I am having trouble understanding common emitter amplifiers.

Can someone please explain how RE1 is limiting the gain of the amplifier in this circuit?

enter image description here

  • \$\begingroup\$ Do you understand that it's the base-emitter voltage that controls the collector current? Consider what happens to that voltage as the emitter current increases. \$\endgroup\$
    – Hearth
    Nov 11, 2023 at 0:07
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    \$\begingroup\$ Increased drop across Re closes the transistor. It´s because this increased drop tries to put the emitter higher in compare with base(the base swing is “still same”). \$\endgroup\$ Nov 11, 2023 at 0:40
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    \$\begingroup\$ Another point of view (easier to understand): The same voltage swing across Re in case of higher Re causes smaller emitter current so also smaller collector current. \$\endgroup\$ Nov 11, 2023 at 0:47
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    \$\begingroup\$ @Marcus Müller, What is so unclear about the question? The guy is just asking about the role of Re in the operation of a common-emitter amplifier stage... and gets very good answers. You have at least proven yourself to be a reasonable person, why are you doing this? Why do the same people do it all the time and the others don't even think of doing it? Ask yourself this question… \$\endgroup\$ Nov 11, 2023 at 12:00
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    \$\begingroup\$ I fully agree to the above comment. There is a clear description of the problem: Understanding the effect of Re1 as far as the gain is concerned. Question to M. Müller: When sombody speaks about "gain" for a common emitter stage, I have no problem to see "what "gain" is in this context" (your words). \$\endgroup\$
    – LvW
    Nov 11, 2023 at 13:21

4 Answers 4


Preface -- The Quiescent State

Let's make one simplification, at first: thevenizing the base resistor divider pair as follows:


simulate this circuit – Schematic created using CircuitLab

You should be satisfied that the circuit on the right is the same as your circuit. Just re-written slightly.

Now let's assume that all of the capacitors are uncharged at \$t=0^+\$, when power is applied to the circuit and that the signal input is held down, tied at ground.

At \$t=0^+\$, \$v_b=v_\text{in}+v_{c_b}=0\:\text{V}\$ and \$v_{e_1}=0\:\text{V}+v_{c_e}=0\:\text{V}\$ and so it should also be clear that between these two voltages it must be that \$v_e=0\:\text{V}\$ as there's no other place for it to be. The base-emitter junction has \$0\:\text{V}\$ across it and so all that can happen are very tiny leakage currents -- way below anything we care about. So \$v_c=V_{_\text{CC}}\$ and, almost unnecessary to say, but \$v_\text{out}=V_{_\text{CC}}\$. That's the starting condition of everything, at power-on.

As time passes (slowly), \$C_b\$ charges through \$R_\text{th}\$ until \$v_b=V_{\text{th}}\$. Not quite, actually. Because there is a small base current and this will lead to a voltage drop across \$R_{\text{th}}\$. So \$v_b\$ will actually be \$v_b=V_{\text{th}}-R_{\text{th}}\cdot I_b\$. But BJT's vary a lot in their behavior, so we really can't say a lot about \$I_b\$, yet. Except that it is hopefully small enough (or \$R_{\text{th}}\$ is small enough) that the voltage drop isn't large -- typically a voltage drop of about 1% of \$V_{_\text{CC}}\$, as a guidance.

An actual BJT can vary enough in its behavior that we cannot really say exactly what \$v_\text{be}\$ will be, but for guessing purposes and assuming a silicon BJT we can say it is likely nearer to \$700\:\text{mV}\$, than not. Might be a little less, or a little more, as any 10-fold change in the collector current will mean about \$60\:\text{mV}\$ difference in our guess about \$v_\text{be}\$. But as you can see, that's less than a 10% error for a factor of 10 in the collector current.

So, it's not an unreasonable starting point and this means also that \$v_e\approx v_b-700\:\text{mV}\$ (less about 1% of \$V_{_\text{CC}}\$ as a rule.)

