# Common Collector Amplifier

- Boris Poupet
- bpoupet@hotmail.fr
- 15 min
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**Presenting the Common Collector Amplifier**

This article deals with another type of bipolar transistor architecture used to amplify signals that is commonly known as** Common Collector Amplifier** (CCA). The CCA can also sometimes be called **emitter-follower amplifier** and we will understand why later in this article.

The first figure below is a simplified electric diagram with no particular biasing circuit presenting the CCA configuration :

The main difference when comparing this architecture with the Common Emitter Amplifier (CEA) is that the output signals are taken on the emitter branch and the collector is always connected directly to the power supply, thus the name “Common Collector”.

It is shown further on that the voltage gain **A _{V}=V_{out}/V_{in}** is approximately equal to 1. Moreover, the phase remains the same during the amplification process, the input and output signals are therefore very similar, hence the name “emitter-follower”. On the other hand, we will see that the current gain

**A**is high but has an upper limit.

_{C}=I_{out}/I_{in}### Equivalent circuit

We can consider the bipolar transistor between the collector and emitter to be an ideal current source of amplification gain β where **I _{out}=β.I_{in}**. It also presents a small resistance given by

**r**known as “

_{e}=25 mV/I_{out }**AC emitter resistance**” or “

**small diode emitter resistance**” and represents the dynamic resistance for small AC signals of the p/n junction of the bipolar transistor.

We see in **Figure 2** an equivalent circuit of the CCA configuration of **Figure 1** considering the transistor such as described above.

### Voltage gain

It is easy to understand that in the configuration presented in **Figure 1**, the voltage gain is approximately equal to 1. Let’s indeed consider the voltage loop between the base and emitter. It comes automatically that **V _{in}=R_{E}.I_{out}=V**

_{out}and thus

**A**

_{V}=1.From **Figure 2 **we can clearly express the input voltage to be **V _{in}=(R_{E}+r_{e}).I_{E }**and the output voltage to be

**V**. When expressing the voltage gain

_{out}=R_{E}.I_{E}**A**, the term I

_{V}=V_{out}/V_{in }_{E}disappears and we get the exact expression of the gain A

_{V }:

From this formula, it comes that **A _{V}<1** but usually

**R**so the approximation

_{E}>>r_{e}**A**is justified. Since the voltage gain is always smaller than 1, V

_{V}=1_{out}<V

_{in }:

**the CCA**

**cannot provide any amplification for the voltage signal**.

### Biasing method

To provide better stability, the base of the bipolar transistor is biased with a voltage divider network such as shown in the following figure.

Since in a CCA configuration, **V _{in}=V_{out }**, the voltage divider network can be represented with a simpler equivalent circuit shown in

**Figure 4**:

Using Kirchoff’s laws or Miller’s theorem in the circuit of** Figure 4**, it is easy to express the output voltage :

For a faithful amplification (no distortion or saturation), the output voltage should respect the condition **V _{supply}=2.V_{out}**. Thus, the bias resistances R

_{1}and R

_{2 }should approximately be equal. However, to be more accurate, the threshold voltage between the base and emitter

**V**needs to be considered. For silicon-based bipolar transistors, the threshold voltage is a constant equal to

_{BE }**V**.

_{BE}=0.7 VThe full expression of the output voltage is hence given by **Equation 2** :

### Input resistance

The CCA amplifier is characterized by a high input resistance. The expression of the base resistance R_{B} comes from Ohm’s law : **R _{B}=V_{in}/I_{in}**. We have previously seen that

**V**, thus :

_{in}=(R_{E}+r_{e}).I_{out}With the condition that **R _{E}>>r_{e }**the final expression of the base resistance can simply be written

**R**

_{B}**=β×R**

**. For this reason, the transistor’s current gain β is the most important factor to set the input resistance of a CCA configuration.**

_{E}When considering a full CCA architecture with the biasing circuit, the total input resistance **R _{in}** satisfies the following formula :

The input resistance of a CCA is always very high which is useful to avoid loading down previous circuits connected to it.

### Output resistance

If we consider the circuit of **Figure 1**, the output resistance is given by the emitter resistance R_{E} which expression is **R _{E}=R_{B}/β**. However, the signal is always taken on a load resistance

**R**in parallel with the emitter branch and in this case the output resistance satisfies :

_{L}The output resistance on a CCA configuration is always very low and for this reason CCAs are used to drive low-resistance loads.

