Welcome back to electronics, this is Dr. Robinson. In this lesson we are going to take a further look at the bipolar junction transistor parameters. In our previous lesson you were introduced to the bipolar junction transistor and our objective for this lesson is to relate the BJT parameters to the characteristic curves. Here are the two equations that describe the operation of the BJT in the active region. And what I want to do is use these equations to solve for the three dominant spice parameters. Beta naught, IS0, and VA. So to begin I'm going to substitute this expression for beta into this equation. IC is equal to beta IB. So I get IC, is equal to beta naught IB times 1 plus VCE over VA. So remember this gives us the linear relationship between IC and VCE. These straight lines in the active region here, of the output characteristics. So what I want to do is find the value for VC E when IC is equal to zero. So I set IC equal to zero. And we know that beta naught, this transistor parameter, is a non-zero quantity. And if we're in the active region, we know that I B is greater than zero. So for this to be true 1 plus VCE over VA is equal to 0 which implies that VCE is equal to negative VA. Now this is linear equation, these are straight lines and when IC is equal to zero we know that we have found the X intercept. So all of these lines let me extend the X axis. All of these lines if we extended these lines, they represent the behavior of the BGAT in the active region. They would all cross at the same point. At the x intercept and at the x intercept we know that VCE is equal to minus VA. So if we had this measured data, these measured output characteristics curves we can solve for the x intercept. To give us the parameter VA the early voltage. Now lets look at this equation again. I'm going to solve this equation for beta not another transistor parameter. So from this equation here I can write that beta not [NOISE] Is equal to IC divided by IB over one plus VCE over VA. So if I'm given a set of output characteristic curves and I pick a single IC VCE pair. Say this point, and we know that this curve is associated with some value of IB. If I take that point and the associated IB, if I know the early voltage NBCE I can plug into this equation to get the parameter beta not. If I measure two points on this same line I can determine the slope of this line and then from what we know about. Straight lines and slopes. I can calculate the early voltage as va is equal to one point on this line or one ic value on this line. Ic divided by the slope of the line minus VCE [NOISE]. Now if we look at this equation the IC equation, we can solve this for ISO if we make the substitution of this expression into here and we can find that ISO. Is equal to Ice to the minus VBE over VT, IC, divided by 1 plus VCE over VA. So here are the three equations for the three dominant transistor parameters. We can solve for VA, then use that value of VA in here to solve for beta not and that value of VA in here to solve for IS not, ISO if we know an ICVBE pair. So for this equation, we need an IC VBE pair. And how do we find that? We measure the transfer characteristic of the BJT. [SOUND] From the transfer characteristic. Let's take a look at how changing the parameters of the transistor changes the characteristic curves of the transistor. So here I have two sets of output characteristic curves where I've kept beta n or beta naught and IS0 the same between the sets but I've changed the early voltage. In this set of curves, I've set the early voltage to ten volts, and in this set of curves, I've set the early voltage to 200 volts. And remember the early voltage gives us an indication of the x intercept of the curves. So this low early voltage, we would expect all these curves to intersect at a short distance, a distance of ten to the left of the origin here. While the early voltage of 200 would extend a distance of, would extend out, and cross the x axis at a distance of 200 volts to the left of the origin here. So, a BJT with a small early voltage has steeper curves in the active region, which means that as VCE changes you can get large changes in IC. But a transistor with a larger early voltage results in less change in collector current with changes in BCE. These two sets of output characteristic curves illustrate how changing the parameter beta changes the curves. On this set of curves the transistor has a beta of 150 and on this set of curves it has a beta, remember the base two collector current gain of 50. So in both sets of curves I've stepped the base current by 20 micro amps. So 0, 20, 40, 60, 80, 100. Micro. And the same on this set of curves. 0, 20, up to 100 microamps. Now, for the large beta transistor, beta of 150. As we go to, say, 20 to 40 microamps, we can read across. And see that I see changes by approximately, say one and a half milliamps. So here we have IC equals 1.5 milliamps for a IB change of 20 microamps. Now on this smaller beta transistor we have again a change of 20 microamps and base current. But we can read across and see that IC changes by less than one milliamps say, say 0.8 milli-amps. So here we have 0.8 milli-amps for the same change in IB, 20 micro-amps. So, the gain in current from base to collector, is smaller for this set of curves. We change by 20 micro-amps the base current, we only change the collector current by 0.8 milli-amps. Whereas for the larger beta transistor, for that same change, we change about 1.5 milliamps. So, if you keep the scales the same, and change the beta of the transistor, what you'll notice is, that these curves get closer together. The distance between two adjacent curves for the stame, same step size, is smaller for the smaller beta than for the larger beta transistor. Here I'm showing two sets of transfer characteristic curves for the NPM NJT. In this curve the saturation current Is is set to approximately 26 palmtops while in this curve the saturation current Is is reduced by a factor of 100. I've kept the scales the same between the curves so you can see the the effect of this change in saturation current. We have a much steeper slope here when the saturation current is higher than we do when the saturation current is lower. So you can consider this to be a more ideal transistor in the sense that. It turns on instantaneously, and this is a more vertical line. Whereas here, we have a more gradual turn on, and our approximation of a fixed voltage in the active region, you can see you would be less accurate for this low IS transistor, than it would be for the higher IS transistor. So, in summary, during this lesson, we determined how the parameters of a BJT affect its characteristics. In our next lesson, we will look at how to use measurements made with a curve tracer, to determine the parameters for an unknown transistor. So, thank you, and until next time.