\$R_{e_1}\$ and \$R_{e_2}\$ form another voltage divider and they have been acting also as a Thevenin equivalent to charge \$C_e\$ as \$v_b\$'s voltage has risen up. So the capacitor \$C_e\$ will charge up to about \$\frac{v_e}{1+\frac{R_{e_1}}{R_{e_2}}}\$.

And that's about where everything will reach after enough time has passed (without signal applied.)

I went through all that to convince you that \$C_b\$ and \$C_e\$ will charge up to some non-zero voltage and then stop charging. So they will hold some non-zero voltage across them once the circuit reaches its equilibrium point or DC operating point. We can't compute the exact numbers because BJTs vary a lot more than resistor values or power supplies do. But we can imagine that they will settle down somewhere near where we designed them to be. So let's go with that.

AC Gain

Now, we get to the important part of your question. But with the above setup, it's now easier to explain.

Suppose a very tiny signal occurs at \$v_\text{in}\$. Let's all it \$\Delta v_\text{in}\$. Because \$C_b\$ already has charged up to \$v_{c_b}\$, this change will be added to \$v_{c_b}\$ to get \$v_b=v_{c_b}+\Delta v_\text{in}\$. This is a sudden change. So sudden that \$C_b\$'s voltage doesn't have enough time to change with the signal.

When \$v_b\$ is suddenly changed like that, the question is, "What about \$v_\text{be}\$, in response?" Well, capacitor \$C_e\$ also has a voltage across it, too. And the change is so sudden that \$C_e\$'s voltage also doesn't have enough time to change with the signal. So \$v_{e_1}\$ doesn't change, yet. It can't as \$C_e\$ is holding it steady for now.

Meanwhile, since \$v_\text{be}\$ may also temporarily be considered fixed too, it must be the case that \$v_e=v_e+\Delta v_\text{in}\$ in response to the base's voltage change. This means that the change in voltage across \$R_{e_1}\$ is \$\Delta v_\text{in}\$ which will cause a change in the current of \$R_{e_1}\$ of \$\Delta i_{e_1}=\frac{\Delta v_\text{in}}{R_{e_1}}\$.

That change in current in \$R_{e_1}\$ means a change in the BJT's emitter current by just that amount. And this impacts the collector current, directly. So \$\Delta i_{c_1}=\frac{\beta}{\beta+1}\cdot\frac{\Delta v_\text{in}}{R_{e_1}}\$. And that collector current change appears as an immediate change in the voltage drop across \$R_c\$. The change will be \$\mid \Delta v_c\!\mid\,=\frac{\beta}{\beta+1}\cdot\frac{\Delta v_\text{in}}{R_{e_1}}\cdot R_c\$.

Now, we can simplify \$\frac{\beta}{\beta+1}\$ and just call it 1, for most BJT cases, without losing sight of the bigger picture.

So the voltage drop change across \$R_c\$ will be \$\Delta v_{_{R_c}}=\Delta v_\text{in}\frac{R_c}{R_{e_1}}\$.

The voltage gain appears to be just \$A_v=\frac{R_c}{R_{e_1}}\$.

Now, just as a final note, since \$\Delta v_{_{R_c}}\$ follows the sign of \$\Delta v_\text{in}\$ but is subtracted from \$V_{_\text{CC}}\$, the collector voltage, \$v_c\$, will decrease when the voltage drop increases, and visa versa.

So the voltage gain is actually written out as \$A_v=-\frac{R_c}{R_{e_1}}\$, because it is in the opposite direction of the signal change.

Secondary Issues

Now, the truth is that when the collector and emitter currents change, it's also true that \$v_\text{be}\$ changes in response. So while I said that the change at the base also appears to be the same as the change at the emitter, the truth of it is that if \$\Delta v_\text{in}\$ is positive, then the collector current slightly increases in response. But \$v_\text{be}\$ also then increases slightly (but only a very tiny amount.) This increase means that the emitter change will just a little bit less than expected and that means that the change in the collector current will be slightly less than I wrote, above. This reduces the actual voltage gain.

