51.2. Next Example with Practical Source Resistance

So yes, so this is this is what I was telling, that all the parameters here we are taking same except the source resistance. So, we are considering the source resistance it is 10 kΩ. Now, let us see what will be the corresponding voltage gain.

In this chunk, the speaker introduces the next numerical example in the context of common base amplifiers. The key change in this example is the inclusion of a practical source resistance of 10 kΩ. This change will impact the voltage gain of the amplifier, as the source resistance now plays a crucial role in the circuit's performance. In previous examples, the source resistance was considered to be zero, simplifying calculations but not reflecting practical conditions.

Think of this like trying to fill a bucket (the amplifier) with a hose (the signal). If the hose is wide (low source resistance), the water (signal) flows easily into the bucket. But if the hose is narrow (high source resistance), it restricts the flow, and the bucket may not fill as quickly or fully, representing the effects of the source resistance on the amplifier's performance.

So, the voltage gain from this point to this point namely emitter to collector we already have seen that voltage gain it is g (r ⫽ R ). And then if I consider from this point to this point primary input to the emitter node it is having attenuation and that attenuation it is.

Here, the speaker refers to the formula for voltage gain in a common base amplifier setup, which is influenced by the small-signal transconductance (g) and the output resistance (R). He points out that the gain from the primary input (where the signal originates) to the emitter is affected by attenuation due to the added source resistance. This basically means that the voltage gain is lower when factoring in the resistive load of the input signal source.

Imagine trying to listen to music from your phone through a set of weak Bluetooth speakers. If the connection strength (analogous to input voltage) is high, you hear the music well. If the connection is weak (due to too much resistance in the source), the sound becomes faint or distorted – this illustrates how a practical source resistance can reduce overall voltage gain in an amplifier circuit.

So, the overall gain now with R = 10 k it is quite small, I mean let me think I did some calculation on that yes. So, this attenuation factor 108.85 getting multiplied by; multiplied by this attenuation factor ...

This segment discusses the overall gain of the amplifier circuit with the 10 kΩ source resistance. The speaker mentions how the previously calculated gain of 108.85 is significantly decreased due to the attenuation caused by the source resistance. This demonstrates a critical point in circuit design: as source resistance increases, it can drastically lower the voltage gain, potentially below a usable threshold.

It’s like trying to talk to a friend across a busy street. If you shout loudly (high initial gain), but a loud truck passes (like source resistance), your voice might not reach your friend effectively, reducing the clarity of your message. Similarly, when the source resistance increases, less of the intended signal reaches the amplifier, thus lowering gain.

Now, what is the effect on the output impedance on the other hand, since we are considering the finite resistance R . So, I let me use different color to explain that, output resistance if I see it is having 2 components one is R and the resistors coming from the active part.

In this part, the speaker explains how the output impedance is affected by both the inherent resistances in the circuit. The total output impedance is determined by the combination of the source resistance (R) and the active component resistance. These components interact, and their relationship can significantly influence the overall output impedance of the amplifier circuit.

Imagine you have two pipes (resistors) connecting to a garden faucet (active component). If one pipe is narrow (high resistance), it will restrict water flow (current). The combined effect of both pipes determines how fast water can flow out. Similarly, in the amplifier, if the output impedance is too high, it reduces the flow of signal strength making it less effective.

So, let me clear this board and then again rewrite with different color ... So, the bottom line here it is rather I should say the information other way, even though R it is say 10 kΩ.

The speaker discusses how upper cutoff frequency relates to both input capacitance and output resistance in the system. It becomes evident that while the source resistance can introduce lower gain, it does not significantly impact the overall upper cutoff frequency as long as the input capacitance remains low. This analysis is crucial for understanding bandwidth in amplifier designs and helps characterize how the circuit will perform under different conditions.

Consider a highway where traffic speed (analogous to frequency) is maintained smoothly. If the roads are wide enough (low input capacitance), even with some cars (source resistance) present, cars can still travel fast (high frequency). This leads to a high upper cutoff frequency, indicating that the design allows for quick signal changes without significant delay or distortion.

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