88.1.4. Lecture – 88

Dear students, welcome back after the break. So, we are going through different numerical examples and now we are going to talk about one common emitter amplifier which is using current mirror and particularly to bias the active load say Q. We are using current mirror and transistor-1; it is the amplifying device then we are assuming that Q and Q they are identical and also we are assuming that whatever this Q and Q are also identical.

In this chunk, we are introduced to the topic of common emitter amplifiers. We learn that a common emitter amplifier is a configuration that amplifies voltage signals using a transistor (here referred to as transistor-1) and a current mirror. A current mirror is a circuit that replicates a current through one active device by controlling the current through another active device of a circuit. The idea here is to keep certain components, specifically transistors Q1 and Q2, identical to ensure predictable behavior in the circuit.

Think of a common emitter amplifier as a microphone in a concert. The microphone captures the sound (the input signal) and amplifies it, much like the transistor amplifies the input voltage. The current mirror acts like a manager who ensures that sound levels are consistent across all speakers (transistors) by adjusting them based on the performance of one particular microphone.

So, to get the I current of transistor-1 and collector current or transistor-4 equal. We want the current flow through transistor-2 should be equal to current flow through transistor-1. And since, Q and Q are identical having the same β value of 100.

This chunk discusses the importance of matching collector currents in a transistor circuit. The collector current of transistor-1 (I_C1) must equal the collector current of transistor-4 (I_C4). This is crucial for the circuit to function correctly, ensuring that the mirrored currents through transistors are equivalent. Since the transistors Q1 and Q2 are identical and have the same beta (β), which represents the current gain, it means they will respond the same way to voltage changes.

Imagine you are at a relay race, where each runner (transistor) must pass the baton (current) to the next with the same timing and speed to ensure a smooth handoff. If one runner is much faster or slower, it could disrupt the race plan, similar to how mismatched currents can cause issues in an electronic circuit.

So, with that we do have this current is also 2 mA and if we assume that the base current loss it is ignorable, then you can say that collector current of transistor-3 it is also 2 mA which is getting mirrored to transistor-4.

In this section, we identify the collector current of the transistors as 2 mA, which is critical for ensuring proper amplification. By assuming that the base current loss is negligible, we conclude that the currents in the relevant transistors (such as transistor-3) also reflect this 2 mA value. This means that transistors in the current mirror configuration effectively replicate the input current levels.

Think of a factory assembly line where each worker (transistor) is responsible for assembling parts (current). If one worker successfully completes their part at a rate of 2 parts per hour, and all are identical and work at the same pace, others will replicate that output without issues, ensuring the production line operates smoothly at that level.

Now, with this information let us try to find the small signal output resistance and voltage gain of the amplifier.

Here, we introduce the calculations for determining the small signal output resistance (R_out) and the voltage gain of the amplifier. The output resistance is derived from the r_o values of the transistors, which characterize how the transistors will respond to changes in output voltage. The voltage gain is calculated based on the transconductance (g_m) and the output resistance, typically resulting in a high amplification factor, indicating the ability of the amplifier to boost signal levels.

Imagine you have a water pump system. The pump’s ability to increase water pressure represents voltage gain. The resistance to the flow, akin to output resistance, determines how efficiently the pump works against potential leaks or blockages. The more effectively a pump can build pressure (higher gain) while managing flow (output resistance), the more reliable and powerful the water delivery system becomes.

As I said that the current flow current flow here and here should be equal and if you see it carefully the DC voltage here it is defined by this V – V drop. So, I should say the voltage here it is 12 V. So, the voltage here it is 12 – 0.6 so, that is 11.4 V.

This section emphasizes the importance of ensuring that current flow remains consistent across the transistors, which in turn affects the DC output voltage of the amplifier. The voltage drops are calculated based on certain parameters, resulting in a DC output voltage of 11.4 V. Understanding these voltage drops and currents within the circuit helps ensure that components operate within their intended ranges, thereby stabilizing overall performance.

Think of a battery in an electronic device. The battery provides a specific voltage (the source voltage), and as current flows through various components, some voltage (energy) gets 'used up' (drops) due to resistance. If the battery voltage is too low or if too much energy is lost, devices may fail to operate properly, similar to how the DC voltage in the circuit needs to maintain a healthy level to function reliably.

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