12
\$\begingroup\$

I'm going to take an example of a simple common emitter amplifier. Forget about biasing and things for now, but focus on the crux of this circuit. As how I understand it, a voltage between the base node and the emitter node is varied which is ultimately amplified by the transistor, causing an inverted (amplified version) of the original signal to appear at the collector node.

Right now, I'm working through a book; Sedra/Smith, Microelectronics.

Throughout the chapter I'm working through, it says that in the active region, Vbe is assumed to be 0.7V. This just does not make sense to me, how can Vbe stay constant when that itself is the input variable for an amplifier stage? This might have started to make sense to me if I was looking at at a CE stage with an emitter resistor (emitter degeneration), where the remaining voltage could be dropped across the resistor. But this is not the case, so enlighten me!

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • 4
    \$\begingroup\$ As a side note: never think of a bipolar transistor as a U to U amplifier. Bipolar transistors are current (iB) to current (iC) amplifiers (iC = hFE*iB). If you put an ideal voltage source in the transistor's base without limiting the current iB, you will fry the transistor. \$\endgroup\$ – Chris Mar 13 '15 at 18:13
  • \$\begingroup\$ Even if you do so (voltage source at base without limiting the current), respecting the limits of the Vbe of the transistor? Isn't the transistor current equation fundamentally Ic=Isexp(Vbe/Vt) (indicating the transistor is more ultimately dependent on the voltage?). I think you are right in saying that the output is current, however I think the input is a voltage. Hence I believe it is a transconductor. \$\endgroup\$ – midnightBlue Mar 13 '15 at 21:34
  • \$\begingroup\$ I guess it's a matter of perspective. You could just replace vBE with rPI*iB and the equation is current dependent. But what really makes carriers inside a bipolar flow are the injected carriers in the base.Plus a lot of people make this mistake:"oh, I'll just put 1V on Vbe and the transistor will be on", only to find out is fried.Vbe is a diode in which you inject a current that avalanches a much bigger one.Now, a CMOS transistor is truly a voltage controlled current source, a transconductor. \$\endgroup\$ – Chris Mar 13 '15 at 23:36
  • \$\begingroup\$ I guess it could be perspective. I actually don't know enough to say. A current that avalanches a bigger one is an interesting way to think about it. \$\endgroup\$ – midnightBlue Mar 14 '15 at 2:40
  • \$\begingroup\$ It isn't a constant 0.7V, and your quotation doesn't say otherwise. It is fairly constant within about +/-10% of that, for small signal NPN transistors, so 0.7V is used as a simplifying assumption, which is what your quotation actually says. For the transistors I usually use it varies between 0.2-0.65V. \$\endgroup\$ – user207421 Jul 2 '16 at 5:39

10 Answers 10

18
\$\begingroup\$

Inverting the collector current equation:

$$i_C = I_Se^{\frac{v_{BE}}{V_T}}$$

yields:

$$v_{BE} = V_T\ln{\frac{i_C}{I_S}}$$

For example, let

$$V_T = 25mV$$

$$I_S = 1 fA$$

$$I_C = 1mA$$

With these values, find that

$$V_{BE} = 0.691V$$

Now, double the collector current and find that

$$V_{BE} = 0.708V$$

Increasing the collector current 100% only increased the base-emitter voltage 2.45%

So, while it is not true that the base-emitter voltage is constant, it is not a bad approximation to consider it constant over a relatively wide range of collector current.

| improve this answer | |
\$\endgroup\$
10
\$\begingroup\$

Vbe in a silicon transistor, acts like a silicon diode would. The Forward Voltage Drop, after a certain amount of current is passed, rises sharply. Increasing the current makes a negligible Vf difference at that point.

enter image description here

Note that the Vf is different for Germanium Diodes, and Transistors, naturally.

| improve this answer | |
\$\endgroup\$
4
\$\begingroup\$

The Ebers-Moll model for the emitter current in a bipolar transistor is:

\$I_e \approx I_{es} e^{\frac{V_{be}}{V_t}}\$

Where \$I_e\$ is the emitter saturation current, \$V_t\ \approx 26mV\$ is the thermal voltage, and \$V_{be}\$ is the base to emitter voltage. For a value of \$I_{es} = 10^{-12}\$ (in the typical range for a small signal silicon device), consider the following Wolfram Alpha plot of the above equation:

Ebers-Moll plot

enter image description here

The Y axis is current and is on a logarithmic scale. You'll notice that for values of \$V_{be}\$ in the range from 0.55 to 0.7 volts, the current through the transistor has an extremely wide range - from microamperes at the low end to an amp at the high end. This is due to the exponential behavior of the governing equation.

