Understanding the base emitter voltage in a common emitter NPN BJT

[JC2000]

TL;DR Summary

I would be grateful if you could check my understanding of the base emitter voltage for a BJT.

$V_{BE}$ is basically the difference in voltage between the base terminal and the emitter terminal.

Normally when silicon is used and the transistor is biased to operate in the active region, $V_{BE}$ = 0.7 V approximately.

The way I understand this is that for an npn BJT, the applied voltage must be at least 0.7 to ensure that the depletion barrier is overcome by the forward bias.

The way this 0.7 V is applied is by connecting a resistor in series with the source voltage ($V_{BB}$) so that there is a voltage drop.

The reason $V_{BE}$ is always around this 0.7 V value and not much more is because the input characteristics of the common emitter BJT suggest that even for a small voltage increase at this point, the spike in current would be large enough so that the power dissipated would damage the device.

Is this reasoning correct?

 

[DaveE]

Yes, basically correct. The BE junction of a BJT is best thought of as just a simple PN (or NP) diode. The base current will in turn effect the amount of collector current in normal operation (i.e. the linear region when the collector current can change). But for the most useful models, the collector current has essentially no impact on the BE junction voltage or current; it just acts like a diode.

 

[Averagesupernova]

But for the most useful models, the collector current has essentially no impact on the BE junction voltage or current; it just acts like a diode.

In school this is one of the things that was drilled into us harder than about any other thing. How to configure transistor biasing to get a stable Q point.

 

[DaveE]

The reason $V_{BE}$ is always around this 0.7 V value and not much more is because the input characteristics of the common emitter BJT suggest that even for a small voltage increase at this point, the spike in current would be large enough so that the power dissipated would damage the device.

This is the right idea. However, in practice the amount of current available is limited by the external circuit (like the series base resistor you mentioned). This is because circuit designers don't want to break the transistors, so they arrange the external circuit to keep all of the currents and voltages within acceptable ranges.

Of course the 0.7V is a rough approximation. It's probably the higher end of the range you'll normally see for low power (signal and logic, maybe 0.5-0.7V or so), and the lower end for high power diodes (maybe 0.7 - 1.1V). It has a significant temperature dependence and there is usually some extra resistance in the structure (bond wires, metalization, bulk Si resitance, etc. ). Anything more precise than the 0.7V estimate that we all use would require you to look up the detailed data for the device you are actually using.

 

[JC2000]

1.Another thing I would like to understand is about the amplification process :

For amplification in a common emitter BJT, the BJT has to be biased so that the operating point is in the active region. This is done by varying $V_{BB}$ and $V_{CC}$ and the resistor values (?)

But isn't $V_{BB}$ also where the input signal comes in (and $V_{CC}$ where the amplified output is generated)? Since $V_{BE}$ is controlled by $V_{BB}$, what happens if the input signal voltage is so low that the depletion potential cannot be overcome by $V_{BE}$? In this case how would amplification occur?!

Have I misunderstood where input/output signals go in and out as well as the role of $V_{BB}$ and $V_{CC}$? If so, how is the biasing done and where is the input signal fed in and where is the output signal received?

2.The second thing I don't feel comfortable with is the variation in base current when it comes to the input characteristics for a common emitter BJT.

The way I see it is that as the voltage at the collector terminal is increased the reverse bias in the np (collector-base junction) increases. However the increase in potential also means that electrons are drawn more strongly from the emitter by the collector and thus fewer electrons are drawn for the base current.

However the explanation I seem to find in places suggests the in the base recombination of holes and electrons occurs, but since the collector voltage is high, fewer electrons recombine in the base region and somehow this drop in recombination in the base causes a drop in the base current! Don't understand how this explanation works.

Thank you for your time and help!

 

[cnh1995]

But isn't VBB also where the input signal comes in (and VCC where the amplified output is generated)

No, Vbb and Vcc are the bisaing voltage sources. They, together with the biasing- resistor network, set the dc operating point (Q point as you said) of the transistor.

In the basic transistor model that you are studying, input is fed across B-E terminals via a coupling capacitor. The output is taken across C-E terminals (through a decoupling capacitor). It can be roughly described as follows:
Take input signal x(t) >> add dc to it>> apply x(t)+dc across BE terminals >> get amplified signal y(t) + dc across CE terminals >> remove dc from amplified signal.
This addition and removal of dc component from signals is done using coupling-decoupling capacitors.

This is just a crude explanation. There are several "units" in an amplifier circuit viz. Input, output, biasing network, coupling-decoupling caps, feedback, stabilizing resistors etc.
You need to identify them and build your equivalent circuit accordingly.

 

[LvW]

Perhaps my answer sounds a bit too simple, but the basic explanation is not very complicated.

A suitable DC bias point is defined by a fixed DC voltage VBE - and, of course a suitable voltage VCE of some volts, which, however, has a minor influence on the most important parameter: Collector current Ic.

