Specifically, I want to know why "r" parameters are considered in AC analysis but not in DC. I am reading from Electronic Devices by Floyd.
For example, in the picture below, VBE is calculated by multiplying IE with r'e, but if it were DC analysis, we would consider VBE approximately 0.6-0.7 Volts.
I am a newbie in electronics, so probably there is something I missed about the analysis of BJT amplifiers.
I'm sure you've encountered the idea of the slope of a curve, before. It's the first thing they teach when learning calculus. Remember this formula?
$$f^{'}\!\!\left(x_0\right)=\lim_{h\to 0} \frac{f\left(x_0+h\right)-f\left(x_0\right)}{h}$$
It's just the local slope of the curve at \$x_0\$.
The dynamic resistance is like that, except that the curve is related to the Shockley diode equation (as it applies to the BJT):
$$I_\text{C}=I_\text{SAT}\cdot\left[\exp\left(\frac{V_\text{BE}}{\eta \:V_T}\right)-1\right]$$
That non-linear curve looks like this:
On the right, I've expanded the view of the quiescent Q-point (the operating point) for the BJT amplifier. (You usually select some point on the curve where the circuit operates when there is no input signal to worry about, around which the circuit is supposed to operate when the AC signal is applied.)
As you can see, for tiny changes nearby the Q-point you can approximate the curve with a simple line (the line that is tangent to the Q-point on the curve.)
In no way is this an actual resistor that works like a resistor with DC applied to it. Real resistors actually are lines and they don't have a voltage across them when there is zero current, for example. Note that this slope intersects the x-axis somewhere to the right of zero? It's just the local slope at the Q-point and for small (AC) changes nearby, you are allowed to assume (for simplification purposes) that it holds for AC changes.
Now, if the changes are large enough then this slope fails. But once this dynamic resistance slope is no longer valid you are, by definition, no longer talking about small signal changes (the usual "AC" assumption) and have moved into the domain of large signal changes.
Of course, in practice people act as if their input signal is "small" enough that the dynamic resistance is always valid when, in fact, it really isn't. It's not uncommon for amplifiers to be operated in such a way that the operation moves up and down the Shockley curve far enough that the local slope value changes enough to matter. In these cases, it's broadly called distortion. This means that the Q-point dynamic resistance slope that was assumed valid, isn't sufficiently valid over the actual operating range. As a result, the output signal will be distorted somewhat (the non-linear curve interacts with it.) How much that may be acceptable is one of those design decisions that engineers make all the time.
So, that's about it. I'll now derive it from the above collector current equation so that you can see how it falls out using derivatives:
$$ \newcommand{\dd}[1]{\text{d}\left(#1\right)} \newcommand{\d}[1]{\text{d}\,#1} \begin{align*} I_\text{C}&=I_\text{SAT}\left[e^{^\frac{V_\text{BE}}{\eta\,V_T}}-1\right]\\\\ \dd{I_\text{C}}&=\dd{I_\text{SAT}\left[e^{^\frac{V_\text{BE}}{\eta\,V_T}}-1\right]}=I_\text{sat}\cdot\dd{e^{^\frac{V_\text{BE}}{\eta\,V_T}}-1}=I_\text{SAT}\cdot\dd{e^{^\frac{V_\text{BE}}{\eta\,V_T}}}\\\\ &=I_\text{SAT}\cdot e^{^\frac{V_\text{BE}}{\eta\,V_T}}\cdot\frac{\dd{V_\text{BE}}}{\eta\,V_T} \end{align*} $$
Since \$I_\text{SAT}\left[e^{^\frac{V_\text{BE}}{\eta\,V_T}}-1\right]\approx I_\text{SAT}\cdot e^{^\frac{V_\text{BE}}{\eta\,V_T}}\$ (the -1 term makes no practical difference), we can conclude:
$$ \begin{align*} \dd{I_\text{C}}&=I_\text{C}\cdot\frac{\dd{V_\text{BE}}}{\eta\,V_T} \end{align*} $$
From which very simple algebraic manipulation produces:
$$ \newcommand{\dd}[1]{\text{d}\left(#1\right)} \newcommand{\d}[1]{\text{d}\,#1} \begin{align*} \frac{\dd{V_\text{BE}}}{\dd{I_\text{C}}}&=\frac{\d{V_\text{BE}}}{\d{I_\text{C}}}=\frac{\eta\,V_T}{I_\text{C}}=r_e^{'} \end{align*} $$
Note: It's perfectly fine to use \$I_\text{E}\$ instead if \$I_\text{C}\$ when computing \$r_e^{'}\$. But in active mode situations where \$r_e^{'}\$ is important for AC analysis, the difference isn't worth worrying about. The actual circumstances for any real parts in a real circuit with wash out any such slight difference, anyway. So don't sweat it.
For ideal resistors, the voltage drop across a resistor is equal to the current through the resistor times the resistor's resistance. This is Ohm's law.
$$V = IR$$
This law, rearranged is
$$R = \frac{V}{I}$$
For a non-linear device, there is no single value of R that relates V and I in a linear way. We can still refer to a "resistance" \$R=\frac{V}{I}\$ as long as we understand that it is merely the ratio of V and I for particular values of V and I, and not a constant.
Small signal "resistance" is something different.
$$r = \frac{dV}{dI}$$
Small signal "resistance" tells us the change in voltage for a small change in current.
For a non-linear device,
$$\frac{V}{I} \ne \frac{dV}{dI}$$
(Except perhaps for some very specific values of V and I).
So, we cannot use r to determine either V or I. That is why r is not used in DC analysis.
Ohh, the answer is simple. Remember that BJT is a non linear device. i.e. collector current varies with base emitter voltage exponential. In DC analysis, we use that exponential equation to determine the voltages at different point of circuit wrt ground. However, sometimes we just assume that \$V_{BE}\$ is 0.6-0.7V, this simplify the calculations. However, in small signal AC analysis, we (approximately) model a BJT (a non linear model) with a linear circuit, and in every linear circuit we can only have linear devices like resistors, caps, inductors, independent source and dependent sources. Diodes, BJT, MOSFET are not linear devices
Note that it is perfectly fine to use exponential model for AC analysis, it is just that it would be very difficult to do that by hand.
When talking about the "small signal model" of a BJT, re (emitter resistance) is present in both AC and DC analysis.
I want to know why "r" parameters are considered in AC analysis but not in DC. Ac analysis is the observation of a signal applied to the circuit. DC analysis (also called bias analysis) is the analysis of the constant DC supply voltages applied to the circuit that effect overall circuit operation.
The device in analysis (the transistor) controls a large current with a small current ( the current flow from emitter to collector is controlled by the emitter to base current. ) The device's basic gain is calculated by the base current multiplied by the gain factor of the device (Beta Hfe). Later on, you will learn that the different circuit arrangements use these basic parameters in calculating the total circuit's gain and dictates how the gain is used (either by voltage gain or current gain) as each circuit topology has its own set of mathematical formulas to calculate the expected results from building that circuit.
The signal applied changes the base current from its static bias point, and the example is trying to explain this without adding the base bias current to simplify formula and explanation. The resistance: r' of e is the intrinsic resistance across the forward biased base to emitter junction in this example. Later you will understand that this intrinsic resistance changes with operating temperature and current limiting is applied (usually an addition of a emitter resistor) to keep the base current from overloading in a self destructive fashion called thermal runaway.
Not all transistors are built the same way and some have integrated resistors and diodes to better support the device in its normal applications its applied to. For example: you might find on a data sheet, a power transistor with a 50 ohm resistor internally between the base and emitter.