For the purpose of analysis, assuming that the \$V_{be}\$ of a small signal silicon transistor for when it is in this range for when in the active region is a reasonable assumption, since if the value of \$V_{be}\$ were significantly smaller only a tiny current would be flowing through the transistor, and if it were much larger, the transistor would have to be passing amperes of current, which is not physically possible for such a device.

Again note that this is only an assumption to facilitate analysis; the \$V_{be}\$ of a specific small signal silicon device in a specific circuit should be in this range if it is in the active region, but the actual value will depend on circuit specifics, device parameters, temperature, and other factors.

The circuit you present is not a good example of a situation to apply this simplification, since as you say, the \$V_{be}\$ of the circuit is the only user-definable parameter. You are free to select any input voltage you wish in this circuit, but since the emitter is connected directly to ground, whatever voltage you apply will be your \$V_{be}\$. There will therefore be only a narrow range of input voltages which will allow the circuit presented to be in the active region; a little to low and the transistor will be cut off, a little too high and an enormous current will flow through the base-emitter junction, causing the collector voltage to pull down due to the load resistor, putting the transistor into saturation.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Okay, so what would happen when the input signal of my simple amplifier go above 0.7V? Are you saying the transistor would be forced to saturation? \$\endgroup\$ – midnightBlue Nov 23 '13 at 23:01
  • \$\begingroup\$ @user1255592 It won't happen at exactly 0.7 volts in a real circuit (probably lower) but if you keep pulling the base voltage up with respect to ground in that circuit, yes that's what will happen. \$\endgroup\$ – Bitrex Nov 23 '13 at 23:03
  • \$\begingroup\$ @user1255592 In a common emitter amplifier with emitter degeneration, the Vbe also varies, but the emitter resistor provides feedback to keep the Vbe excursion in a very small range, and the transistor remains in the active region. In such a circuit it is reasonable to use the "0.7" volt approximation, since the deviation from this value due to the signal is very small (though it must occur for the transistor to amplify.) \$\endgroup\$ – Bitrex Nov 23 '13 at 23:10
  • \$\begingroup\$ Thanks for the reply! Thatis starting to make sense, so what would the typical on voltage for this configuration of the transistor be? Around 0.5V? Is this a good reason why we use the emitter resistor? I keep hearing that adding emitter resistor = make circuit more linear. By linear, do they mean this widening of the input voltage range? EDIT: I think you just simultaneously answered my question! \$\endgroup\$ – midnightBlue Nov 23 '13 at 23:10
  • \$\begingroup\$ So then, how much would you say the input would vary in a simple common emitter with degeneration? Is it correct to say that the only play I have is between 0.5V to 0.7V? So then, is it a good idea to say that a good base DC biasing voltage is 0.6V? \$\endgroup\$ – midnightBlue Nov 23 '13 at 23:17
3
\$\begingroup\$

The Fermi level is the average energy of mobile electrons (or holes) in semiconductor material. The Fermi levels are expressed in electron volts (eV), and may be viewed as representing the voltage seen by the electrons.

Intrinsic silicon (and germanium) has the Fermi level halfway between the top edge of the valence band and the bottom edge of the conduction band.

When you dope the silicon to P-type, you add a lot of holes. Now you have a lot more available carrier states down near the top of the valence band, and this pushes the Fermi level down close to the valence band edge. Similarly, when you dope N-type, you add a lot of electrons, which creates a lot more available carrier states up near the conduction band, and pushes the Fermi level up close to the conduction band edge.

For the doping levels typically found in a base-emitter junction, the difference in Fermi levels between P and N sides is about 0.7 electron-volts (eV). This means that an electron travelling from N to P dumps 0.7 eV of energy (in the form of a photon: This is where light-emitting diodes get their light: the materials and doping are chosen such that the difference in Fermi levels across the junction gives rise to photons at the desired wavelength, as determined by Planck's equation). Similarly, an electron moving from P to N must pick up 0.7 eV somewhere.

In short, Vbe is essentially just the difference in Fermi levels on the two sides of the junction.

This is Semiconductors 101 material, in that you have to understand this before you go any farther. The fact that it is 101 does NOT mean it is simple, or easy: It takes two semesters of calculus, two semesters of chemistry, two semesters of physics, and a semester of differential equations, to lay down the prerequisite groundwork for the semiconductor theory class that explains all of the above in gory detail.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Gracefully explained. Thank you kind sir for your insight. This has opened my eyes to the material science of semiconductors. And has given me a better fundamental understanding of the movement of energy. I will definitely follow this up with some studies. Do you have any recommendations of resources for that? \$\endgroup\$ – RedDogAlpha Sep 14 '17 at 2:28
  • \$\begingroup\$ Take a competent semiconductor materials and devices class at a good engineering school. Plan on, as I said, two semesters of calculus, two semesters of chemistry, two semesters of physics, and a semester of differential equations. I got lucky: I took the class from a guy who (a) loved the material (b) loved to teach (c) was REALLY good at teaching. I later found out that the word on him was that you worked TWICE as hard for the grade in his class as any other, and it was worth the effort. \$\endgroup\$ – John R. Strohm Sep 15 '17 at 1:34
1
\$\begingroup\$

The base emitter junction is a PN junction or you can consider that as a diode. And the voltage drop across a silicon diode when forward biased is ~0.7V. That is why most of the books write \$V_{BE} = 0.7V\$, for an NPN silicon transistor with forward biased emitter junction at room temperature.