Hence, this current Ic depends exponentially on the base-emitter voltage only Ic=f(Vbe).
Therefore, any variation of Vbe - caused by an input signal voltage vin=d(Vb) at the base - will vary the collector current Ic by an amount d(Ic)=d(Vbe)*gm. According to Ohms law, this variation in Ic (signal current d(Ic)=ic) will cause a corresponding signal voltage d(Vc)=d(Ic)*Rc at the collector node.

The transconductance gm=Ic/Vt (Vt: temperature voltage) is identical to the slope of the exponential function Ic=f(Vbe) - measured at the DC operational point.

(Note that the base current IB was not mentioned - it determines the input resistance only and has no influence on the voltage gain)

 

[JC2000]

No, Vbb and Vcc are the bisaing voltage sources. They, together with the biasing- resistor network, set the dc operating point (Q point as you said) of the transistor.

So Vbb and Vcc are used to fix the operating point. Vbe and Vce are the voltage differences between the terminals (base emitter and collector emitter respectively). The reason Vbe and Vce are not the same as Vbb and Vcc is because of resistors used in the circuit. Practically, Vce can have a much greater range of values as it is simply dependent on Rc and Vcc. Vbe on the other hand hovers in the 0.5-1.1 V range (depending on the application) because it can essentially be viewed as a diode.

1. Why is a dc component added to the input signal?
2. In the diagram below of a common emitter BJT used as an amplifier, how is Vbb kept independent of Vin?

3. Could you explain this bit as well :

The second thing I don't feel comfortable with is the variation in base current when it comes to the input characteristics for a common emitter BJT.

The way I see it is that as the voltage at the collector terminal is increased the reverse bias in the np (collector-base junction) increases. However the increase in potential also means that electrons are drawn more strongly from the emitter by the collector and thus fewer electrons are drawn for the base current.

However the explanation I seem to find in places suggests the in the base recombination of holes and electrons occurs, but since the collector voltage is high, fewer electrons recombine in the base region and somehow this drop in recombination in the base causes a drop in the base current! Don't understand how this explanation works.

Thanks!

 

[cnh1995]

So Vbb and Vcc are used to fix the operating point. Vbe and Vce are the voltage differences between the terminals (base emitter and collector emitter respectively). The reason Vbe and Vce are not the same as Vbb and Vcc is because of resistors used in the circuit.

Yes, provided all these voltages are dc.

1. Why is a dc component added to the input signal?

To allow BE junction to accept/process the alternating input signal. The whole point of setting the Q-point is to enable the amplifier to accept and amplify an alternating signal.
A normal p-n junction only conducts in one direction. Add dc to it, and it will carry/conduct an alternating signal too, since the resultant voltage is unidirectional and well above its knee voltage.

You need to be fluent in Thevenin/Norton and Superposition theorems. They are the key to develop and understand equivalent circuits for analog systems.

I didn't understand your 3rd question fully. Can you elaborate?

 

[JC2000]

Thanks for the help!

2.In the diagram, what I don't understand is why have they not shown a constant voltage source $Vbb$ for biasing? The $V_b$ in the diagram refers to $V_{BE}$, right? That apart, what I am confused about is how do they ensure that $V_{in}$ does not affect the biasing $V_{BB}$?!

3. In the third question I am unable to understand why an increase in $V_{CE}$ causes a decrease in $I_B$.

4. Also, I don't understand how the amplifier action works.

When biased in the active region the collector draws electrons from the emitter and a minuscule fraction of those electrons go towards the base terminal giving rise to the base current. Thus the relation can be derived that a small base current corresponds to a very large emitter or collector current.

What I fail to see is where the input signal comes into the picture? Does the input signal cause an increase in the base current and thus an increase in the collector and emitter currents? If so, I don't understand how an increase in base current causes an increase in the emitter or collector currents. I understand mathematically that since α or β have high values, a slight increase in IB would mean a large increase in the collector or emitter currents. However practically it seems that the base current is simply a part of the collector current which branches away and so an increase in base current could be allowed by having a smaller emitter current...

 

[cnh1995]

The amplifier in the diagram is a practical one. There is only one source (Vcc) used for dc biasing. The Vcc>> R1>>R2 network is called voltage divider bias (one of the biasing techniques).
Also, the RE in the circuit is for thermal stabilization. I think you are considering too many things at once which is resulting in confusion.
I would not recommend that diagram if you want to learn amplifiers from scratch. Also, it is difficult to teach amplifiers in a forum thread.
Currently, you seem to be confused with operation of a transistor as a semiconductor device.
1) Forget amplifiers. Understand the working principle of a BJT.
2) Understand its regions of operation (active, saturation, cutoff).
3)Understand the input and output curves.
Here you will work only with dc. You don't need circuit representations for these 3 points, so no question of how many biasing batteries, resistors or where to connect them, ac or dc etc.
4) Work on dc biasing and its techniques.
5) Study the small signal amplification process. It will be lot easier if you have a good grasp on first 4 points. Here you will see the role of coupling and decoupling capacitors.
6) Work on small signal equivalent circuits. This requires thevenin, superposition, dependent sources etc and a bit of math. Here you will understand the amplifier action in terms of circuit theory, feedback techniques and their effects, and a lot of other cool stuff.
7) You can study thermal stability and techniques to achieve it etc.

