The common emitter amplifier is a cornerstone of analog electronics, offering significant voltage gain and serving as a fundamental building block in many circuits. Utilizing the 2N3904 transistor, this project focuses on creating an amplifier capable of boosting a 0.1V, 1000 Hz sine wave input to a 1.5V peak-to-peak (P2P) output, powered by a 5V supply from a Red Pitaya.

**Objective**

To design, calculate, and assemble a common emitter amplifier that achieves the desired amplification. This involves calculating appropriate resistor values for transistor biasing and ensuring the circuit operates effectively within its intended parameters.

**Materials and Setup**

- 2N3904 NPN transistor
- Resistors and capacitors for the amplifier circuit
- Red Pitaya for power supply (5V) and as a signal generator/oscilloscope
- Breadboard and connecting wires

### A typical common emitter amplifier schematics:

**Calculating the components**

Determine the desired gain:

**Calculate the emitter resistor (Re) and collector resistor (Rc):**

The gain of the common emitter amplifier is approximately Rc/Re. Letâ€™s express Rc in terms of Re:

$Rc = \text{Gain} \times Re = 15 \times Re$Using the rule of thumb we can choose the voltage across Re (Vre) to be around 10% of the power supply voltage (Vcc). In this case,

$V{Re} = 0.1 \cdot 5V = 0.5V.$Using Ohmâ€™s Law, we can calculate Re:

$Re = \frac{V_{Re}}{I_{c}} = \frac{0.5\text{V}}{5\text{mA}} = 100\Omega$Now that we have Re, we can calculate Rc:

$Rc=15Ã—100Î©=1500Î©=1.5kÎ©$**Calculate the bias resistors (R1 and R2):**

Calculate the base current (Ib) using the rule of thumb that Ib should be around 1/10 of Ic:

$I_{b} = \frac{I_{c}}{10} = \frac{5\text{mA}}{10} = 0.5\text{mA}$Calculate the voltage across R2 using the base-emitter voltage (Vbe=0,7 typically for NPN transistors) and the Re voltage :

$V_{R2} = V_{be} + V_{e} = 0.7\text{V} + 0.5\text{V} = 1.2\text{V}$Using Ohmâ€™s Law, we can calculate R2:

$R2 = \frac{V_{R2}}{I_{b}} = \frac{1.2\text{V}}{0.5\text{mA}} = 2.4\text{kÎ©}$Since the closest available value is 2.2 kÎ©, we can use that for R2.

Calculate the voltage across R1 (VR1):

$V_{R1} = V_{cc} - V_{R2} = 5\text{V} - 1.2\text{V} = 3.8\text{V}$Using Ohmâ€™s Law, we can calculate R1:

$R1 = \frac{V_{R1}}{I_{b}} = \frac{3.8\text{V}}{0.5\text{mA}} = 7.6\text{kÎ©}$Since the closest available value is 10 kÎ©, we can use that for R1.

**Calculating Capacitors and their functions:**

Using the given values of R1 = 10 kÎ©, R2 = 2.2 kÎ©, Rc = 1.5 kÎ©, and Re = 100 Î©, along with the given collector current Ic = 5 mA, we can calculate the actual capacitance values needed for Cin, Cout, and Ce, so our 1000Hz signal gets amplified correctly.

**Cin (input coupling capacitor):**

Cin is used to couple the input signal (AC component) to the amplifier while blocking any DC voltage from the input source. The value of Cin should be chosen such that it provides a low-impedance path for the input signal frequency while maintaining a high impedance for the DC component. To ensure a low reactance at the input frequency (1 kHz), the accurate capacitance value can be calculated using the next steps:

We first calculate the parallel combination of R1 and R2:

$R_{in} = \frac{R1 \times R2}{R1 + R2} = \frac{10\text{kÎ©} \times 2.2\text{kÎ©}}{10\text{kÎ©} + 2.2\text{kÎ©}} \approx 1.83\text{kÎ©}$Now, we can calculate Cin:

$C_{in} = \frac{1}{2\pi f R_{in}} = \frac{1}{2\pi \times 1000 \times 1.83\text{kÎ©}} \approx 86.8\text{nF}$A standard value close to the calculated value is 100 nF.

**Cout (output coupling capacitor):**

Cout is used to couple the output signal (AC component) from the amplifier to the load while blocking any DC voltage from the collector. The value of Cout should be chosen similarly to Cin, considering the output impedance of the amplifier (which is approximately Rc) and the desired output frequency range.

$C_{out} = \frac{1}{2\pi f R_{c}} = \frac{1}{2\pi \times 1000 \times 1.5\text{kÎ©}} \approx 106\text{nF}$A standard value close to the calculated value is 100 nF.

**CE (emitter bypass capacitor):**

CE is used to bypass the AC signal around the emitter resistor Re. This improves the amplifierâ€™s gain at higher frequencies by reducing the negative feedback. The value of Ce should be chosen such that it provides a low-impedance path for the AC signal at the desired frequency range while maintaining a high impedance for the DC component. For Ce, we can use the formula with Re:

$C_{e} = \frac{1}{2\pi f R_{e}} = \frac{1}{2\pi \times 1000 \times 100\text{Î©}} \approx 1.59\text{ÂµF}$A standard value close to the calculated value is 1.5 ÂµF or 2.2 ÂµF.

Based on these calculations, the actual capacitance values for Cin, Cout, and Ce can be approximated as 100 nF for Cin, 100 nF for Cout, and 1.5 ÂµF or 2.2 ÂµF for Ce. These values should provide good performance in the 1 kHz frequency range.

**Assembling and Measuring**

So after some calculations, we can assemble the circuit on a breadboard using the approximate resistor and capacitor values since we were limited to choosing our desired values from the kit.

Re: 100 Î©

Rc: 1.5 kÎ©

R1: 10 kÎ©

R2: 2.2 kÎ©

Cin=Cout=100nF

Ce=2.2uF

Letâ€™s assemble the circuit on the breadboard in the configuration from the schematics above and connect the Red Pitayas signal generator OUT1 to the circuit input(Cin). Then connect the IN1 to the circuit output (Cout). For help, you can refer to the picture below:

Now letâ€™s run the oscilloscope app, set the OUT1 to 1000Hz sine signal with 0.05V (0.1V P2P) and adjust the IN1 oscilloscope settings until you get a clear view of the signal. We can also use the MEAS function to display the IN2 P2P value. We obtained the results below:

From the measured output voltage, we can now calculate the exact gain we got by using our calculated component values.

$\text{Gain} = \frac{V_{out}}{V_{in}} =\frac{1,435}{0.1}=14.35$**Conclusion**

In conclusion, the common emitter amplifier using a 2N3904 transistor was designed to amplify a 0.1V, 1000 Hz sine wave input to a 1.5V P2P output with a 5V power supply. The calculated resistor values were R1 = 10 kÎ©, R2 = 2.2 kÎ©, Rc = 1.5 kÎ©, and Re = 100 Î©, resulting in a target gain of 15. However, the measured output voltage was 1.435V P2P, which indicates a slightly lower gain than the intended design.

There can be several reasons for this discrepancy, such as component tolerances, temperature variations, and the non-ideal behaviour of the transistor. Nonetheless, the amplifier circuit was successful in significantly increasing the input signal amplitude, and the measured output of 1.435V P2P is close to the target value of 1.5V P2P. Overall, the amplifier demonstrates good performance and can be considered satisfactory for many applications.

Written by AndraÅ¾ Pirc

This teaching material was created byÂ Red Pitaya https://www.redpitaya.com/