# Transistor Load line doesn't intersect characteristic curves....

• nfi
In summary, the load-line does not intersect the characteristic curves for a BC108 transistor. The results of the experiment suggest that it will be difficult to design an amplifier based on these results.

#### nfi

Homework Statement
Plot the characteristic curves of a BC108 Transistor
Relevant Equations
Rb=100k, Rc=1k, VCC=15V
Hello,

I've carried out an experiment to plot the characteristic curves (Ic vs Vce) for a BC108 transistor and then attempted to find where the load-line intersects those curves. Below are my results:

...as you can see, the load-line doesn't intersect the characteristic curves at all.

When carrying out the experiment, the values of Ic shown were recorded as Vce was varied from 0V to the final value of Vce shown - it wasn't possible to drive Vce any further for each curve, despite the supply being 15V.

Am I doing something wrong here, or is this to be expected? I was hoping to find the upper and lower value of Ib to design an amplifier, but this isn't going to be possible with these results. Can I just extrapolate by extending the characteristic curves? i haven't tried changing Rc for a smaller value - would this improve the results?

Any helpful suggestions would be very welcome,

Thanks :)

Are you sure of your arithmetic? 15V/101,000 ohms =?

EDIT: Assume common emitter circuit. Is the Rb the base bias resistor and is there no emitter resistor?

Last edited:
I think you've already determined that $V_{CC} = V_{R_C} + V_{CE}$.

In other words, $V_{CE} = V_{CC} - V_{R_C}$

(This is of course assuming common emitter configuration with the emitter connected directly to ground [$R_E = 0 \ \Omega$].)

[Edit: By that I mean to say that I don't find anything wrong with your data so far, regarding that.]

In your experiment, I'm guessing that you adjusted the supply voltage between 0 and 15 volts when making the $V_{CE}$ measurements for a given $I_B$.

But if you want each of the characteristics curves to all extend out to 15 V, sometimes you'll have to increase the $V_{CC}$ to be greater than 15 V, to account for the voltage drop across $R_C$, when taking measurements.

[Edit: in other words, I think the problem was stopping your measurements when the $V_{CC} = 15 \ \mathrm{V}$. I think you need to increase $V_{CC}$ to higher voltages: Increase $V_{CC}$ to whatever it takes to bring $V_{CE}$ up to 15 V, or whatever voltage you want the curves to extend to. (Take care not to burn anything up though.)]

Last edited:
berkeman
If the resistor values are given as constraints, then the advice that @collinsmark gave above is the way to go.

Here is another tip that you may soon find useful though.
A sometimes non-obvious point is that the Slope of the Load Line is related to the value of Rc.

For instance the maximum Vc is when there is no Collector current. In that case Vc=Vcc, or Vc=15V on the X-axis of your graph.

The minimum Vc is when the transistor is saturated, or full on. The Ic will then be limited by Vcc and the value of Rc (RLOAD). On your graph that is 15mA on the Y-axis. The maximum current follows the usual Ohms Law.

That being said, the Load Line can be moved around on the graph to intersect the curves by changing Vcc (moving the Load Line left or right), or by changing Rc (changing the Load Line slope).

Hope it helps.

Cheers,
Tom

p.s. as long as the transistor maximum voltage and current are not exceeded, and since the curves almost reach the Load Line, many of us would just extrapolate the Curves and call it "Close Enough". (But you have to burn up a few before you learn when you can get away with it!)

collinsmark

## 1. What is a transistor load line?

A transistor load line is a graphical representation of the operating characteristics of a transistor. It shows the relationship between the voltage and current at the collector terminal of the transistor.

## 2. Why is it important for the transistor load line to intersect the characteristic curves?

It is important for the transistor load line to intersect the characteristic curves because this intersection point represents the operating point of the transistor. This point determines the voltage and current levels at which the transistor will function and can affect its performance.

## 3. What does it mean if the transistor load line doesn't intersect the characteristic curves?

If the transistor load line doesn't intersect the characteristic curves, it means that the transistor is not properly biased. This can lead to inefficient operation or even damage to the transistor.

## 4. How can the transistor load line be adjusted to intersect the characteristic curves?

The transistor load line can be adjusted by changing the biasing voltage or current. This can be done by adjusting the values of the resistors in the transistor circuit.

## 5. Are there any other factors that can affect the intersection of the transistor load line and characteristic curves?

Yes, other factors such as temperature and variations in the transistor's characteristics can also affect the intersection of the transistor load line and characteristic curves. It is important to consider these factors when designing a transistor circuit.

• Electrical Engineering
Replies
68
Views
3K
• Electrical Engineering
Replies
6
Views
1K
• Engineering and Comp Sci Homework Help
Replies
1
Views
1K
• Precalculus Mathematics Homework Help
Replies
5
Views
876
• Electrical Engineering
Replies
4
Views
7K
• MATLAB, Maple, Mathematica, LaTeX
Replies
3
Views
2K
• Electrical Engineering
Replies
3
Views
5K
• Electrical Engineering
Replies
13
Views
2K
• Engineering and Comp Sci Homework Help
Replies
1
Views
5K
• Calculus and Beyond Homework Help
Replies
15
Views
2K