Homework Assignment #1

DATE ASSIGNED: 21 Jan 03

DATE DUE: 30 Jan 03

Chapter 3 Problems #3.(21,22,56)

Chapter 4 Problems #4.(29,33,37,51,63,95)

Chapter 5 Problems #5.(55)

Write a short description (<1 page) of what is meant by "small signal analysis". You need to touch on the following concepts: large signal vs. small signal, linear and nonlinear systems, differential/incremental quantities, and superposition. 

Notes, Hints and Suggestions:

Problem 4.29

The key to solving this problem is to remember what the definitions of the Early voltage and the output resistance are.

Remember that for an ideal transistor, the collector current is established by setting the collector-base current. With that fixed v_BE, the i_CE will, ideally, remain fixed and independent of any changes in v_CE. But this is not the case for real transistors where i_CE does in fact change with changes in v_CE. We quantify this effect by looking at the derivative of i_CE with respect to v_CE (at a fixed value of v_BE) and define the reciprocal of this derivative to be the output resistance.

r_o = dv_CE / di_CE =  v_ce / i_ce

Notice that if you plot i_CE against v_CE then (1/r_o) is simply the slope of this curve. The Early voltage, V_A, is simply defined as the negative of the point where this sloped line would intersect the voltage axis.

So, with this in mind, if I give you any two points along one of these lines - in other words, the values of the collector current that correspond to any two values of the collector-emitter voltage as long as I hold the base-emitter voltage fixed, then you have two points on a straight line and by finding the slope and the x-intercept (not the y-intercept) you have the output resistance and the Early voltage.

Looking at it another way, if you invert the current and voltage axes, the you have the following relationship for a given value of v_BE:

v_CE = r_o*i_CE - V_A

This relationship holds regardless of how you choose to graph it.