But \$V_{BE}\$ for a particular transistor is not constant. It varies with temperature and current through the junction.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ OP was asking specifically when there is no base resistor. \$\endgroup\$ – sherrellbc Jul 20 '14 at 17:45
1
\$\begingroup\$

Good question. The oft-quoted Vbe of 0.7V is only an approximation. If you measure the Vbe of a transistor that is actively amplifying it will show a Vbe of 0.7V or thereabouts on a multimeter, but if you could zoom in on that 0.7, as you can with an oscilloscope, you would see tiny variations around it, so at any one instant in time it might be 0.6989V or 0.70021V as the input signal that sits on that bias - the one you want amplified - fluctuates about that bias point.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

Firstly, you should notice that \$v_{BE}\$ is not the only input variable of a transistor. For instance, if you manage to keep \$v_{BE}\$ constant somehow, you can still change the transistor coletor current by varying \$v_{ce}\$ (Early effect).

Now, I believe that you are confusing \$v_{BE}\$ with its cc component \$V_{BE}\$. That is, \$v_{BE}=V_{BE} + v_{be}\$. In active region, \$V_{BE}\$ is constant by definition and should be extremely close to \$0.7V\$ for silicon devices. Therefore, now the input of the active mode transistor is ca component \$v_{be}\$ which is freely to vary under the constraints of the polarization circuit.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ To make it clear: Vbe is, of course, not constant because it is the input quantity which controls the output quantity (current). With other words - changing the output current resp. the output voltage (created across the coll. resistor) in a typical amplifier stage REQUIRES that the input voltage changes. \$\endgroup\$ – LvW Jul 20 '14 at 16:31
  • \$\begingroup\$ What is ca and cc components? I wrote this question forgetting about small signal/large signal 'components' because that confuses me too. If we get a prolonged higher voltage input, at what point do you call it a large signal input and when do we call it small signal. What if we had a very large input signal swing, that cannot fit the small input range required for this analysis. \$\endgroup\$ – midnightBlue Jul 25 '14 at 12:40
  • \$\begingroup\$ LvW that's why I wrote this question! I find it confusing that books teach the Vbe is constant when it is the input variable. @user3084947 how can we alter Vce without altering the supply rails or changing the resistors? \$\endgroup\$ – midnightBlue Jul 25 '14 at 12:45
  • \$\begingroup\$ @midnightBlue In order to understand what is ca or cc componente, you should study signal processing theory, specifically, generative models based on sinusoidal oscilations such as Fourier series. \$\endgroup\$ – André Cavalcante Jul 26 '14 at 14:53
0
\$\begingroup\$

Your question is excellent.

Transistors, in theory only, are fully closed for any Ube < 0.7V and are fully open for any Ube > = 0.7V. In some low power transistors, this idealised Ube can be 0.6V or 0.65V.

In practice, Ube can range from 0V to 3V even more for high power transistors. In practice, transistors do get slightly open for any Ube > 0 and continue to increase their openness with the increase of Ube.

However, as mentioned, the dependence of Ice or, better said, Rce from Ube is heavily non linear after a given point and, thus, the increase of Ice does not lead to a huge increase of Ube, yet, there is such.

Below 0.7V, the increase of Ice can be somewhat linear and this depends on the transistor.

The maximal Ube at the maximal Ice is easily 2.5V to 3V for huge power transistors and Ice's greater than 25A.

One thing for sure : in analogue applications, the dependence of Ice from Ube must definitely be considered, mainly for high power or high current transistors.

Have a look at 2N5302 which has Ube = 3V at Ice = 30A and Uce = 4V.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ Welcome to EE.SE! You might consider making your answer more readable using MathJax formatting for your variables with subscripts. \$\endgroup\$ – user2943160 Jul 2 '16 at 1:31
  • \$\begingroup\$ "Transistors, in theory only, are fully closed for any Ube < 0.7V and are fully open for any Ube > = 0.7V." To me, tis statement sounds rather confusing and/or misleading (see the well-known Shockley-equation, used in the Ebers-Moll transistor model). \$\endgroup\$ – LvW Jul 2 '16 at 7:57
0
\$\begingroup\$

At end of this posting, you'll know how to compute voltage gain of a bipolar.

