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# base impedance of a bjt amplifier

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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?

bjtshareedit  follow  flagedited 13 hours agoCircuit fantasist5,53711 gold badge99 silver badges2626 bronze badgesasked 17 hours agoSayan5555 bronze badges

• 1It’s just ohm’s law; Z or R = V/I. – Andy aka 17 hours ago
• yes but the voltage we are applying is between the base and emitter, then where is the emitter – Sayan 17 hours ago
• 1It’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. – Andy aka 17 hours ago
• 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? – Sayan 16 hours ago
• @Sayan no, the material doesn’t determine the base impedance. If silicon was a perfect conductor (zero ohms) the base impedance would still be roughly the same. Instead, the base impedance is created by the diode’s very very thin, charged, and insulating layer called the “depletion layer.” Find explanations about diodes, and the nature of diode impedance. The moving electrons have thermal velocity/KE, so are able to get across this charged layer. So, the thin insulating layer at the pn junction behaves NOT like an insulator, but instead like a volt-controlled, var. resistor. Same in any diode – wbeaty 11 hours ago
• The input impedance (for AC signals) is shown on the datasheet of a 2N3904 little transistor on a graph with differences caused by current. – Audioguru 6 hours ago

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## 3 Answers

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 vbe/ibvbe/ib and give it the shorter name of zin(base)zin(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 (vbevbe), and the input current is the base current of the transistor (ibib), sozin=vbeibzin=vbeib

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).shareedit  follow  flagedited 15 hours agoanswered 17 hours agoThe Photon111k33 gold badges138138 silver badges263263 bronze badges

• 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? – Sayan 16 hours ago
• @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′ere′ or do you want a more physical description of why it exists? Or both? – jonk 12 hours ago
• @Sayan no, the bulk silicon material itself isn’t the cause of the base impedance. – wbeaty 11 hours ago
• @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.) – wbeaty 10 hours ago
• @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 10 hours ago

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.

This is a functional explanation of the phenomenon 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.shareedit  follow  flagedited 9 hours agoanswered 10 hours agoCircuit fantasist5,53711 gold badge99 silver badges2626 bronze badges

• What do you mean by “dual transistor” behaviour? – tlfong01 just now   Edit

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!)shareedit  follow  flagedited 13 hours agoanswered 13 hours agoPete W12644 bronze badges New contributoradd a comment

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