This reduction can be predicted by the use of a new variable called \$r_e^{\:'}=\frac{V_T}{I_q}\$ where \$V_T\$ is about \$26\:\text{mV}\$ and \$I_q\$ is taken to be the quiescent state's collector current, whatever that works out to be. This is approximate, itself, because \$r_e^{\:'}\$ actually isn't a constant, but itself varying with the collector current as it changes with the signal. But it's a good enough approach.

So the final voltage gain is actually taken to be closer to \$A_v=-\frac{R_c}{R_{e_1}+r_e^{\:'}}\$. That modification can be important at times. Especially when \$R_{e_1}\le 10\cdot r_e^{\:'}\$. (Which is definitely always true when \$R_{e_1}=0\:\Omega\$ in the case of AC-grounded CE stages.)

This is one reason why understanding the quiescent state itself helps in understanding the realized voltage gain.


For a first level approximation for AC gain make these assumptions for a nearly perfect transistor:

  1. The transistor has very high gain.
  2. As an emitter follower, the gain is equal to 1.
  3. \$i_e = i_c\$
  4. \$C_E\$ and \$C_b\$ are AC shorts (not a property of the transistor).

For simple AC analysis:
$$i_e = {v_{in} \over R_{e1}}$$

$$v_{out} = i_c R_c $$

Combining the above equations and Item 3 above:
$$v_{out} = {v_{in} \; R_c \over R_{e1}} $$

Thus, the gain is: $${v_{out} \over v_{in}} = {R_c \over R_{e1}} $$

This above completely ignores the emitter resistance which is given by:

$$ r'_e = {V_T \over I_e}$$ Where \$I_e\$ is the DC emitter current.

\$V_T\ = {kT \over q}\$, where k is the Boltzmann constant, T is absolute temperature, and q is the charge on an electron. \$V_T\$ is approximately 25 mV at room temperature.

\$r'_e\$ is added to \$R_{e1}\$ which gives us a better approximations of the gain: $${v_{out} \over v_{in}} = {R_c \over {R_{e1}+r'_e}}$$

This is an overly simplified model for AC gain of the common emitter amplifier, but it is accurate enough to get an understanding of the gain.

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    \$\begingroup\$ Since the TO has problems to UNDERSTAND the circuit, we should explain to him what the quantity re in your gain expression really is. it is NOT the "emitter resistance" ! It is nothing else than the inverse transconductance gm of the transistor (re=1/gm). At first, it belongs not to the emitter and, secondly, it is not a resistance at all! And the transconductance is the slope of the exponenetial function Ic=f(Vbe) \$\endgroup\$
    – LvW
    Nov 11, 2023 at 9:33

Think about Q1 having very high gain so you could neglect the base current .Think also that the output loading is very low ie say a standard high impedance scope probe .If you can for a moment accept these statements then it follows that collector current equals emitter current to a good approximation. Now it is clear that gain is limited by vthe ratio of Rc and Re1 because Ce bypasses RE .


Here is my (short) derivation of the gain formula, which answers your question (applying Ohms rule and a simple signal voltage sum at the input):

  • Gain A=Vout/Vin=-icRc/(vbe+ie*Re1)....... (Note that we can set ie=ic, which is the "normal" and allowed simplification; Re2 is bridged by Ce and has no effect for signal voltages)

  • Dividing the above expression by vbe and setting ic/vbe=gm (transconductance gm is the slope of the exponential function ic=f(vbe) ), we arrive at: A=-gmRc/[(1 + gmRe1]

  • For Re1=0 we arrive again at the known gain expression (without feedback): Ao=-gmRc

  • Comment: Some authors use the abbreviation re=1/gm and write A=-Rc/[(re + Re1)]. However, one should know that re - in spite of the dimension volts/ampere - is NOT a resistance. It nothing else than a dynamic/differential quantity which depends on the quiescent DC current (slope of the Ic=f(Vbe) function). Note that in the above derivation all quantities with small letters (i, v) are small-signal values only.


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