BJT Common Emitter Amplifier

Overview

The BJT common emitter amplifier is a general-purpose BJT-based amplifier that it typically used for voltage amplification. It offers great voltage gain and ok current gain. The input impedance is moderate but unfortunately it has high output impedance. The output is inverted with respect to the input. It is commonly followed with a buffer circuit such as a common-collector amplifier to reduce the output impedance. The common emitter amplifier find use in audio and RF applications.

The MOSFET analogue to the BJT common emitter amplifier is the common source amplifier.

Properties:

Voltage Gain	High
Current Gain	Medium
Power Gain	Medium
Input Impedance	Medium
Output Impedance	High
Phase Shift	180°

How A Common Emitter Amplifier Works

• \(R1\) and \(R2\) are used to provide a DC bias point for the base of the transistor, using the standard resistor divider technique (to be exact, you also have to take into account that the transistor draws some current from the output of the resistor divider, but generally you can ignore that).
• \(C1\) is used to AC couple the input signal to the DC bias point – it’s value is chosen so that it appears as a short for the AC signal frequencies of interest but blocks DC.
• \(R_E\) adds emitter degeneration¹ and makes the amplifier gain more stable with variations in \(\beta\). \(C_E\) is the emitter bypass capacitor and is used to bypass \(R_E\) so that the AC signal essentially sees the emitter connected directly to ground.
• \(R_C\) is the collector resistor which helps set the voltage gain of the amplifier. Sometimes this is called the load resistor², however this can be confusing, as typically the “load” is placed after the output AC coupling capacitor.
• \(R_L\) is the load resistance. You may see this and \(C_{OUT}\) omitted from some diagrams of the common emitter amplifier.
• \(C_{OUT}\) is the AC coupling capacitor on the output, which blocks the DC component, similarly to \(C_{IN}\).

Gain Of A Common Emitter Amplifier

The voltage gain of a common-emitter amplifier (by definition) is:

$$\begin{align} A_V = \frac{v_{out}}{v_{in}} \\ \end{align}$$

Remember that \(v_{in}\) and \(v_{out}\) are lower case and represent changes in the signal (i.e. deltas, and ignore their DC levels). Now, assuming \(i_c \approx i_e\), the change in voltage at the output is:

$$\begin{align} v_{out} = - i_e R_C \\ \end{align}$$

And the change in voltage at the input is:

$$\begin{align} v_{in} = i_e (r_e + R_E) \\ \end{align}$$

Note that we have to take the internal emitter resistance \(r_e\) into account here, as the emitter bypass capacitor is going to remove the \(R_E\) term further down, leaving only \(r_e\).

Substituting these equations for \(v_{in}\) and \(v_{out}\) into the gain equation gives:

$$\begin{align} A_V &= \frac{- i_e R_C}{i_e (r_e + R_E)} \nonumber \\ &= -\frac{R_C}{r_e + R_E} \\ \end{align}$$

Remember that the value for \(r_e\) is dependent on the emitter current at the DC bias point:

$$\begin{align} r_e &= \frac{25mV}{I_E} \\ \end{align}$$

Thus, for our signal frequencies at which the \(C_E\) capacitor shorts out external resistor \(R_E\), the emitter resistance is just \(r_e\) and the gain becomes:

$$\begin{align} A_V &= -\frac{R_C}{r_e} \nonumber \\ &= -\frac{I_E R_C}{25mV} \\ \end{align}$$

Common Emitter Amplifier Design Process

How do you design a common emitter amplifier? Let’s do a worked example to progress through the design steps.

Assumptions

• \(V_{CC}\) is \(12V\)
• We’ll be using the venerable BC548BTA NPN transistor from onsemi in our amplifier.
• We’re trying to get as much gain as possible (a noble quest).

Steps

1. Choose collector current: Chose a suitable DC collector current for your amplifier. A reasonable choice would be \(I_C = 10mA\) (max. \(I_C\) for the BC547B is \(100mA\)).
   
   

2. Determine the emitter resistor \(R_E\): As a rule of thumb, 10% of \(V_{CC}\) is normally dropped across \(R_E\)^{3 4}:

   $$\begin{align} V_{R_E} &= 0.1V_{CC} \nonumber \\ &= 0.1*12V \nonumber \\ &= 1.2V \nonumber \\ \end{align}$$

   
   

   And then:

   $$\begin{align} R_E &= \frac{V_{R_E}}{I_{R_E}} \nonumber \\ &= \frac{1.2V}{10mA} \nonumber \\ &= 120\Omega \nonumber \\ \end{align}$$

   
   