This list is never ending...

Meanwhile, I request analog experts here to share their views.

 

[LvW]

3. In the third question I am unable to understand why an increase in VCE causes a decrease in IB
4. Also, I don't understand how the amplifier action works.

* I have tried to explain the basic principle of voltage amplification in my post #7.
Is there anything you do not understand? In this case, it is best to formulate a specific question.
Only then, it is possible to give you a clear and correct answer.

* As far as question no. 3 is concerned:
Why are you interested in secondary problems? A transistor is - in reality - a rather complex device. However, for most applications you can forget such rather unimportant effects, which are not necessary to understand the basic function of voltage amplification.
So - why are you asking for IB variations (caused by VCE variations)?
Yes - there is a certain influence (responsible for the slight positive slope of the output characteristics Ic=f(VCE) with IB=const). And it is, of course, possible to physically explain this relation. But only for some specific applications it is really necessary to take this effect into consideration (Early effect).
As a consequence, the BJT is not an IDEAL current source - but this does not affect its basic amplification principles.
Again, I have shown in post#7 that Ib (resp. the so-called current gain B or beta) plays no role for voltage amplification because the BJT is a voltage-controlled device.
Yes - of course, there is a DC input current IB and an signal input current ib=d(IB) which adds to IB - but the voltage amplification between base and collector is determined by the product d(Vbe)*gm only.

 

[Tom.G]

See this old post from 2018. The details of 'how a transistor works' are on pgs. 25-27 of the GE Transistor manual, with a needed prerequisite beginning on pg. 17. I really recommend starting on pg.1 though, which covers the atomic structure and conduction theory in a very straightforward manner.

https://www.physicsforums.com/posts/5976335/

In case that post disappears, here are the links it contains.
http://www.introni.it/pdf/GE - Transistor Manual 1964.pdf
https://books.google.com/books/about/RCA_transistor_manual.html?id=o05rAAAAMAAJ

Cheers, and learn much!
Tom

 

[JC2000]

Thank you for your response! I think as cnh1995 pointed out, I think I ought to go back to the basics and systematically understand each bit before posting specific questions as I would be better placed to understand the answers if I am clear about the basics! Nonetheless, thanks for the help! I will come back to the post once I go over the basics once more.

 

[JC2000]

See this old post from 2018. The details of 'how a transistor works' are on pgs. 25-27 of the GE Transistor manual, with a needed prerequisite beginning on pg. 17. I really recommend starting on pg.1 though, which covers the atomic structure and conduction theory in a very straightforward manner.

https://www.physicsforums.com/posts/5976335/

In case that post disappears, here are the links it contains.
http://www.introni.it/pdf/GE - Transistor Manual 1964.pdf
https://books.google.com/books/about/RCA_transistor_manual.html?id=o05rAAAAMAAJ

Cheers, and learn much!
Tom

Thanks! I think this is very much needed!

 

[LvW]

I think I ought to go back to the basics and systematically understand each bit before posting specific questions ...

Just one important recommendation: Be careful and critical when selecting knowledge sources. Do not blindly follow everything you will find in the internet.

Just one counter example: "Basic Electronics, Chapter 1: Conduction in Semiconductors, Prof. Sudhinra, Bangalore"

Quote: "Due to increase in temperature following parameters will change: ...Vbe increases at a rate of 2.4mV/deg. With increase in temperature the base current IB will increase and since IC=B*IB, IC is also increased."

This is complete garbage...
* VBE does not increase because it is kept stiff externally (as stiff as possible) due to the base voltage divider.
* The tempco d(Vbe)/d(T)=-2.4 mV/deg is NEGATIVE and means the following: The current Ic is strongly temperature-dependent (increases with temperature) - and if Ic is to be kept constant, we must externally REDUCE the voltage Vbe by -2.4mV/deg. (In practice this is done by a suitable emitter resistor).
* Hence, the relation Ic=B*IB plays no role at all (the BJT is a voltage-controlled device as already indicated by the tempco d(Vbe)/d(T) for Ic=const.).

 

[JC2000]

Just one important recommendation: Be careful and critical when selecting knowledge sources. Do not blindly follow everything you will find in the internet.

Just one counter example: "Basic Electronics, Chapter 1: Conduction in Semiconductors, Prof. Sudhinra, Bangalore"

Quote: "Due to increase in temperature following parameters will change: ...Vbe increases at a rate of 2.4mV/deg. With increase in temperature the base current IB will increase and since IC=B*IB, IC is also increased."

This is complete garbage...
* VBE does not increase because it is kept stiff externally (as stiff as possible) due to the base voltage divider.
* The tempco d(Vbe)/d(T)=-2.4 mV/deg is NEGATIVE and means the following: The current Ic is strongly temperature-dependent (increases with temperature) - and if Ic is to be kept constant, we must externally REDUCE the voltage Vbe by -2.4mV/deg. (In practice this is done by a suitable emitter resistor).
* Hence, the relation Ic=B*IB plays no role at all (the BJT is a voltage-controlled device as already indicated by the tempco d(Vbe)/d(T) for Ic=const.).

!

Thanks for the heads up!