# Why do we need 4 resistors for a common emitter amplifier?

I'm studying the 4 resistors bias configuration, I'm trying to understand why we need everyone of these resistors.

If I understood correctly, Re is used to make the Q-point stable, while R1 and R2 are used to determine Vbe, but I can't understand why we need RL.

I also would like to ask why in the common base configuration we have a bypass capacitor in parallel with R2 and in the common collector we have a capacitor between the collector and the ground. Why we need to bypass these resistors in these cases? I understood that in the common emitter we bypass Re because it reduces the gain, is it the same thing in the others configurations?

Thanks

Answer We need RL otherwise the output voltage would always be equal to the supply voltage, this is true for the common emitter amplifier,but if we take the emitter voltage as output voltage(common collector configuration) we don't need RL anymore. Moreover the DC component of the output voltage depends on RL

About the bypass capacitors, we use them because some resistors reduce the gain(but we need these resistors during the bias).

Thanks to eveyone.

• No RL means no output. Jun 12, 2020 at 14:56
• I think this is true for common emitter and common base, but it is not a good explanation for common collector and common emitter with degeneration reristor
– Juà
Jun 12, 2020 at 15:02
• Your title says common emitter and this means you need an output at the collector. Jun 12, 2020 at 15:04
• RL is whatever the amplifier is driving. If it wasn't there, you wouldn't need the amplifier.
– user16324
Jun 12, 2020 at 16:43
• RL is the genuine load... and the voltage drop VRL across it is the genuine output voltage. There is only a "small" problem - the load is not grounded. Then we apply little trick taking the complementary voltage VCC - VRL = VC instead the original load voltage VRL. In other words, we connect the load between the collector and ground. Since RL is no more a load, we rename it "RC". It continues to "produce" voltage but we do not use it as an output... Jun 13, 2020 at 11:27

RL is not required to bias the transistor collector current, however without RL you will get zero output, since the output would be connected directly to Vcc if RL is replaced by a short.

Typically you want to make RL as high as possible to maximize the voltage gain, but not so high that the output clips high or low with extremes of input voltage and taking bias tolerances into account. That means that RL is usually selected so that the collector voltage (with no input) is more than half of Vcc.

• @Circuitfantasist They long predate the BJT. You will find the phase-splitter used far, far more often (I think) in tube circuits. (There's a reason for why that is. But I'll leave it for another time.) In any case, once all these various ways of "looking at something" arrive in mind, it's very difficult not to see all of them at once, afterwards. And this often allows very quick novel ideas to arrive when considering how to minimally modify a circuit to provide some new and powerful feature without impairing the rest.
– jonk
Jun 13, 2020 at 20:15
• @Circuitfantasist Just a few comments. I think every informed challenge written in a comment to an answer should be responded to by the answerer. Some frequent writers here do not do this and I consider it both offensive and disrespectful. I tend to dismiss their opinions, in time, as the value of their answers are far, far less as a result. Second, this whole voting thing bothers me. I've said so many times. Is my answer any better because my number is bigger? No!!! What matters is what I write and how I handle objections/questions and if I improve what I write over time as a result.
– jonk
Jun 13, 2020 at 21:43
• @jonk, Exactly... Let me just add, when it comes to the human side of the problem, that I never imagined that people in the world are so equal in their shortcomings. There is no difference between the colleague next to me and the colleague from SE, RG, or WP on the other side of the globe. Nothing can change human nature... Jun 13, 2020 at 22:03
• @Circuitfantasist Re: points; A numeric system of some kind should NOT accumulate. Instead, it should automatically have a percentage subtracted from it every single month, or week. (There may be some "minimum earned value," I suppose, which respects the fact that someone has contributed a lot at one time.) I think of it like I do a wealth tax (which I think is proper -- I disagree with income taxes, but not wealth taxes.) Every year, the "pile" is eroded. If you want to keep it, you must make it work. Not sit idly by.
– jonk
Jun 13, 2020 at 22:16
• @Circuitfantasist And well-placed criticisms to answers and good commentary dialog should count for something. Not sure what. But an active, engaged dialog is very valuable. And the answerer should get extra credit for incorporating good value from the comment thread back into the answer. Not sure how that is achieved. But perhaps by allowing the answer to be broken into sections, with each section being able to be counted for separate points (up to a limit) and new sections starting fresh on that limit. I think it would take discussion to work out a good system, though. It's called "design."
– jonk
Jun 13, 2020 at 22:19

The wealth of SE EE are all these young people who ask incredible questions... and who make us try to answer them in an incredible way. Here is my "incredible story" about this circuit of four resistors and a transistor.

