I'm having a hard time understanding an emitter follower. What I don't understand is why there is no voltage gain but only current gain. From my understanding when the transistor is turned on, the CE acts like a piece of wire, so the Vout should be equal to V+. But no the Vout follow the Vin less 0.7V. Can anyone explain it to me?


  • \$\begingroup\$ Other than CE acting as a diode? \$\endgroup\$ Jul 1, 2013 at 4:30

2 Answers 2


The 0.7V drop is from the Base-Emitter junction being a PN junction (in an NPN transistor), which is the same as a diode (a silicon diode has a forward drop of ~0.7V).
A bipolar transistor is either NPN or PNP.

The reason it has current gain is that the base current turns the transistor on, allowing current from the collector (which is connected to V+) to flow to the emitter.

The reason it doesn't have voltage gain is due to the "negative feedback" effect from Re.
Let's run through an example of why the emitter stays 0.7V below Vb and does not reach V+.

Let's say we have this setup, and Q1 has a current gain of 200:


Now say we apply 3V to the base.
We know that the transistor begins to turn on when the base is ~0.7V higher than the emitter, so at this point current starts to flow from V+ into the collector and out through the emitter through Re to ground.
Now here's the important bit - when the current flows through Re a voltage appears across Re.

Now for arguments sake let's say the transistor "tries" to turn on fully and since we have a rising current flowing through Re, the voltage across Re rises also.

What happens when the voltage across Re reaches 2.3V?

Well, you should see where this is going now - the base is still at 3V. When the emitter was at 0V, the base-emitter (b-e) voltage was >0.7V and the transistor was on. Now, however, the b-e voltage is at 3V - 2.3V = 0.7V! so if the voltage across Re rises any further, the transistor would turn off. So the circuit has a natural limiting mechanism, and what happens is that it always sits at ~0.7V below the base voltage. It would not matter if the current gain is infinite, the emitter voltage cannot rise above this point without "turning itself off".

Here is a simulation of the above circuit, with the base voltage gradually ramped up from 0V to 3V:

CE Simulation

Here's another simulation with a capacitor added in to prevent the emitter voltage from rising too quickly, so we can see how the transistor turns on fully at first to charge the cap as quickly as possible, then (almost) turns off again as the cap reaches ~2.3V and only the resistor current is left as things settle:

CE with cap


Ce with Cap simulation

  • \$\begingroup\$ A good idea for understanding circuits - deliberately delaying the transistor operation so that we can see what it does over time. It can also be applied to op-amp circuits. \$\endgroup\$ Nov 15 at 11:10
  • 1
    \$\begingroup\$ 19 points for a wrong explanation and 1 point for the correct explanation (below). Funny! \$\endgroup\$
    – LvW
    Nov 18 at 13:59

The motive

What made me go back 10 years and write this fancy circuit story? Perhaps the dissatisfaction with today's prevalent prosaic questions concerning mostly details of specific circuit implementations...

I understand that the world is like that and thinking has given way to doing... but I secretly hope that there are still even a few curious people who would be interested in the philosophy behind this great circuit solution....

How do we understand circuits?

To truly understand an unfamiliar circuit, concrete explanations (such as "current flows through the base-emitter junction", "collector voltage decreases", etc.) are not sufficient. With them "we will see the trees" but to "see the forest behind them", we need answers to more general questions like: "What problem does the particular circuit solve? What was it created for? What is the most general idea on which it is based? Where can we see it in life? What is the conceptual block diagram of the circuit? With what conceptual elements is it built?"

From this point of view, electrical circuits are not built on purely electrical ideas but on much more general non-electrical ideas... and to understand them, we have to discover what those ideas are.

The challenge of "simple" circuits

(Sometimes) there is such a paradox in circuits - complex circuits are easy to explain but simple circuits are hard to explain. And really how to explain a circuit consisting of only one transistor? What to explain here?

Just such a one-transistor circuit is the emitter follower, and I will accept the challenge to reveal its idea.

The problem

Most of the tasks we solve in life come down to what I figuratively call "active copying" - we make a quantity that is an exact but powerful copy of another (original) quantity. The power is taken from an additional source (power supply).


simulate this circuit – Schematic created using CircuitLab

There are many such popular examples such as maintaining a constant speed, distance and direction in the car, temperature in a room, speech level, etc. Here is a funny example illustrating the idea of this phenomenon: Imagine that, for some reason, an elephant decided to copy the movements of an ant-:) It does exactly the same thing as it but it can be incredibly loaded.

Basic idea

The "serious" but not so useful for our purposes name is negative feedback. We implement it by producing an output copy quantity Y, comparing it by subtraction to the original input quantity X, and changing Y until it equals X.


simulate this circuit


I have made this circuit many times with various real devices, but now I have taken advantage of the nice CircuitLab features.

Variable voltage source

Manually-controlled voltage source: Probably already in the 19th century they realized this idea (without suspecting they had made a "voltage follower":-) So why don't we do it now? For this purpose, we take another variable voltage source Vout and connect it in the opposite direction through a zero indicator NI (sensitive voltmeter) to the input voltage source Vin.


simulate this circuit

Thus the two voltages are subtracted and their difference is shown by the zero indicator. We adjust Vout so that NI shows 0 V. As a result, its voltage will follow the input voltage, and if we connect a load RL, it will draw current (green arrows) from Vout and not from Vin.

