52.1. Analog Electronic Circuits

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 part, we are introduced to the common base amplifier. It’s a type of transistor amplifier configuration typically used for high-frequency applications or where a low input impedance is desired. The speaker emphasizes the importance of using practical components rather than ideal conditions, highlighting how real circuits often rely on a single power supply instead of separate ideal voltage sources. The reference to Thevenin equivalent resistance also indicates a move toward simplified circuit analysis using theoretical models.

Think of a common base amplifier like a music amplifier. Instead of using multiple power supplies to turn it on, it often uses just one outlet. Just like how we want the sound to be easily heard from one speaker rather than having a complex setup, a common base amplifier simplifies its design to get the job done efficiently.

So, 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 connected to ground which generates the voltage here.

This chunk discusses how a practical voltage is established at the base of the BJT (Bipolar Junction Transistor) using a configuration of resistors which creates a potential divider. The voltage V used in this practical setup isn’t just a single value; instead, it must be calculated based on the values of R and R, demonstrating the consideration of non-ideal conditions in circuit design.

Imagine you are pouring juice into a glass. If you pour slowly, the juice fills the glass up to a certain level, which can represent the voltage divider. Here, the juice represents the voltage generated by the resistors, making sure it doesn’t overflow (or exceed) the base voltage needed for proper transistor operation.

Let us try to see the operating point of the transistor by considering R , R , and so on. Now, 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 section, the focus shifts to calculating the operating point of the transistor, utilizing the values of resistors and the voltage applied. The operating point, often referred to as the Q-point, is crucial as it determines the performance characteristics of the amplifier. It helps ensure that the transistor operates in the desired region, avoiding cutoff or saturation under AC signal conditions.

Think of the operating point like setting the right temperature on your oven for baking. If the temperature is too low (cutoff), your cake won’t bake, and if it’s too high (saturation), it will burn. Just like understanding the right settings for baking, knowing how to find the operating point ensures that the transistor performs effectively within its limits.

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

After establishing the operating point, it's essential to find the small signal parameters, which are crucial for analyzing the transistor's response to small input voltage variations. The parameter g represents transconductance which indicates how effectively a change in the input voltage can generate a change in output current, while r is the small-signal output resistance.

Imagine you are flooding a field with water, where g represents how quickly the water can flow through tiny holes in the ground, and r stands for how resistant the ground is to this flow. The quicker water (current) can spread, the more effectively your field grows, similar to how these parameters impact the amplifier's operation.

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.

This chunk addresses the output swing – the range of output voltage that the amplifier can deliver, given the supply voltage and circuit resistance. Understanding output swing is critical in ensuring that the signal can be amplified without distortion. Additionally, it mentions input impedance, which affects how much of the input signal is actually delivered to the amplifier's input without being lost.

Think of output swing like the range of motion in a swing set. The supply voltage is like the height of the swing; it defines how far the swing can travel forward and backward (output swing). Similarly, if there are too many kids trying to push on one side (low input impedance), not everyone can enjoy the swing ride!

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 fed into the input terminal in the form of current.

This part discusses how in practical applications, instead of introducing input signals as voltages, it's often more effective to use current sources. Common base amplifiers typically behave better as current amplifiers. In cases where input impedance is low, using current instead of voltage helps minimize issues related to signal attenuation.

Consider the difference between water (current) flowing through a garden hose (amplifier) versus the pressure (voltage). If you send too much water too quickly (voltage), it might spill out and not reach the garden; whereas if you let just the right amount flow steadily (current), every droplet nourishes the plants effectively.

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