1. Base-emitter input. The only way we can control a transistor is by applying input voltage (about several hundred millivolts) to its base-emitter junction. For some reason, most frequently, we present this voltage as a small difference between two relatively high single-ended (referenced to ground) voltages. Thus the base-emitter junction is floating and we have a few ways to control the transistor.

To investigate them in the laboratory, my students mount a circuit similar to yours - Fig. 1, where the two input voltages are "produced" by 1 k potentiometers… and the collector current is visualized by an LED. Moving the potentiometer sliders, they have the feeling that the base and emitter (voltages) "move" up and down.

Fig. 1. A set-up for investigating various transistor stages on the blackboard (Vcc = 12 V). Voltages are visualized by bars (in red) with proportional height; current paths are visualized by loops (divider currents in green, base current in blue and collector current in violet) with proportional thickness.

I suggest you mount the circuit on a prototyping board - Fig. 2, and reproduce the next experiments. It is preferable to use (geometrically and electrically) linear potentiometers. If you have two more multimeters (voltmeters) besides V1 and V2, you can connect them in parallel to Rc and to the output OUT (Vc). Of course, you can insert an ammeter(s) too.

Fig. 2. A set-up for investigating various transistor stages on the prototyping board

2. Base input. First, you can set (by the help of the potentiometer P2) constant voltage Ve at the emitter and then vary (by the help of the potentiometer P1) the base voltage Vb. Only be careful to keep Vb with a few hundred millivolts above Ve. You will see that when increasing V1, the LED gradually begins to glow (Ic increases)... VRc increases… but Vout decreases. The name of this arrangement is "common-emitter stage".

Maybe you notice that Ve slightly varies in the same direction with Vb because the emitter current changes and P2 is not a perfect voltage source (its Thevenin's resistance is not zero). Here it is undesired effect but later we will use it. To make Ve "stiff" enough, connect a big (> 1000 microF) "bypass capacitor" in parallel to P2 output (between the emitter and ground)... and observe the result when quickly wiggle P1 slider. As they say, the emitter is "AC grounded".

3. Emitter input. But with the same success, you can set (now by the help of the potentiometer P1) constant voltage Vb at the base and then vary (by the help of the potentiometer P2) the emitter voltage Ve. Now be careful to keep Ve with a few hundred millivolts below Vb. Now you will see that when increasing V2, the LED gradually begins to glow dimmer (Ic decreases)... VRc decreases… but Vout increases. The name of this arrangement is "common-base stage".

Here you may notice that Vb slightly varies in the same direction with Ve because the base-emitter junction conveys Ve variations... and the emitter "pulls down" the base through the base-emitter junction. As above, to make Vb "stiff" enough, connect a big "bypass capacitor" in parallel to P1 output (between the base and ground)... and observe the result. Now the base is "AC grounded".

4. Both base and emitter input. If you are curious enough, continue with these exciting experiments by varying both Vb and Ve. First set such a difference Vb - Ve that the LED glows with a dim light (around 650 mV). Then grab the potentiometer sliders with both hands and start moving them simultaneously:

...in the same direction, with the same rate. Very interesting - both voltages simultaneously change but their difference Vb - Ve, Ic, VRc and Vout do not change. They name this "common mode" and, as a rule, introduce it when explaining op-amps. But you met it right now, at transistor circuits. Indeed, you could get to know it even from the bridge circuits (the so-called "balanced bridge").

