves.

Â 0, 20, up to 100 microamps.

Â Now, for the large beta transistor, beta of 150.

Â As we go to, say, 20 to 40 microamps, we can read across.

Â And see that I see changes by approximately, say one and

Â a half milliamps.

Â So here we have IC equals 1.5

Â milliamps for a IB change of 20 microamps.

Â Now on this smaller beta transistor we have

Â again a change of 20 microamps and base current.

Â But we can read across and see that IC changes by less than one milliamps say,

Â say 0.8 milli-amps.

Â So here we have 0.8 milli-amps for the same change in IB, 20 micro-amps.

Â So, the gain in current from base to collector,

Â is smaller for this set of curves.

Â We change by 20 micro-amps the base current,

Â we only change the collector current by 0.8 milli-amps.

Â Whereas for the larger beta transistor, for that same change,

Â we change about 1.5 milliamps.

Â So, if you keep the scales the same, and change the beta of the transistor,

Â what you'll notice is, that these curves get closer together.

Â The distance between two adjacent curves for the stame, same step size,

Â is smaller for the smaller beta than for the larger beta transistor.

Â Here I'm showing two sets of transfer characteristic curves for the NPM NJT.

Â In this curve the saturation current Is is set to approximately 26 palmtops

Â while in this curve the saturation current Is is reduced by a factor of 100.

Â I've kept the scales the same between the curves so

Â you can see the the effect of this change in saturation current.

Â We have a much steeper slope here when the saturation current is higher than we

Â do when the saturation current is lower.

Â So you can consider this to be a more ideal transistor in the sense that.

Â It turns on instantaneously, and this is a more vertical line.

Â Whereas here, we have a more gradual turn on, and our approximation of

Â a fixed voltage in the active region, you can see you would be less accurate for

Â this low IS transistor, than it would be for the higher IS transistor.

Â So, in summary, during this lesson,

Â we determined how the parameters of a BJT affect its characteristics.

Â In our next lesson, we will look at how to use measurements made with a curve tracer,

Â to determine the parameters for an unknown transistor.

Â So, thank you, and until next time.

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