52.3.3. Voltage Gain and Input Impedance

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, A B that 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 introduces the concept of voltage gain using a common base amplifier circuit. The supply voltage is mentioned as being 12V, and resistances R_A and R_B are both 100 kΩ. The Thevenin equivalent voltage at the base node is calculated to be 6V, and the Thevenin resistance is given as 50 kΩ. Additionally, the base terminal current is specified as 5 µA, which is crucial for understanding how the circuit reacts based on supply and input parameters.

Think of a water park where the supply of water (akin to supply voltage) needs to maintain steady pressure through various slides (the electrical circuit). The Thevenin voltage is like knowing the exact level of water in a tank at the top of a slide, making it critical to understand how much pressure (base terminal current) you need to maintain the ride quality and safety.

So, if we consider the IR drop with respect to Thevenin equivalent voltage, so, at the base terminal the DC voltage is coming 5.75 V. On the other hand, the collector current is 0.5 mA. The R value it is 6 kΩ. So, the drop across this 6 kΩ it is 3 V.

Here, we calculate the voltage drop at the base terminal after considering the equivalent circuit parameters. The DC voltage is deduced to be 5.75V at the base terminal when accounting for the IR drop. The collector current flows at 0.5 mA and induces a voltage drop across the resistor R_C, calculated as 3V using the formula V = I × R.

Imagine you’re riding a bike down a hill. The steepness of the hill (voltage) increases your speed (current), but you also have to navigate curves and traffic (resistors) that may slow you down. Understanding the overall potential energy as you navigate becomes akin to calculating the voltage drop across each part of the circuit.

So, this is output signal swing in the negative side. So, this is the possible output swing in the negative side till the transistor it is entering into saturation region. On the other hand, if we consider the positive swing, now we consider output swing in the positive direction.

This section discusses how to compute output swing for both negative and positive scenarios. The output swing is evaluated based on the DC voltage and saturation limit. The negative swing is calculated by evaluating the maximum collector voltage allowed before hitting the 0.3V threshold for saturation. Similarly, the positive swing calculates based on the DC voltage maximizing the collector output limit based on supply voltage.

Think of a swing set in a playground. The maximum height you can swing back and forth (output swing) is limited by how high the swing was put (supply voltage). If you swing too low, the swing will stop altogether (negative swing), but if you try to climb the vertical bar (positive swing), the swing will be limited by how far back you can push off from the ground.

Coming to other aspect that 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.

This chunk highlights the importance of input impedance in a common base amplifier circuit. It’s noted that the inherent low input impedance can lead to signal attenuation if the source resistance is not proper. It implies careful design considerations are required when dealing with practical source resistances.

Consider trying to pour syrup from a thicker bottle (representing high input impedance) into a small cup. If the cup’s opening (source resistance) is too small, the syrup will not flow smoothly and may cause drips (attenuation). The design must ensure the bottle and cup are appropriate for each other to allow a steady flow of syrup (electrical signals).

Now we can compare each of these terms and you may recall the value of g ; g is ℧ and r is 5.2 kΩ and R is 10 kΩ.

In this part, the concept of current gain is evaluated. The gain is derived from comparing the small signal parameters, focusing on the transconductance g_m and the associated resistances in the circuit. These comparisons are essential to elucidate how effective the circuit will be in amplifying the current.

Picture a relay race where the baton must be efficiently passed from one runner to another (current gain). The runners (electronic components) need to be of equal speed and strength. If one leg in the relay is slower (higher resistance), it affects the overall speed of the team (current gain) and their success in completing the race efficiently.

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