Lets examine a table of Vbe versus Collector Current, for an imaginary bipolar:

VBE Ic

0.4 1uA

0.458 10uA Notice 58mV more Vbe gives exactly 10X more current.

0.516 100uA

0.574 1mA

0.632 10mA

0.690 100mA [transistor is HOT, so the current may runaway and melt the transistor (a known risk with bipolars biased with constant base voltage)]

0.748 1AMP transistor is HOT

0.806 10Amps transistor is HOT

Can we actually operate a bipolar transistor over 1uA to 10Amps collector current? Yes, if its a power transistor. And at higher currents, this fine table -- showing 58 milliVolts more Vbe produces 10X more current --- loses accuracy because the bulk silicon has a linear resistance and curve tracers will show that.

How about smaller than 58mV changes? Vbe Ic 0.2 volts 1nanoAmp (approx. 3 factors of 58mV below 1uA at 0.4v) 0.226 2.718 nanoAmp (the 0.026v of physics gives E^1 more I) 0.218 2.000 nanoAmp 0.236 4.000 nanoAmp 0.254 8.000 nanoAmp (you'll find N*18mV in voltage references)

OK, enough tables. Lets view the bipolar transistor similar to vacuum tubes or MOSFETS...............as transconductors, where changes in Input Voltage cause changes in Output Current.

Bipolars are fun to use, because we know EXACTLY the transconductance for any bipolar, if we know the DC collector current (that is, with no input AC signal).

For shorthand, we lable this the 'gM' or 'gm', because vacuum tube databooks used the variable "mutual transconductance" to explain how the Grid voltage controlled Plate current. We can honor Lee deForest by using gm for this.

The gm of a bipolar, at 25 degrees Centigrade, and knowing kt/q is 0.026 volts, is -------> Ic/0.026 and if the Collector current is 0.026 amps (26 milliAmps), the gm is 1 amp per volt.

Thus 1 millivolt PP on the base causes 1milliAmp PP collector AC current. Ignoring some distortion, which you can predict using Taylor Series. Or Barry Gilbert's writings on IP2 and IP3 for bipolars.

Suppose we have 1Kohm resistor from collector to +30 volt, carrying 26mA. The Vce is 30 - 1K*26ma = 30 - 26 = 4 volts, so the bipolar is in "linear" region. What is our gain?

Gain is gm*Rcollector or 1amp/volt * 1,000 ohms or Av=1,000x.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Unfortunately, the DEFINITION ofthe tranconductance gm is not given. It is the slope of the exponential Ic=f(Vbe) characterisrtics gm=d(Ic)/d(Vbe). Because of the exponential form the result is gm=Ic/Vt. \$\endgroup\$ – LvW Jan 31 '17 at 10:28
0
\$\begingroup\$

Your question is:

how can Vbe stay constant when that itself is the input variable for an amplifier stage?

The easy answer is that, well, it isn't:

  1. \$V_{BE}\$ does not remain strictly constant in the active region, but for the purpose of the DC analysis of the circuit we can safely assume that it is. Most of the answers to your question have focused on developing (pretty well) the physical explanation behind this assumption. However, I think you're looking for something else
  2. \$V_{BE}\$ may be your "input variable", but from the BJT perspective what's relevant is \$I_b\$. Remember: the BJT is a current amplifier device. Of course you can derive a voltage gain, but only after proper biasing and loading.

But now I'll try to answer what I believe is your actual doubt. I think that you're mixing up concept from DC analysis and small signal analysis of the circuit.

What you call "input variable" has in fact an AC component on top of a DC component:

AC+DC components

The DC component is there just for biasing the base. That's the "constant \$V_{BE}\$" you're referring to. BUT (and this is the important part), the AC component is the signal that we actually want to amplify. And, of course, it's not constant at all.

I think that now you can see where your confusion comes from. Don't worry, it's a pretty common confusion. I've always thought that most teachers and books don't do a good job in explaining how to think in terms of DC analysis vs. small signal analysis and which assumptions should be applied in each one.

Summing it all up:

  1. When analysing the DC circuit, we ignore the AC signal (actually, we set it to zero) and assume \$V_{BE}\$ to be constant at 0.7V. If we want to be more precise, we can compute the actual \$V_{BE}\$ value consistent with the actual \$I_b\$. This will set the quiescent point of the amplifier (the DC values which the AC signals fluctuates around).

  2. When analysing the small signal circuit, we ignore the DC voltages (actually, we set all of them to zero) and just focus in the AC signal, which isn't constant. Note how \$R_c\$ becomes grounded in the circuit diagram below because \$V_{cc}\$ has been set to zero for analysis purposes. Also note the subtlety: the AC signal is often referred to as \$v_{BE}\$, whereas the DC bias is \$V_{BE}\$.

CE small signal circuit

Note: you can find the source for the diagram above here.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.