# Common Base Amplifier

- Boris Poupet
- bpoupet@hotmail.fr
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### Presenting the Common Base Amplifier

In this article, we present the last topology of amplifiers for bipolar transistors known as the **Common Base Amplifier** (CBA). In **Figure 1 **below, the electric diagram of a CBA is presented, no particular bias circuit or decoupling capacitors are shown here.

Some specifications need to be highlighted for CBAs :

- The base is linked to the ground of the circuit, hence the name “Common Base”.
- The input signal is delivered to the emitter branch of the bipolar transistor.
- The output signal is taken to the collector branch of the bipolar transistor.

Further in the article, we will see that in many ways the CBA behaves in opposition with respect to the Common Collector Amplifier (CCA).

### Equivalent circuit

An equivalent circuit of **Figure 1 **can be drawn considering the collector branch to be an ideal current source and the p/n junction between the collector and emitter branches to behave like a small diode resistance **r _{e}**

**=25 mV/I**.

_{out}

It can already be anticipated from **Figure 2 **that since **I _{in}=I_{out}+I_{B}** (from Kirchhoff’s laws), the current gain

**A**of a CBA configuration is

_{I}=I_{out}/I_{in }**A**. Therefore, the current gain of a CBA configuration is strictly lower than 1, so this type of amplifier

_{I}=1-(I_{B}/I_{out})<1**cannot amplify currents**. However, we will see further in the tutorial that the voltage gain is high.

### Current gain

We have already seen in the previous paragraph that the current gain A_{I} is strictly lower than one. To get the exact formula of A_{i}, we consider as mentioned previously that **I _{in}=I_{out}+I_{B}**. Moreover, we define

**I**with β the transistor’s current gain. Notice that here

_{out}=β×I_{B}**I**in contrary to the previous amplifier configurations CEA and CCA.

_{B}≠I_{out }The output current satisfies I_{out}=A_{I}×I_{in}=β×I_{B }, when isolating A_{I}, it comes :

Dividing the numerator and denominator by I_{B}, the term I_{out}/I_{B}=β appears and we get the exact expression of the current gain for a CBA configuration :

As an example, a bipolar transistor of gain **β=200** has a current gain **A _{I}=200/201=0.995≅1**. Hence, the current gain of a CBA configuration

**can always be approximated by 1**without committing too much error. Since the current gain is equal to 1, the output current I

_{out}follows the input current I

_{in}, hence the other name commonly given to this configuration

**current follower/buffer**.

### Input resistance

As seen from the input in the emitter branch, the total input resistance is **R _{E}//r_{e}** where the symbol “//” denotes the fact that the emitter and small diode resistor are in parallel.

However, the emitter resistance R_{E} is always much higher than the small diode resistance r_{e}, hence it comes :

The input resistance of a CBA configuration is therefore equal to the small diode resistance r_{e} between the emitter and collector branches, this value of impedance is very small.

### Output resistance

On a real CBA configuration, a load R_{L} is placed in parallel with the collector resistance R_{C}. The output resistance is thus given by **R _{out}=R_{C}//R_{L}**. If the load is chosen such as R

_{L}>>R

_{C}, the output resistance simplifies to

**R**.

_{out}=R_{C}### Voltage gain

It is considered in the following, such as proved previously that **A _{I}≅1**. The voltage gain of a CBA configuration is thus given by the ratio

**A**where

_{V}=V_{out}/V_{in }**V**and

_{out}=R_{C}×I_{out}**V**. It comes afterwards that :

_{in}=(R_{E}//r_{e})×I_{in}Since the collector resistance satisfies **R _{C}>>r_{e}**, the voltage gain of a CBA configuration is very high. We can moreover highlight that the voltage gain of a CBA configuration is the same as for Common Emitter Amplifiers except that the sign is here positive :

**the output voltage signal is in phase with the input voltage signal**. This formula is valid if the load R

_{L}is considered to satisfy R

_{L}>>R

_{C}. However, in the general case, the expression of the voltage gain is :

### Example : Voltage, Current and Power gains

In this section, we consider a real CBA configuration presented in **Figure 3** with a voltage divider network to bias the base which is composed of two resistor R_{1} and R_{2}. Moreover, a load R_{L} is in parallel with the collector resistance R_{C}. Notice that a decoupling capacitor is added between the base and ground to make this diagram correct, but for the sake of simplicity, its value is not given and won’t be taken into account for the following calculations. Finally, the bipolar transistor’s current gain is **β=100**.

