51.1.5. Voltage Gain Calculation

So, let us go to the numerical example of common base amplifier. What we do have here it is the basic circuit given here and you can see that we do have ideal bias as well as more practical bias of the collector terminal. And also at the base node of this circuit we do have a DC voltage along with Thevenin equivalent resistance of the DC source. And since it is common base configuration and we want this base node should be AC ground to ensure that we are also connecting C large capacitor there.

This chunk introduces the setup of a common base amplifier, mentioning the ideal and practical biases in the circuit. The base node is kept at AC ground through a large capacitor, which allows AC signals to pass while blocking DC. This capacitor ensures that the base does not influence the AC performance.

Think of the capacitor as a door that allows people (AC signals) to come in and out freely while keeping the room (the circuit) at a stable temperature (DC voltage).

If we see that this I it is given as 1 mA and β is 100. So, we may approximate that collector current I = I and which is same as 1 mA, so this is the DC current. And then the voltage at the base it is given as 6 V and the emitter voltage here it can be obtained by considering 6 V here and V all of 0.6 V. So, we can say at the emitter node we do have 6 ‒ 0.6 that is 5.4 V.

To find the operating point, you need the collector current (1 mA) and the DC voltage at the base (6 V). The emitter voltage is then calculated by subtracting the base-emitter voltage drop (0.6 V from a BJT) from the base voltage. This yields an emitter voltage of 5.4 V, which is critical for determining how the transistor operates within the desired region.

Imagine a water tank where the base voltage is like the height of the water level, and the BJT drop (0.6 V) is akin to a pipe that reduces the height. The effective water height (emitter voltage) is what determines how high you can pump the water (current) down the line.

So, using those parameters so we can find the corresponding voltage gain. So, let us see how we do get the voltage gain. So, the voltage gain starting from the emitter terminal to the collector terminal and if I call that voltage gain it is A , you may recall from our previous days discussion it was ( ). In fact, if you do if you compare and 1 definitely, we can approximate this by only.

The voltage gain (A_v) of the amplifier is determined by the ratio of the output voltage change to the input voltage change. This relationship simplifies to the output resistance and the transconductance when calculated correctly. In this amplifier, as noted, the gain is approximately 108.85 due to favorable parameters.

Consider a loudspeaker. If you shout (input signal) and the speaker amplifies your voice to ten times louder (output signal), the loudness ratio is like the voltage gain — showing how effectively your input produces a larger output.

Now, next thing is that we can try to find the output impedance. The output resistance it is resistance of this R this path and then the other resistance coming from the active device in parallel. So, that is in parallel with whatever the impedance will be seen.

The output impedance is significant in understanding how much the amplifier will load the previous stage. It's calculated based on the resistances in the circuit and the characteristics of the transistor. The aim is to keep the output impedance low to enable better voltage transfer to the load.

Think of output impedance like the width of a pipe. A narrower pipe (higher impedance) restricts water flow, while a wider pipe (lower impedance) allows for better flow to the garden (load).

if we see the capacitance at this node for this signal, of course this coupling signal coupling capacitor it is quite large. So, whenever we are talking about input capacitance it is coming the emitter to whatever the ground node AC ground node will be considering that is the net capacitance.

In examining the input capacitance, the capacitors define how the amplifier will respond to high-frequency signals. It's important because it sets an upper cutoff frequency, essentially determining the range of frequencies the amplifier can handle effectively.

Imagine a sponge that holds water (capacitors). If the sponge is too porous (too much capacitance), it can soak up frequencies that you might not want, like unwanted noise. The right amount of porosity allows you to keep only the useful 'signals'.

--