### Current gain

Let’s consider a real CCA with a biasing circuit such as presented in **Figure 3**. The current gain A_{C} is expressed by the ratio **A _{C}=I_{out}/I**

_{in }. The calculation of A

_{I}depends on the values of the bias resistances :

- If the parallel resistance
**R**is much greater than the base resistance R_{1}//R_{2}_{B}, most of the current flows to the base and therefore I_{in}≅I_{B }so that**A**_{C}=β. - If
**R**is lower or the same magnitude as R_{1}//R_{2}_{B}, the current gain needs to be determined by the formula**A**with_{C}=I_{out}/I_{in }**I**and R_{in}=V_{in}/R_{in }_{in}can be determined from**Equation 4.**

To sum up, if R_{1}//R_{2}>>β×R_{E }, **A _{C}≅β**. If not,

**A**. The transistor gain β is thus the maximum current gain achievable by a CCA configuration.

_{C}=I_{out}/I_{in}### Example : Voltage, Current and Power gains of Common Collector Amplifier

Having in mind the expressions and characteristics of the voltage gain, the input and output resistances, let’s consider the following circuit with a transistor gain of **β=200** and determine the voltage, current and power gains.

The output resistance is given by the emitter resistance R_{E }in parallel with the load resistance R_{L }:

The base resistance is given by R_{B}=β.R_{out}=200×500=**100 kΩ**. The input resistance is thus given by the base resistance R_{B }in parallel with the bias resistances R_{1 }and R_{2} :

It can already be highlighted that R_{in}>>R_{out} such as mentioned previously. Let’s now calculate precisely the voltage gain A_{V}_{.} It is first considered that A_{V}=1 so that **Equation 2 **is valid and therefore we can express the voltage drop across R_{E} such as :

The current across the same resistance is thus given by I_{E}=4.3 V/1 kΩ=**4.3 mA**. The small AC resistance can then be determined by r_{e}=25 mV/4.3 mA=**5.8 Ω**. Since R_{E}=1 kΩ, we can confirm the hypothesis R_{E}>>r_{e}.

Finally, the voltage gain is expressed by A_{V}=500 Ω/505.8 Ω=**0.989≅1**.

In this example since R_{1}//R_{2}=**10 kΩ**<<R_{B}=100 kΩ the approximation A_{C}≅β is not valid and the current gain needs to be calculated with the formula A_{C}=I_{out}/I_{in}.To compute the current gain A_{C}, it is needed to determine the input and output currents I_{in }and I_{out} :

- The output current is simply I
_{out}=V_{out}/R_{out }and since V_{out}≅V_{in }**, I**_{out}=1 V/500 Ω=**2 mA**. - The input current is given by I
_{in}=V_{in}/R_{in}=1 V/9.1 kΩ=**110 μA**.

Finally, the current gain is expressed by A_{C}=2 mA/110 μA=**18.2**.

The power gain is given by **A _{P}=A_{V}×A_{C}=18**. However, since R

_{E}=R

_{L}, the power given to the load is only half :

**A**.

_{P,load}=9### The Darlington pair

We have seen that the transistor’s current gain β is the factor limiting the total input resistance and the output current **I _{out}=β.I_{in}**. Indeed, if β increases, the base resistance

**R**increases and therefore

_{B}=β.R_{E}**R**increases as well.

_{in }A simple way to overcome that limitation without having to purchase an expensive specific high gain transistor is through the **Darlington Pair** (DP) presented below in Figure 6 :

A DP consists in connecting together two bipolar transistor. The emitter of the first transistor is connected to the base of the second and both collectors are shorted to the power supply. We suppose the first transistor to have a gain **β _{1 }**and the second is

**β**.

_{2}In this configuration, first of all, the input current is being amplified, the output of the first transistor is **I _{out,1}=β_{1}.I_{in}**. This same current I

_{out,1}becomes afterwards the input current of the second amplifier which is a CCA configuration. The final output current is

**I**.

_{out,2}=β_{2}.I_{out,1}=β_{1}.β_{2}.I_{in}Finally, the total current gain of a DP is equal to **Ω=β _{1}.β_{2}** which leads to a very high output current. Moreover, the base resistance can as well be expressed

**R**which leads to a very high input impedance.

_{B}=Ω.R_{E }### Conclusion

As a conclusion, we have seen that the Common Collector Amplifier does not amplify voltage signals since it’s voltage gain is strictly lower than 1 but usually can be approximated to 1, his nickname “emitter-follower” comes precisely from this behavior since the phase is also conserved. However, the current gain of a CCA is high with a upper limit equal to the transistor’s current gain β and depends on the values of the bias resistances. Moreover, we have seen that the input resistance is high and the output resistance is low. This characteristic makes the CCA configuration to be useful as a voltage buffer : the CCA can be interposed between a high and low impedance block to prevent any undesired loading. In the final section, we show an example of architecture called a Darlington pair that can overcome the limitation set by **β** for the input resistance and the output current.

In the next article, we present the last topology of the bipolar transistor : the Common Base Amplifier.

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