# base impedance of a bjt amplifier I am currently reading Malvino's book as my text book. As stated in the picture "At low frequency (maybe DC) this impedance is purely resistive and defined as ....", I am not able to understand how did they derive the formula, also, is there any intuitive "feeling" for the base impedance? let me ask the question more precisely,first, do they mean the base impedance (maybe dynamic) as the impedance at the base "material" as shown in the picture below? • It's just ohm's law; Z or R = V/I. Dec 18, 2020 at 15:02
• yes but the voltage we are applying is between the base and emitter, then where is the emitter Dec 18, 2020 at 15:04
• It's still just ohm's law - the emitter is maybe 500 to 800 mV DC below the base terminal. The base-emitter is forward biased with DC and the formula in your question is ohm's law for the AC signals superimposed on that DC bias. Dec 18, 2020 at 15:07
• the question has been edited, can you just answer what I am thinking about the base impedance is corrrect or not? i'm thinking the base material as my resistance, am i correct? Dec 18, 2020 at 16:20
• @Sayan, Ah, your comment: "do they mean the base impedance (maybe dynamic) as the impedance at the base "material" as shown in the picture below?" seems confusing. The base material only is NOT the impedance. You need to consider the p-n junction as the impedance, NOT just the isolated base material. We don't have the Malvino's book. Perhaps you can use the following article and refer to the diagrams there for easier discussion. Common Emitter Amplifier - Electronics Tutorials electronics-tutorials.ws/amplifier/amp_2.html. Dec 19, 2020 at 10:57

The key words are "is defined as". This is a definition. There's no proof, it was simply chosen to simplify further analysis to take the term $$\v_{be}/i_b\$$ and give it the shorter name of $$\z_{in(base)}\$$.

there any intuitive "feeling" for the base impedance?

An impedance is a ratio of a voltage to a current.

The input impedance should then be the ratio of the input voltage to the input current.

And if the circuit being discussed (which you haven't shared) is a common-emitter amplifier, then the input voltage is the base-emitter voltage of the transistor ($$\v_{be}\$$), and the input current is the base current of the transistor ($$\i_b\$$), so

$$z_{in}=\frac{v_{be}}{i_b}$$

is exactly what we should expect for this circuit.

do they mean the base impedance (maybe dynamic) as the impedance at the base "material" as shown in the picture below?

First, yes impedance is a dynamic parameter. That is, it describes the behavior of a circuit element in a linearized AC analysis.

But it is a circuit-level parameter defined in circuit theory, not a device level parameter defined in device physics. With some approximations you can derive the impedance from the physics, but you can't point to some small region within the device and say you're measuring the impedance there.

The base-emitter voltage is defined between the base and emitter terminals, and the base current is the current into the base terminal. The physics (for example carrier distributions, etc.) in the emitter region will affect the base-emitter voltage as much as the physics in the base region will. The physics in the collector region will even have an effect (for example when the collector current contributes to a voltage rise in the parasitic resistance in the emitter region).

• the question has been edited, can you just answer what I am thinking about the base impedance is corrrect or not? i'm thinking of the base material as my resistance, am i correct? Dec 18, 2020 at 16:20
• @Sayan I really didn't like Malvino's book much (yes, I have it here.) I won't go into the reasons right now, though. The definition they are talking about can be readily derived from a simplified version of the DC Ebers-Moll model (level 1.) There is also a bulk impedance at the base, emitter, and collector, which are added into the level 2 Ebers-Moll model, but their equation doesn't recognize it. Are you looking to see the derivation of $r^{ '}_e$ or do you want a more physical description of why it exists? Or both?
– jonk
Dec 18, 2020 at 20:28
• @Sayan no, the bulk silicon material itself isn't the cause of the base impedance. Dec 18, 2020 at 21:00
• @jonk Heh, As an EE, Malvino's book screwed me up badly, because of its wrong explanation of how BJTs work. Unforunately a friends father gave me a copy when I was a kid, so those BJT-misconceptions had many extra years to "fester." A similar thing happened to Win Hill of AOA fame, preventing him from doing transistor designs: cr4.globalspec.com/comment/720374/Re-Voltage-vs-Current (His famous book "Art of Electronics" attacks those Malvino misconceptions, especially brutallay the AOA's lab manual.) Dec 18, 2020 at 22:46
• @wbeaty I was involved in teaching students (volunteer activity) at a local Community College using that text. I also attended the lectures and the teachers were TERRIBLE and, I think, as equally confused as the students! In my activity of trying to help the students understand the material and pass the course, I grew to truly HATE that book!! It's bad on so many levels, including the problem sets which are poorly handled and where you have to guess at them in order to know how to interpret them. (A skill the students lack.) That it is in its 8th edition provides still less excuse for it.
– jonk
Dec 18, 2020 at 22:53

can you just answer what I am thinking about the base impedance is corrrect or not? i'm thinking the base material as my resistance, am i correct?

