Welcome back after the short break. What we are discussing is Common Base Amplifier and we will be going for Common Gate amplifier, but before going I have another example based on the common base where we are talking about practical circuit of this base bias. Instead of having you know ideal separate voltage source here along with the Thevenin equivalent resistance R , practical circuit wise we may have only one supply voltage and from that we need to generate whatever the voltage we like to generate here. So, we can have a potential divider constructed by say R and R connected to ground which generates the voltage here and also we may have a situation that instead of having this ideal current source, we may have some practical component either active device or passive bias circuit.

In the next slide we have the example, here as I said that the voltage for the base we are generating in this base voltage by V and then the potential divider constructed by R and R and the value of R and R are given here. The other information about the BJT as well as the other parameters are remaining same which are enlisted here the R intentionally we are making slightly different value or I should say we have made some change here and at the emitter instead of having ideal bias, we have R and its value is given here close to 10 kΩ.

Let us try to see the operating point of the transistor by considering R , R , and so on. A B Now, in this case V also I have changed. So, instead of 10 V it is 12 V and R and R are both 100 kΩ. So, we can say the voltage source coming to the base is 6 V.

So, if I consider this loop and if I see the potential drop, 6 ‒ 0.6 = I × 50 k + 101I × 10.306 k. So, that gives us I = . And that gives us close to 4.95 µA of base current. Now, if we multiply this base current with β, we can get the collector current and the emitter current is of course, (1 + β)‧I .

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, this 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.

So, 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. In fact, in practical purposes when we consider R in the range of kilo ohms or say 10 kΩ, it will be having significant large amount of attenuation. So, instead of considering the circuit in the form of feeding the signal in the form of voltage in practical purposes we consider let the signal be feed into the input terminal in the form of current.

First of all on the stimulus part we need to replace current source. Signal current source is i and it may be having a finite conductance and this is conductance is . Now, this signal it is going here and again through this capacitor the signal is arriving to the emitter node. Now, once we have this i we are feeding at the emitter node and then we like to see how much the current we will be getting here particularly in unloaded condition.

So, what we have to cover here this numerical example and in this numerical example what is the important point here it is in case if we are not using the C what kind of affect it may come and this is important for common base amplifier.