...in the opposite directions, with the same rate. Now both voltages simultaneously change and their difference Vb - Ve, Ic, VRc and Vout vigorously change. They name this "differential mode" and also introduce it when explaining op-amps but you could meet it when investigating bridge circuits ("unbalanced bridge"). This circuit is the prototype of a transistor differential stage (aka "differential pair" or "long-tailed pair").

It is interesting that, in the circuits above, when Vb - Ve > 0.7 V, the base-emitter junction practically connects (like a bridge) the two input voltage sources that can cause interesting effects. If Vb - Ve <= 0, the base-emitter junction is cut off and there is no any connection between them.

5. Both base and "self-emitter input". But let’s finally return to your 4-resistor circuit and carry out the most interesting experiment. Disconnect the upper end of P2 and you will see that the circuit continues operating... but only controlled by P1 (Vb). How is it possible?

Now the transistor itself changes the emitter voltage (by changing the current through the lower part of P2) so that it follows the base voltage (as though the transistor acts as R2). Hence the name of this circuit - "emitter follower". Of course, it is a follower if you take the emitter voltage as an output. But you can take Vc as an output... and then it will be a common-emitter amplifier with negative feedback (the so-called "emitter degeneration"). Note that it acts both as a follower and an amplifier... but if you want to use only the follower, then it would be better to fix Vc. So, connect a big "bypass capacitor" to the output (between the collector and ground)... and observe the result. Now the collector is "AC grounded".

6. Both emitter and collector output. But why not use both voltage drops VRe and VRc as outputs? They are created by the (almost) same current Ic (Ie) that connects them like an "electric transmission" so VRc/Rc = VRe/Re... and VRc/VRe = dVc/dVe = Rc/Re...You can think of this configuration as of a voltage divider supplied by constant current. Indeed it is an odd divider... but still divider... a divider with two outputs. As in the classic divider, two resistors are connected in series... but only they are separated by the current source (transistor).

VRe is referenced to ground but VRc to Vcc. So we take its complement Vc instead… but it is inverted. Thus we obtain both direct and inverse voltage. Usually, Rc = Re; so VRc = VRe. The name of this circuit is "phase splitter".

7. Current source (sink). If you consider the collector current as an output, the circuit acts as a (voltage-controlled) current source driving an LED as a load. The transistor keeps a constant voltage drop VRe across the constant resistor Re; so the emitter current Ie (Ic) is constant. The resistor Rc is already not necessary since the current is set by Re (and Vb); so you can remove it (try it and compare the two circuits). But if you leave Rc, its resistance will be compensated by a decreased voltage drop across the transistor. You can vary Re since the lower part of P2 is a variable resistor.

8. AC amplifiers. You can continue with these interesting experiments by applying AC voltage (with low frequency to see the LED flashing... or connect an oscilloscope). Connect the input source through a capacitor in series to the base (AC common-source amplifier) and emitter (AC common-base amplifier)... but that is another story...

I am almost done with my story... True, it turned out to be quite long... but keep in mind that this is a story about all kinds of transistor stages combined in one circuit.

I hope my story would be useful to you not only with its content but also with the way circuit ideas are presented. With it I wanted to show that circuitry can be entertaining and fascinating... and not just a craft.

$$\R_L\$$ is what reduces the voltage at the output. When the transistor is completly off, no current is flowing through $$\R_L\$$ (assumming the following circuit has a high input impedance). This results in an output voltage equal to the positive supply voltage.

Now, the way this common emitter amplifier works is by creating a current through the transistor (and $$\R_e\$$) when the input signal is rising. This current also has to flow through $$\R_L\$$, and by that resulting in a voltage drop of the output.

$$V_{out} = V_{cc} - I_{R_L} \cdot R_L$$

If $$\R_L\$$ wasn't there, the output would always be equal to the supply voltage, until the transistor draws more current than the supply can handle or the transistor burns up (whatever happens sooner).

