52.2. Common Base Amplifier

Welcome back after the short break. What we are discussing is Common Base Amplifier and we will be going for Common Gate amplifier, but before going I have another example based on the common base where we are talking about practical circuit of this base bias. Instead of having ideal separate voltage source here along with the Thevenin equivalent resistance R, practical circuit wise we may have only one supply voltage and from that we need to generate whatever the voltage we like to generate here.

In this chunk, we introduce the common base amplifier and its practical applications. The base amplifier typically relies on ideal voltage sources and resistances. Here, we see that in practical situations, engineers often work with a single voltage supply and must derive the necessary biasing voltages from it. This involves using components like resistors to create a potential divider, which will provide the required voltage to effectively operate the amplifier.

Think of a common base amplifier like a chef who needs a specific ingredient (voltage) to perfect a dish (the amplified signal). Instead of having access to all the ingredients (ideal sources), the chef only has one pantry item (the supply voltage) and uses different kitchen tools (resistors) to derive what is needed from that single item.

Let us try to see the operating point of the transistor by considering R, R, and so on. In this case V also I have changed. So, instead of 10 V it is 12 V and R and R both are 100 kΩ.

In this part, the focus is on determining the operating point of the transistor in the amplifier circuit. The operating point is crucial because it defines the DC conditions under which the amplifier operates. By changing the voltage to 12V and adjusting the resistors (both being 100 kΩ), the calculations hinge on understanding the relationship between these components to establish how the transistor can be made to work optimally under these conditions.

Imagine tuning a musical instrument. Changing the tension (voltage) and ensuring the right thickness of the string (resistors) is akin to setting the operational point of a transistor — it's about getting everything precisely right for optimal output, much like making sure your guitar sounds perfect before a performance.

So, if I multiply this base current with β, we can get the collector current and the emitter current is of course, (1 + β)‧I.

This segment explains the relationship between base current, collector current, and emitter current in a bipolar junction transistor (BJT). The symbol β (beta) represents the current gain of the transistor. The emitter current can be determined by the formula (1 + β) times the base current (I). This principle is key to understanding how well the common base amplifier can amplify signals.

Consider using a water pump. The base current represents the amount of water you start with, while β is like the efficiency of the pump — it determines how much more water you can push out into the system. The total water flow (emitter current) you achieve is determined by your initial input, multiplied by the efficiency of that pump.

Once we obtain the operating point of the transistor, then again we can go for calculating the value of the small signal parameters namely g.

In this chunk, the focus is on calculating small signal parameters once the operating point is established. Small signal parameters, like transconductance (g), help in determining how the amplifier behaves in response to small input fluctuations around the operating point. This behavior is critical for analyzing amplifier performance under actual working conditions.

Think of this process as fine-tuning a speaker to adapt to the acoustics of a room. Once you know how the sound system is set (the operating point), you can adjust the equalizer settings (small signal parameters) to ensure it performs optimally for varying music inputs, providing clear and balanced sound.

This gives us the DC voltage, at the output node is 9 V and at the base is 5.75 V.

This portion analyzes the output swing of the common base amplifier, where the output voltage and base voltage are calculated. The output swing defines the range of output voltage changes that the amplifier can handle without distortion. Here, it's shown that from the 9 V output voltage, the amplifier can swing downwards to approximately 5.45 V before it risks entering saturation, while it can theoretically swing upwards towards the supply voltage.

Imagine a swing set on a playground; the height of the swing (output voltage) can vary from a maximum (12 V) to a minimum (5.45 V). The swing must operate smoothly within this range, just like an amplifier must handle input signals without distortion, ensuring children can enjoy safe, joyful rides.

If I say that we are shorting this node here; the corresponding v = 0. And if you recall the small signal model here...

This chunk discusses the current gain of the common base amplifier, detailing how input and output currents relate in terms of their gain. It highlights that the current gain is typically about 1, meaning that the input current approximately matches the output current under many conditions, although there are corrections for more precise calculations. This reflects the ability of the amplifier to effectively transfer signal strength without amplification.

Consider a relay system where one small switch can control a large gate. The input current (the small switch) controls the output current (the gate operation), effectively maintaining balance as it opens or closes. The common base amplifier operates similarly, maintaining the flow of current with minimal loss during transfer, akin to how the switch governs the gate's movements.

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