Behavioral voltage source: To get rid of this tedious work, we can assign it here, at CircuitLab, to a "behavioral voltage source". Its voltage is controlled by the difference between the two voltages multiplied by a very large factor ("gain") of 10E9 (10G). The imperfect 1 kΩ voltmeter with the figurative name "RL1k" serves as a "visualized load".


simulate this circuit

So when we apply an input voltage, the source starts to increase its voltage but the difference decreases; the voltage slows down and stops changing once it equals the input voltage. As we can seen in the graphs below, Vout constantly follows Vin.

STEP 2.2a

STEP 2.2b

Voltage-controlled voltage source: CircuitLab gives us another, even more professional option to make this conceptual schematic using a voltage-controlled voltage source VCVS (the fancy name of a voltage amplifier:-) This is how op-amp followers are made but that is another topic.


simulate this circuit

Variable resistor

We can make a variable voltage source by connecting a variable resistor Rce in series with a constant voltage source Vcc (power supply). We also need to add a load resistor. As such, I suggest using a "bad" voltmeter with a greatly reduced resistance of 1k; let's name it figuratively RL1k. Thus it will serve as a "voltage-visualized resistor". Actually, Rce and RL1k act as a voltage divider with variable transfer ratio K = RL1k/(Rce + RL1k).


simulate this circuit

We can apply the same trick as above to simplify the circuit by replacing the resistor Rce and the "ideal" ammeter Ic with an imperfect ammeter with variable resistance to visualize the current.

Variable current source

For some reason, however, regulating elements (transistors, tubes, etc.) act not as linear (ohmic) resistors but as current stabilizing non-linear resistors called "current sources". They are not sources in the literal sense of the word because they do not "produce" but only consume power.

Manually-controlled current source: So let's use such an element from the CircuitLab library. As it is commonly said, it creates a current Ic that flows through the load resistor RL1k and creates a voltage drop Vout = Ice.RL1k across it.


simulate this circuit

In this "negative feedback game", we adjust Ic to reach a zero voltage difference Vbe; as a rezult, Vout =Vin.

Behavioral current source: As above, we can automate the circuit operation using a behavioral current source. Its current is controlled by the difference between the two voltages multiplied by a very large factor (10 M transconductance).


simulate this circuit

By passing its current through the load RL1k, the current source adjusts the voltage drop Vout across the load so that to zero the difference voltage Vbe. As a rezult, Vout exactly follows Vin.

STEP 4.2a

STEP 4.2b

STEP 4.2c

Voltage-controlled current source: And here CircuitLab gives us an option to make this conceptual schematic using a voltage-controlled current source VCCS (the fancy name of a transconductance amplifier).


simulate this circuit

Emitter follower

Real voltage followers are made by transistors. There a transistor "observes" the voltage difference Vbe by its base-emitter junction and adjusts its collector current to make the voltage across the emitter load equal to the input voltage.

Unbiased transistor: But the BJ transistor is a "nasty" device because its base-emitter junction "eats" about 0.7 V of the input voltage. As you can see from the circuit diagram below, when we apply the same 1 V input voltage, only 368 mV is left for the load...


simulate this circuit

... and the output voltage graph is lowered by this downward offset.

STEP 5.1

Increased input voltage: Since 0.650mV is lost in the base-emitter junction, we think, then let's increase Vin by that much to compensate the loss.


simulate this circuit

Biased transistor: We can implement it without changing Vin by connecting a biasing voltage source that adds its Vbias voltage in series to the input source.


simulate this circuit

As a result, the input and output voltage graphs almost match.

STEP 5.3


To see the point of using the emitter follower, we need to compare it with the more primitive common-emitter stage.

Common-emitter stage

By its very nature, the transistor collector-emitter output part behaves is a controllable current source. So, when we insert a load resistance RL in series to this part (between the collector and Vcc), and begin changing it, the transistor changes its "Rce resistance" in the opposite direction thus trying to keep constant the total resistance RL + "Rce".


simulate this circuit

As a result, the current stays almost unchanged but the (output) voltage drop VRL = Ic.RL across RL linearly changes. Let's sweep RL to see this in a graphical way.

STEP 6.2a

STEP 6.2b

Emitter follower

But when we move the load RL to the emitter, the transistor behavior changes dramatically - from a current source it becomes a voltage source. This is where the negative feedback mechanism comes into play and the transistor begins keeping constant not the current but the voltage drop across the load. So, when we insert a load resistance RL in series to the collector-emitter part but between the emitter and ground, and begin changing it, the transistor changes its "Rce resistance" in the same direction thus trying to keep constant the resistance ratio RL/(RL + "Rce").


simulate this circuit

As a result, the (output) voltage across RL stays almost unchanged but the current changes. Let's sweep RL to see this in a graphical way.

STEP 6.1a

STEP 6.1b


In an emitter follower, using negative feedback, a transistor is converted from a "current source" to a "voltage source".


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