• We can also think of the combination of a transistor and a load RL in parallel as of a "current divider" supplied through the collector resistor Rc. The transistor shunts the load and regulate the current through it by diverting a part of the load current... Jun 13, 2020 at 12:26

Resistors R1 and R2 are there to establish base voltage VB and base current IB. While it’s true Re does help to make the Q-point stable via a form of negative feedback. It also helps to establish emitter current IE which is approximately equal to collector current IC And also establish a emitter voltage VE. RL, which is essentially a collector resistor, is needed to establish collector current IC and voltage at collector VC

As for common base configuration, a bypass capacitor is placed across R2 so that at signal frequencies, it’s equivalent to a low reactance path to ground. This would essentially ground the base terminal for ac signals (mid frequencies). You are correct in that for the common emitter amplifier, Re reduces the gain when left unbypassed. If we bypass it then we would increase the gain of amplifier.

Also bypassed emitter resistor is only used in common emitter amplifiers. There no bypassing emitter resistors in common base or common collector( at least to my knowledge).

• Thanks, then bypass capacitors are always used because bypassing resistors we increase the gain? in any configuration?
– Juà
Jun 12, 2020 at 14:58
• Bypassing isn’t always used. You may want to bypass to obtain high gain or you may want to design a low gain amplifier and leave it unbypassed. It depends on the application. Jun 12, 2020 at 15:00
• @Juà, Bypass capacitors simply AC short-circuit (decrease the impedance of) this element to which they are connected in parallel. You have to consider what this leads to - whether to increase or decrease the gain... Jun 15, 2020 at 3:39

The resistor drops the voltage based on current through it. A 1k resistor drops 1V when 1mA current passes through it. The same 1k resistor drops 2V when 2mA passes through it. Thanks to almighty Ohm's Law.

The thing with transitor is that the current through its collector can be varied by changing the base current. So whenever you have higher current passing through base, it results in higher current through collector and higher drop at RL resistor and vice versa. This allows the collector voltage (output) to change according to the the base current (input).

If you don't keep the resistor, you will short your collector to the supply voltage. Meaning, your collector voltage(output) can't change for changing base Current (input).

Try visualizing what I have mentioned by building the circuit and testing it.

• Simply speaking, (all sorts of) transistors are devices with current output... but in the most cases we need a voltage output. That is why, we insert a resistor in the current path to convert it to voltage. The question is "Where do we insert the resistor?" BTW besides the usual places, if the power supply is floating, we can insert the load between the negative terminal of the power supply and ground:) Then the load will be grounded as we want... Jun 13, 2020 at 12:13

[You've mislabeled your schematic -- the collector bias resistor is called $$\R_C\$$, and is not the load resistor. The load resistor represents the output.]

I'll come at this from another angle. A transistor acts like a current source. (You can see this in the small-signal hybrid-pi model, where the BJT becomes a voltage-controlled current source with a transconductance $$\g_m\$$.) But a common-emitter amplifier is supposed to be a voltage amplifier. So what do we do?

Well, if you have a current source and you want a voltage, you can connect the current source to a resistor ($$\R_C\$$). The resistor voltage is proportional to the current, so your linear voltage-controlled current source becomes a linear voltage-controlled voltage source -- an amplifier!

Without $$\R_C\$$, your load is driven directly by the transistor, so your output is a current:

simulate this circuit – Schematic created using CircuitLab

With $$\R_C\$$, your output comes from the transistor/$$\R_C\$$ "voltage divider". If $$\R_L >> R_C\$$, the output voltage is independent of the load:

simulate this circuit

Another way to think of this is that the BJT current source and $$\R_C\$$ form a Norton equivalent-style circuit, which can be transformed into a Thevenin equivalent:

simulate this circuit

Here you can easily see that $$\R_C\$$ is the output resistance of the amplifier.

• I highly appreciate your "transistor/RC voltage divider" viewpoint because this is the lowest possible level of presentation. But it is a quite odd divider - a "non-linear" or "dynamic" divider where the lower resistor is a "voltage-controlled dynamic resistor". Imagine the transistor base is held at constant voltage… and the supply voltage Vcc varies. Then, the "divider" output voltage (Vc) will follow the "input" supply voltage… and the "divider" will act as a "follower". Thus Vcc variations directly appear at the collector. BTW this is a well-known voltage-shifting technique in op-amps. Jun 15, 2020 at 15:09