3. Find the collector resistor \(R_C\): We are dropping \(1.2V\) across the emitter resistor. That leaves \(10.8V\) to be dropped across the collector resistor and the BJT. Assuming a saturation voltage of 200mV, this gives the BJT \(10.6V\) of swing. For maximum symmetrical output, we want to drop half of this \(10.6V\) across the collector resistor:

   $$\begin{align} R_C &= \frac{V_{R_C}}{I_{R_C}} \nonumber \\ &= \frac{0.5*10.6V}{10mA} \nonumber \\ &= 530\Omega \nonumber \\ \end{align}$$

   
   

4. Find the base current: Calculate \(I_B\) using the approximate gain:

   $$\begin{align} I_B &= \frac{I_C}{\beta} \nonumber \\ &= \frac{10mA}{200} \nonumber \\ &= 50uA \nonumber \\ \end{align}$$

   
   

5. Determine the base voltage \(V_B\): \(V_B\) is just the emitter voltage plus the diode \(V_BE\) drop:

   $$\begin{align} V_B &= V_E + V_{BE} \nonumber \\ &= 1.2V + 0.7V \nonumber \\ &= 1.9V \nonumber \\ \end{align}$$

   
   

6. Calculate resistor divider values: Chose \(R1\) and \(R2\) to set the output of the resistor divider to match this base voltage. We also want to make sure the current flowing through the resistor is 10x the current that will be sucked out of it into the base of the transistor, that way we can ignore the loading of the BJT when calculating the resistor values.

   $$\begin{align} I_{R2} &= 10 \cdot I_B \nonumber \\ &= 10 \cdot 50uA \nonumber \\ &= 500uA \nonumber \\ \end{align}$$

   Now we can easily calculate the value of \(R2\):

   $$\begin{align} R2 &= \frac{V_{R2}}{I_{R2}} \nonumber \\ &= \frac{1.9V}{500uA} \nonumber \\ &= 3.8k\Omega \nonumber \\ \end{align}$$

   And \(R1\):

   $$\begin{align} R1 &= \frac{V_{R1}}{I_{R1}} \nonumber \\ &= \frac{12V - 1.9V}{500uA} \nonumber \\ &= 20.2k\Omega \nonumber \\ \end{align}$$

7. Calculate input AC coupling capacitor: The rule of thumb is to make sure the impedance of the capacitor is 10x less that the AC impedance of the resistor divider at the lowest frequency of interest⁵. Our lowest frequency of interest is \(20Hz\).

   $$\begin{align} R_{in} &= R1 || R2 \nonumber \\ &= \frac{R1 \cdot R2}{R1 + R2} \nonumber \\ &= \frac{20.2k\Omega \cdot 3.8k\Omega}{20.2k\Omega + 3.8k\Omega} \nonumber \\ &= 3.20k\Omega \nonumber \\ \end{align}$$ $$\begin{align} Z_{C_{in}} &= \frac{R_{in}}{10} \nonumber \\ &= \frac{3.20k\Omega}{10} \nonumber \\ &= 320\Omega \nonumber \\ \end{align}$$ $$\begin{align} C_{in} &= \frac{1}{2\pi f Z_{C_{in}}} \nonumber \\ &= \frac{1}{2\pi \cdot 20Hz \cdot 320\Omega} \nonumber \\ &= 25uF \nonumber \\ \end{align}$$

8. Calculate emitter bypass capacitor: The same rule of thumb applies to \(C_E\), except this time it’s impedance should be 10x smaller than \(R_E\):

   $$\begin{align} Z_{C_E} &= \frac{R_E}{10} \nonumber \\ &= \frac{120\Omega}{10} \nonumber \\ &= 12\Omega \nonumber \\ \end{align}$$ $$\begin{align} C_E &= \frac{1}{2\pi f Z_{C_E}} \nonumber \\ &= \frac{1}{2\pi \cdot 20Hz \cdot 12\Omega} \nonumber \\ &= 663uF \\ \end{align}$$

9. Calculate the gain:

   $$\begin{align} A_V &= -\frac{I_E R_C}{25mV} \nonumber \\ &= -\frac{10mA * 530\Omega}{25mV} \nonumber \\ &= -212 \\ \end{align}$$

   Or in dB:

   $$\begin{align} A_{V(db)} &= 20\log(A) \nonumber \\ &= 20\log(212) \nonumber \\ &= 46.5dB \\ \end{align}$$

10. Done!

The finished schematic, along with voltage sources ready for simulation is shown below.

Given the large gain of \(46.5dB\), I didn’t want to saturate the output so I chose an input sine wave signal with an amplitude of \(10mV\) at a frequency of \(1kHz\). The simulated input and output voltages are shown below (note the change in the y-axis scale - the input is in \(mV\) and the output is in \(V\)).

The simulated frequency response shown below is close to what we expect. The simulated gain of around \(42dB\) is close enough to our calculated gain of \(46.5dB\) considering all the approximations we made! The phase shift is \(180^{\circ}\) for most of our signal bandwidth.

References