52.1.5. Common Base and Common Gate Amplifiers (Contd.): Numerical Examples (Part B)

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 you know 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. So, we can have a potential divider constructed by say R and R connected to ground which generates the voltage here and also we may have a situation that instead of having this ideal current source, we may have some practical component either active device or passive bias circuit.

In this section, the speaker introduces a practical circuit example based on the common base amplifier. The focus is on how real circuits do not always use ideal components (like separate voltage sources or perfect current sources) but rather practical configurations that combine resistors to create desired voltage levels. This sets the stage for understanding how to analyze circuits where components have non-ideal characteristics, which is crucial in electronics design.

Think of it like cooking without having all the perfect ingredients. Sometimes you have to make do with what you have, like using a different spice than the recipe calls for. In electronics, just as in cooking, adapting to what's available is key to successfully achieving the desired outcome.

In the next slide we have the example, here as I said that the voltage for the base we are generating in this base voltage by V and then the potential divider constructed by R and R and the value of R and R are given here. The other information about the BJT as well as the other parameters are remaining same which are enlisted here the R intentionally we are making slightly different value or I should say we have made some change here and at the emitter instead of having ideal bias, we have R and its value is given here close to 10 kΩ.

This chunk discusses specific parameters and configurations used in the common base amplifier, specifically mentioning the use of a voltage divider created by resistors to set the base voltage. It emphasizes the changes made in component values (like the emitter resistor) to reflect practical scenarios rather than ideal ones, which is essential for understanding real-world applications of amplifiers.

Imagine you're tuning a musical instrument. You may have to adjust the tuning pegs multiple times to get it just right for a specific song. Similarly, in designing circuits, engineers must fine-tune component values to achieve optimal performance in real-world situations.

Let us try to see the operating point of the transistor by considering R , R , and so on. A B Now, in this case V also I have changed. So, instead of 10 V it is 12 V and R and R are both 100 kΩ. So, we can say the voltage source coming to the base is 6 V.

In this section, the speaker focuses on determining the operating point of the transistor based on the new parameters, including changed supply voltage and resistor values. Understanding the operating point is critical as it affects how the amplifier functions — essentially ensuring that the transistor operates in the correct region for amplification.

Think of it like finding the best speed for a car in different driving conditions. If you change the road (like changing the voltage supply), you need to find the optimal speed (operating point) for that condition to ensure a smooth and efficient ride.

So, if I consider this loop and if I see the potential drop, 6 ‒ 0.6 = I × 50 k + 101I × 10.306 k. So, that gives us I = . And that gives us close to 4.95 µA of base current. Now, if we multiply this base current with β, we can get the collector current and the emitter current is of course, (1 + β)‧I .

This segment involves solving for the base current, collector current, and emitter current in the transistor. The calculations show the relationship between these different currents and their significance in the performance of the common base amplifier. It illustrates a hands-on approach to understanding current flow in a transistor and the importance of keeping track of these parameters.

Imagine a water system where the flow of water (current) in a pipe changes based on the pressure (voltage) and pipe diameter (resistance). Similarly, in an amplifier, the base current flows through the transistor, affecting the overall output like water flowing through a system affects the overall output pressure.

The next thing we are going to talk about the output swing on the circuit and as we have discussed that the supply voltage is 12 V and R and R both are equal to 100 kΩ. So, this gives us the Thevenin equivalent voltage. At the base node it is 6 V and the Thevenin equivalent resistance is 50 kΩ. Now, the base terminal current of 5 µA as we have discussed before.

This chunk describes how to determine the range of output voltage swings from the common base amplifier. It explains how to calculate the maximum and minimum voltage values the output can reach and discusses the factors that limit the swing, including transistor saturation. Understanding output swing helps in designing circuits that can handle desired signal variations without distortion.

Consider a swing set in a park. The height of the swing can vary; however, it is limited by the height of the chains and the ground beneath. Similarly, the output voltage swing is constrained by the transistor's limits and supply voltages.

So, the input impedance of this circuit is expected to be low. If the source resistance is significant, then the input signal will be experiencing significant amount of attenuation. In fact, in practical purposes when we consider R in the range of kilo ohms or say 10 kΩ, it will be having significant large amount of attenuation. So, instead of considering the circuit in the form of feeding the signal in the form of voltage in practical purposes we consider let the signal be feed into the input terminal in the form of current.

This section highlights the concept of input impedance in the context of common base amplifiers, explaining that a low input impedance results in signal attenuation when a significant source resistance is present. The text suggests feeding the signal as a current instead of a voltage to mitigate losses, framing the common base amplifier as more of a current amplifier.

Imagine trying to pour a thick liquid (like syrup) through a narrow straw (low impedance). The syrup won't flow easily, resulting in a significant waste of the syrup. Similarly, input impedance affects how effectively a signal can be processed in an amplifier circuit.

First of all on the stimulus part we need to replace current source. Signal current source is i and it may be having a finite conductance and this is conductance is . Now, this signal it is going here and again through this capacitor the signal is arriving to the emitter node. Now, once we have this i we are feeding at the emitter node and then we like to see how much the current we will be getting here particularly in unloaded condition.

The text discusses the concept of current gain in common base amplifiers by analyzing the behavior of the circuit when a signal current is introduced. It emphasizes understanding the unloaded condition where the output is directly connected to ground, allowing for observing the amplification of current without additional loads. This is crucial for assessing the efficiency of the amplifier.

Think about a person trying to lift weights. If they lift freely (unloaded condition), they can gauge their strength effectively. But if they lift while carrying a heavy backpack (loaded condition), it becomes difficult to understand their true strength. Similarly, analyzing current gain in an unloaded state helps us see how effectively the amplifier can perform without extra burdens.

So, what we have to cover here this numerical example and in this numerical example what is the important point here it is in case if we are not using the C what kind of affect it may come and this is important for common base amplifier.

The section concludes the discussion on the numerical example, highlighting that there are practical implications to ignoring certain components (like capacitors) and their effects on the performance of the common base amplifier. This reinforces the importance of understanding all elements within a circuit to ensure proper functionality.

Just like forgetting to include certain ingredients in a recipe can lead to a failed dish, neglecting specific components in an electronic circuit design can result in improper performance. It emphasizes the need for careful consideration and inclusion of all necessary parts when designing.

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