Electrical – Two-Stage Common Emitter Amplifier Is Clipping Unexpectedly

Transistor amplifiers always have a certain amount of distortion, since the amplifying devices are nonlinear. In this case the current through the transistor depends exponentially on the input voltage.

A smaller signal will exhibit less distortion, so reducing the amplitude could be a first step. To ensure a proper operating point the output signal should be probed at the collector of Q1.

For a further reduction of the distortion feedback is required, which in this case could be a simple emitter resistor.

Feedback decreases distortion but decreases the gain as well. A multistage amplifier with global feedback could be used, but an opamp based design is usually easier and offers good performances since a large gain is available for feedback.

When constructing a low-gain amplifier like this, you need to pay attention to the voltage swing at both the emitter and collector of the transistor.

Let's say that you want to maintain a minimum voltage of 500 mV across the transistor in order to keep it from saturating. This means that the remaining 8.5V of your power supply will be distributed as 4.64V across the collector resistor and 3.86V across the emitter resistor.

This means that the collector can swing at most from 9V - 4.64V = +4.36V (minimum, near saturation) to 9V (maximum, near cutoff). Therefore, you want to set your Q point to close to the middle of this range (about 6.68V) in order to maximize the available signal swing.

This corresponds to an emitter voltage of 1.93V at the Q point, so you should design a bias network that has a Thevenin equivalent voltage of about 2.58V (V_E + 650 mV).