If you do not want to trouble with the physics of semiconductors, still there is a chance to answer this question by the concept of dynamic resistance. Just imagine that there is a variable "resistor" RBE inside the transistor, between the base and emitter. When you change the input voltage VBE across it, its resistance changes as well... and this affects the rate of change of the base current IB.

For example, imagine that you begin increasing VBE. At some point (about 0.6 V), the resistance begins decreasing. Then, in Ohm's ratio IB = VBE/RBE, the numerator increases while the denominator decreases. As a result, the current depends on both variables and increases more vigorously than if the resistance was constant... as though, there is a lower static resistance between the base and emitter seen by the input voltage source. But this is an illusion - there is no lower static resistance... there is only a decreasing dynamic resistance.

## Graphical representation

EDIT. The operation of the input circuit can be illustrated graphically. For this purpose, the three elements of the input circuit part are divided in two parts - an imperfect voltage source VIN with internal resistance R and the base-emitter junction (diode). The resistor R is included for convenience- it enlarges the input voltage range, protects the junction and makes the operating point more stable.

The IV curve of the real voltage source is a vertical line (in black) that is shifted to the right with VIN and inclined to the left with an angle determined by R; the IV curve of the base-emitter junction (in red) is the well-known exponential diode curve. Since the same voltage is applied across both parts and the same current flows through them, their IV curves are imposed on the same coordinate system - Fig. 1. The intersection (operating) point A represents the present voltage VA and current IA... and this is the graphical solution of the circuit equation. Fig. 1. Diode as a dynamic resistor - a graphical representation (the diode IV curve is idealized). A resistor R is connected in series to the input voltage source for convenience

We can stop here as everyone does when presenting this circuit. This means taking the diode curve for granted. But I suggest to continue by imagining how this low differential resistance RDIFF can be obtained with the help of a varying static resistance RST. Thus we will visualize my initial explanations at the top where I suggested to replace the complex diode with a humble resistor. For this purpose, we replace the complex diode IV curve with the humble resistor IV line. It passes through the coordinate origin and has a slope determined by the current resistance RST.

Now, when we begin to increase VIN and its IV curve moves to the right, initially RST IV curve stays immovable and the current increases proportionally to VIN. But when VIN approaches the diode threshold (VD in the figure), RST begins decreasing simultaneously with the VIN increasing. Its IV curve begins rotating counterclockwise and the operating point changes its trajectory up.

And here the trick is - the input voltage source "sees" not the linear RST IV curve (as would be the case if it were static) but the trajectory of the cross (operating) point A... and this is the exponential diode curve.

If we want, we can conduct such an attractive experiment only by three elements - variable voltage source, constant resistor R and variable resistor (rheostat) RST... and, of course, someone that varies RST in such a way... Then the exponential diode IV curve will be an illusion created by a linear resistor.

This is a functional explanation of the phenomenon "dynamic resistance" that gives you an intuitive notion about the base-emitter junction behavior. With the same success, you can use it to explain the LED and Zener diode behavior... then, the dual transistor behavior... and finally, the negative resistance phenomenon.

• What do you mean by "dual transistor" behaviour? Dec 19, 2020 at 8:53
• @tlfong01, Thanks for the response; it means a lot for me. Diodes behave as voltage stabilizers. They keep up an (almost) steady voltage across themselves when the current through them varies. So, their IV curves have an (almost) vertical section. Contrary, transistors behave as current stabilizers. They keep up an (almost) steady current through themselves when the voltage across them varies. So, their IV curves have an (almost) horizontal section. I am not sure if "dual" is the right word in English... Dec 19, 2020 at 10:26

It's the "small signal impedance" between base terminal of the transistor, and emitter terminal of the transistor.

(change in V_be) / (change in I_b)

Note the value depends on the bias point, the larger current flowing from the collector into the emitter, I_c.

Note that it is in reality a complex value, i.e. there is a phase angle between the voltage and the current ... one of them can lead/lag the other. Leave that aside for the time being.

I would refer you to the explanation of transistors in the classic textbook Horowitz & Hill. (however be warned that despite being the all-time classic, it's not a great text for a beginner!)