# Common Emitter Amplifier Design

### - common emitter amplifier design - a simple approach to determining the component values for the performance required.

### Transistor Circuits Section Includes

The common emitter amplifier is one of the most common transistor amplifier configurations.

Fortunately the common emitter amplifier design is quite easy to realise.

It is easy to design the common emitter amplifier using a straightforward approach to the calculations. In this way it is possible to quickly choose the component values required.

Some of the design requirements and circuit constraints will depend upon the particular application. For example an AC coupled amplifier will differ from that required for being driven by logic, etc. To overcome this examples of an AC common emitter amplifier design, and a simple logic buffer using a common emitter design will be shown.

## Simple logic common emitter amplifier design

This very simple design for a logic buffer or common emitter amplifier design is about as simple as any design can be.

**Simple logic buffer using a common emitter amplifier design**

A common emitter amplifier acting as a buffer for a logic IC is very easy to design.

Although not the only way to design the stage, the following process flow could be used.

The choice of the transistor, marked Q1 in the diagram will depend upon a number of factors:*Choose transistor:*

- Power dissipation anticipated.
- Switching speed required - for switching applications choose a switching transistor, not another form of transistor with a high bandwidth, ft.
- Current gain required.
- Current capability required.
- Collector emitter voltage.

The determination of the collector resistor, R2 is achieved by determining the current required to flow through the resistor. This will depend upon elements such as the current that the circuit needs to deliver. It may also be that an LED indicator is required in series with the collector resistor. The current should be determined to give the required light output. The value of the resistor can be determined using Ohms law, knowing the current flowing through the resistor and the voltage across it.*Calculate collector resistor:*The base current is the collector current divided by the value of β or hfe which is virtually the same. Ensure that there is sufficient current drive to turn the transistor ON for the lowest values of β even at low temperatures where values for β will be lower. Care should be taken not to drive excessive current into the base as switching may take longer as a result because excess stored charge needs to be removed.*Determine base resistor value:*Once the design has been competed it is necessary to re-evaluate some of the initial decisions and estimates in case the final design has changed anything.*Re-evaluate initial assumptions:*

## Simple AC coupled common emitter amplifier design

The basic circuit for an AC coupled common emitter amplifier design is shown below.

**Basic common emitter amplifier design**

This circuit is not widely used because it is difficult to define the exact operating point of the circuit as a result of variations in the values of β encountered. Again the flowing process flow could be used:,/p>

The choice of transistor will depend on factors including the power dissipation anticipated, collector emitter voltage, bandwidth and the like.*Choose transistor:*The value of this should be chosen so that the collector sits at about half the supply rail for the required current. The resistance value can be determined simply using Ohms law. The current value should be chosen to give resistance / output impedance that will be acceptable for the following stage.*Choose collector resistor:*Using the β figure for the transistor, determine the base current. Then using Ohms law and a knowledge of the supply voltage and the fact that the base will be 0.5V (for silicon) above ground, calculate the resistor.*Choose base resistor:*Using a knowledge of the input and output impedances, determine the value of capacitor to equal the impedance at the lowest frequency of use. (Xc = 2π f C where C is in Farads and the frequency is in Hz).*calculate decoupling capacitors:*Revisit all calculations and assumptions to ensure they are all still valid in the light of the way the circuit has developed.*Revisit calculations:*

## Comprehensive AC coupled common emitter amplifier design

**Common emitter amplifier complete with bias resistors, decoupling, etc**

The common emitter amplifier design is relatively straightforward. The following design flow can be used as a basis.

As before, the transistor type should be chosen according to the anticipated performance requirements.*Choose transistor:*It is necessary to determine the current flow required to adequately drive the following stage. Knowing the current flow required in the resistor, choose a collector voltage of around half the supply voltage to enable equal excursions of the signal up and down. This will define the resistor value using Ohms law.*Calculate collector resistor:*generally a voltage of around 1 volt or 10% of the rail value is chosen for the emitter voltage. This gives a good level of DC stability to the circuit. Calculate the resistance from a knowledge of the collector current (effectively the same as the emitter current) and the emitter voltage.*Calculate the emitter resistor:*It is possible to determine the base current by dividing the collector current by β (or hfe which is essentially the same). If a range for β is specified, work on the cautious side.*Determine base current:*This is easy to calculate because the base voltage is simply the emitter voltage plus the base emitter junction voltage. This is taken to be 0.6 volts for silicon and 0.2 volts for germanium transistors.*Determine the base voltage:*Assume a current flowing through the chain R1 + R2 of around ten times that of the base current required. Then select the correct ratio of the resistors to provide the voltage required at the base.*Determine base resistor values:*The gain of the circuit without a capacitor across the emitter resistor is approximately R3/R4. TO increase the gain for AC signals the emitter resistor bypass capacitor C3 is added. This should eb calculated to have a reactance equal to R4 at the lowest frequency of operation.*Emitter bypass capacitor:*The value of the input capacitor should equal the resistance of the input circuit at the lowest frequency to give a -3dB fall at this frequency. The total impedance of the circuit will be β times R3 plus any resistance external to the circuit, i.e. the source impedance. The external resistance is often ignored as this is likely to not to affect the circuit unduly.*Determine value of input capacitor value:*Again, the output capacitor is generally chosen to equal the circuit resistance at the lowest frequency of operation. The circuit resistance is the emitter follower output resistance plus the resistance of the load, i.e. the circuit following.*Determine output capacitor value :*In the light of the way the circuit has developed, re-assess any circuit assumptions to ensure they still hold valid. Aspects such as the transistor choice, current consumption values, etc.*Re-evaluate assumptions:*

Although there may appear to be a large number of stages to designing the common emitter amplifier, they are all relatively straightforward and do not take long. The calculations and requirements may change the design flow slightly in order to meet the requirements for a particular circuit design.

* By Ian Poole*

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