The current gain of this CBA configuration is simply given by :

Before determining the voltage gain of this configuration, the first step is to calculate the total input and output resistances, and for that, we need the value of the small diode resistance.

The voltage drop in the collector resistance R_{C} is given by :

Therefore, the current across the collector resistance is I_{C}=V_{C}/R_{C}=0.97 V/5 kΩ=**194 ****μA**. From this value, it comes that the small diode resistance is r_{e}=25 mV/194 μA=**129 Ω**.

The input resistance is hence given by :

Since in this configuration **R _{L}<R_{C}**, the parallel resistance

**R**needs to be considered as the total output resistance :

_{L}//R_{C}Finally, the voltage gain is determined from **Equation 4** :

In theory, the voltage can here be amplified by a factor 34.65. However, as seen during the Introduction to Electronic Amplifiers tutorial, the output voltage is limited by the power supply votage. Therefore, the output voltage here reaches only 2×V_{supply}=20 V peak to peak instead of 2×34.65=69.3 V peak to peak and a rather important saturation effect will be observed.

It is interesting to check if our calculations are correct in determining both input and output currents :

- The input current is given by V
_{in}/R_{in}=1 V/114.2 Ω, thus**I**._{in}=8.76 mA - The output current is given by V
_{out}/R_{out}=34.65 V/4 kΩ, thus**I**._{out}=8.66 mA

We can see that the input and output currents are approximately equal and the ratio I_{out}/I_{in}=0.99 gives again the current gain previously calculated.

If we consider that the output voltage can indeed be amplified by a factor 34.65, the power gain A_{P} of this configuration is given by **A _{I}×A_{V}=0.99×34.65**, thus

**A**. However, since

_{P}=34.3**R**, only a fourth of the power is delivered to the load :

_{L}=4×R_{C}**A**.

_{P,load}=8.57Using the simplified expression from **Equation 3** R_{C}/r_{e}=5 kΩ/129 Ω gives a voltage gain **A _{V}=38.8**. The simplified value of the current gain is

**A**which in turn gives a power gain A

_{I}=1_{P}=38.8 instead of 34.3 for the real value. The error E

_{P}for the power gain is therefore :

### Conclusion

In this tutorial, we dealt with many aspects of one of the three elementary topology of amplifier known as the **Common Base Amplifier** (CBA). We have seen that such a configuration cannot amplify currents since its current gain is approximately equal and strictly lower than 1, hence the name “current buffer/follower” often given to CBAs. However, we have seen through theory and an example that the voltage signal can be highly amplified and its voltage gain is only limited by the power supplied in the collector branch. As opposition to the **Common Collector Amplifier**, the input resistance of a CBA configuration is low and its output impedance is high. This feature makes CBAs very useful to interpose between low load inputs and high load outputs such as in radio frequency circuits. Finally, we have seen through an example how to practically calculate the voltage, current and power gains of a CBA configuration.

As a general conclusion, we have seen during this tutorial three elementary configurations of bipolar transistor-based amplifier : the **Common Emitter Amplifier** (CEA), the **Common Collector Amplifier** (CCA) and the **Common Base Amplifier**.

We summarize and give in the following a comparison of these different configurations:

- In absolute value, the voltage gain is the same for CEA and CBA configurations. However the CEA shifts the signal of a 180° phase since it has a sign “-“, therefore, the CEA inverts the signal.
- The input resistance is : approximately the same for CEA and CCA configurations.
- The output resistance is : the same for CEA and CBA configurations.
- The voltage gain is : high for CEA and CBA, ≅1 for CCA.
- The current gain is : high for CEA and CCA, ≅1 for CBA.
- The power gain is : very high for CEA, high for CBA, medium for CCA.

Finally, due to their different characteristics, the applications of these three configurations are also different :

- The CEA, due to its high voltage and current gains is used as a universal amplifier.
- The CCA, due to its high input and low output resistances is used as a step-down impedance adapter. It is also used as a current amplifier and an oscillator.
- The CBA, due to its low input and high output resistances is used as a step-up impedance adapter. It is also used as a voltage amplifier, an oscillator or a high frequency amplifier thanks to its good behavior in frequency.