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Common Emitter Amplifier

In the last post, we studied the transistor model for small-signal analysis, and there I had briefly explained that why we are using h-parameter for the analysis of the BJT transistor. In this post, you will be studying voltage divider bias in little depth and at the same time, we will be studying some basic concepts to design an amplifier so that we will understand the working of amplifier in-depth, and in the next post, we can design a proper and well prepared basic amplifier.

In my post on different biasing techniques of transistors, I didn't explain the voltage divider bias in detail because I thought of explaining it while starting amplifiers so that you can grasp the concept very well. So let's start with our post.

Diagram Of Voltage-Divider Bias:-

Now consider,

current I1 flows through R1

current I2 flows through R2

current Ic flows through Rc

current IE flows through RE

By considering this we conclude the following thing


So as you can see in this derived expression that if IE increases due to heating then I2 decreases which in short means that base voltage (Vb) decreases. Due to this reason, the voltage divider bias is also known as a self-bias circuit.So now if we are designing amplifier with the help of this transistor so IE(total) = Ie(dc) + ie(ac). So you can see that due to the introduction of ac signal IE(total) increases due which the base voltage Vb decreases and if base voltage decreases then there might be a situation that the input ac signal might be clipped due to which we can't get proper amplification of it at the output. 

For Example 

1)when Ac signal is introduced:-Let's say by proper biasing we had set our Vb to 2 volts (Vb=2v) and IE(total) = Ie(dc) so at starting the input ac signal would be amplified properly.

2)When ac signal reaches towards Emitter terminal:- As soon as we introduced ac signal in our amplifier assume that the value of our Ac signal is 0.6 volts peak to peak

So now IE(total) = Ie(dc) + ie(ac) i.e. IE(total) increases after some-time due to which Vb decreases let's say the value of Vb decreases from 2v to 1v (ASSUMED)

Now,for positive half cycle of Vin or ac-signal the value of Vb would be Vb(+) = Vb + (0.3) = 1 + 0.3 = 1.3v 

Now,for negative half cycle of Vin or ac-signal the value of Vb would be Vb(-) = Vb + (-0.3) = 1 - 0.3 = 0.7v


and we know that for the transistor to operate in active or saturation region Vbe should be greater than 0.7v else our transistor will operate in the cutoff region so the transistor won't be able to pass some negative part of the ac signal that means we are not getting a proper amplified signal at the output.

[NOTE:-Vb won't change drastically from 2v to 1v but I had just mentioned it so that you can understand it very well. This kind of situation mainly happens only if the transistors become too old.]

If you want to understand this concept in depth make sure to watch the video given beow.

So to overcome this problem we are connecting a capacitor in parallel with Re so that the ac current or ie(ac) will pass through the capacitor and Ie(dc) will pass through the resistor RE due to which Vb will not decrease and we will get the proper amplified signal of the applied input signal at the output. Also, the dc signal which we are providing to the transistor ideally should be a straight line signal but practically it also offers some noise so to clear that noise as well we use a capacitor since noise has a high frequency. 

I hope now all things are clear so that we can move towards some formulas or some concepts required to design an amplifier in my next post I will show you how to design an amplifier

As I discussed in my earlier posts that to design an amplifier 4 things are important i.e. it's Input impedance (Zi), Output impedance(Zo), Voltage gain(Av) & Current gain(Ai). I had already discussed how to calculate Ic, Ib, Vce, or in-short dc analysis of transistors in my earlier post on biasing techniques of the transistor. So please visit that post if you are not familiar with it.

Remember that during the small-signal analysis of the transistor all the dc sources act as an open circuit [Superposition Theorem] & all the capacitors are replaced by short-circuits

Therefore RE is also replaced by short-circuit since the voltage drop across RE is zero. So transistor in small-signal analysis can be redrawn as shown below

The value of hie=2.7k & hfe/β=110 (for bc547) at room temperature

So as you can see above that the input impedance (Zi) = Rth || hie

The output impedance (Zo) = Rc

Voltage Gain of Amplifier :- Av = (Vo/Vi) = -hfe.Ib.Rc / hie.Ib = (-hfe.Rc) / hie.

Current Gain of Amplifier :- Ai = (Io/Iin) = (hfe.Rth) / (Rth+hie)

Note that we will get maximum amplification only if the q-point is in the middle of the load line.
 Vce(q) = Vcc/2     &    Ic(q) = Ic(max)/2
That means there should be 50% of the voltage drop across Vce and the rest  50% of the voltage should be dropped across Rc and Re (i.e. collector resistor and emitter resistor respectively).
Therefore VRc = 40%Vcc     &     VRe = 10%Vcc
The concept would be more clear by watching the image and the video given below.

So I conclude the whole post by just posting a single video on my you-tube channel I hope it will surely help you 

 I hope that you understood all the concepts that we discussed in this post in the next post we will discuss how to design an amplifier.


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