54.3.5. Current Gain for Common Base Amplifier

So, we do have the same circuit configuration which we have discussed in our analysis part and what we are looking for here it is, instead of find we should say it is given. So, probably what are the things are given, probably this part it is given at least the voltage swing and the either the voltage gain or most important thing is the input impedance. So, and we need to find the value of different components here. In fact, all of these components it will not be there we need to find rather this components. We do have this other information device related information so given to us and also the supply voltage is given to us then maybe the load capacitance also. And let to start with we do have the supply voltage of 12 V and then let you consider again the output swing it is a ± 4 V which means the peak to peak it is 8 V.

In this chunk, the focus is on the initial conditions and requirements for designing a common base amplifier. We are given specific parameters such as a supply voltage of 12V and an output swing of ±4V, which means the amplifier needs to handle a total output variation of 8V. The goal is to determine other circuit components based on these given parameters, thus ensuring the design meets performance specifications.

Think of designing an amplifier like setting up a water system. You need to know the maximum amount of water you want to go through the pipes, just like knowing how much voltage swing is required. If you know you can only use a certain pipe size (the supply voltage of 12V), you need to ensure that your system can handle the demand (the ±4V output swing) without overflowing or not delivering enough water.

So, the drop across this resistance it will be it should be at least 4 V and let we keep this voltage 5 V. So, that gives us this DC voltage again 7 V and then to gate the so this gives us the +ve swing +ve side it is ensured ‒ve side to get the ‒ve side swing the voltage at the base node of the transistor should be sufficiently low namely the base voltage here it should be less than (7 V ‒ 4 V) here. So, I should say this voltage should be < 3 V.

Here, the voltage drops and their contribution to the circuit's operation are discussed. The design requires that there is at least 4V across a specific resistor. By keeping one voltage at 5V, we calculate the necessary base voltage for the transistor to operate effectively in both positive and negative swings. The voltage needs to be carefully monitored, ensuring the base voltage is optimized for the correct operation of the transistor.

Imagine adjusting the settings on a kitchen faucet to get just the right amount of water flow. You need to find the correct balance between how much water pressure (voltage) is needed to fill a pot (the output swing) without overflowing. Just like too much pressure can cause splashes, having the wrong base voltage can cause problems with the amplifier's performance.

So, we can say this is 3 is to 1 right. So, I hope in my previous exercise I have done correctly namely the R and R for common base I have done the correct calculation; maybe you can check that, but anyway let us proceed with this example. So, what we have it is it is 3 : 1 ratio. And in case if we want to ignore the effect of I the current flowing through this circuit we can consider B it is almost say 10 times higher than I at least 10 times higher than I.

The focus here is on the calculation of resistor values R and R based on their ratio. A 3:1 ratio for these resistors is recommended to ensure proper operation of the amplifier. Furthermore, the chunk discusses how to approach the design ignoring the base current effects, allowing designers to simplify calculations by assuming an ideal scenario where input current is significantly larger than the output.

This is similar to planning how many people can sit at a dinner table. If you know for every three seats of one type you should have one seat of another (the 3:1 ratio), it makes it easier to plan. Likewise, simplifying calculations by ignoring smaller effects can yield clearer designs, much like ignoring minor details when arranging seating at a big dinner.

So, once we obtain this I to achieve the input impedance now we obtain I also and then C E I we are already obtained. So, from that we can calculate R = 7 V not 7 V, 12 ‒ 7 rather 5 V, . So, that gives us a 2.5 kΩ.

This part of the analysis involves calculating the collector current based on the obtained parameters. Knowing the input impedance allows us to find relating current values and subsequently enables the calculation of the required resistance for optimal operation of the amplifier. The process of logically deriving one parameter from another ensures accuracy and usability in the final design.

Think of this as measuring how much flour you need to bake a cake. Once you know how much cake batter you’re aiming for (the input impedance), you can figure out how much flour (the resistance) to use based on each ingredient's proportion. Accurate measurements result in a much tastier cake.

And you may recall the gain it = g (R ⫽ r ). And if you consider 2 mA of current value of r = so, that is 25 kΩ. On the other hand, R it is 22.5 kΩ so, we do have g here it is and (R ⫽ r ) which it is which = right. In fact, if you see here it is this part it is it is coming.

In this concluding section, the overall gain of the amplifier is determined by combining the parameters for gain analysis. The chunk emphasizes understanding how each variable relates and how to calculate the gain based on the chosen resistances and current. The recognition of achieved values relative to expectations is pivotal for any successful design.

Think of calculating the effectiveness of a marketing campaign. You want to see how many customers you reached (the gain). By assessing the amount spent (the current) and the responses received (the resistances), you can measure how effective your campaign was in a clear mathematical way.

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