### Common Emitter Amplifier Designing

In my last post, I explained to you why we are using voltage divider bias and some formulas to design an amplifier so in this post I will show you how to design a basic amplifier. Being a biomedical engineer I would like to discuss its application from a biomedical point of view. Recently we heard the news of neuralink chip and I think this is one of the very successful inventions or ideas from a biomedical point of view. So let's say if you want to see and analyze your heartbeat signal and by analyzing it you are using it in some other application. Now the voltage of your heartbeat signal is in a few milli-volts and if you want to analyze that signal you can't analyze it properly so there you can make use of an amplifier.

I'm taking voltage divider bias to design an amplifier because it provides good stability which I had discussed in my earlier posts. We are using 9V of battery and Bjt-transistor (bc547) having current gain(hfe) or β = 110 and the input impedance of transistor or hie=2.7kΩ

[We have to design an amplifier which will amplify signal having a frequency above 20hz and the input signal is of 10mVp-p signal to 1Vp-p]

Given:-

Vcc = 9v and f = 20hz

β/hfe = 110 and hie = 2.7kΩ .....................(From the datasheet of bc547)

we also know that Vce = Vcc/2 so for this condition to satisfy VRc = 40%Vcc and VRe = 10%Vcc

where VRc is the voltage drop across resistor Rc & VRe is the voltage drop across resistor Re

Process Of Designing:-

Since we are designing an amplifier where we have to raise 10mv of signal to 1V

So Av = Vo/Vin = 1/(10*10^-3) = 100

Now ,

In my last post I had mentioned a formula that

Av = -hfe*Rc/hie

From here we get

Rc = Av*hie/hfe=100*2.7k/110...(since the value of resistor can't be negative therefore we neglect it)

Rc = 2.45kΩ

Rc ~ 2.5kΩ

We know that

VRc = 40%Vcc=0.4*9=3.6v

Ic = VRc/Rc=3.6/2.5k=1.44mA

Ic = β*Ib

Ib = 13uA

Similarly Vre = 10%Vcc

VRe = 0.1*9 = 0.9v

And Ic~Ie = 1.44mA

Re = VRe/Ie = 0.9/1.44mA

Re = 625

Re ~ 620Ω

Now,

(Remeber that I2 is always 1/10th time of Ic)

I2 = 1/10*Ic = (1/10)*1.44mA = 0.14mA

I1 = Ib+I2 = 0.15mA

V2 = Vbe+VRe = 0.7+0.9 = 1.6v
V1 = Vcc-V2 = 9v-1.6v = 7.4v

R2 = (V2)/I2 = (1.6)/0.14mA = 11.4k ~ 11.5kΩ
R1 = (V1)/I1 = (7.4)/0.15mA = 49.33k ~ 51kΩ

Now we will be dealing with capacitors the values of capacitors should be selected such that it will attenuate the frequencies which are below 20hz while it provides almost constant gain for frequencies above 20hz.
so, here are formulas to calculate Cin, Cout & Ce
Cin = 1/2*pi*Xin*f
now here we know f=20hz but Rin is unknown to us
Xin = R1||R2||hie = 51k||11k||2.7k = 2kΩ
Cin = 4uf
Remember that we had discussed noise in the last post so to clear it we select impedance of Ce that is Xe = RE/10
Xe = 620/10 = 62Ω
Ce = 1/2*pi*Xe*f = 128uf ~ 130uf
For Cout the impedance of Xc = RC+RL  since in our case we are not using and load resistor therefore RL=0
Xc = Rc = 2.4kΩ
Cout = 1/2*pi*Xc*f = 3.3uf

NOTE:- There are some standard values of resistor and capacitor and we have to use it according to the design we got R1 as 49.33k so we took 51k since there is no such resistor of 49.33k while resistor 51k is the standard value resistor.

Let's Implement It Practically:-

At  10Hz if we are giving 10mv of signal to our ce amplifier we are getting a gain of 15 as you can see below.
We know that
Av = Vo/Vin
Av = (80+75)mV/10mV
Av = 15.5

At 50Hz we are getting a gain of 68.1 as you can see below
We know that
Av = Vo/Vin
Av = (340+335)mV/10mV
Av = 68.1

At 100 Hz we are getting a gain of  94.5 as you can see below
We know that
Av = Vo/Vin
Av = (455+490)mV/10mV
Av = 94.5

At 120Hz we are getting a gain of  100 as you can see below

We know that
Av = Vo/Vin
Av = (485+515)mV/10mV
Av = 100

VIDEO:-
I embedded 2 videos in this section the 1st video will give you a short idea of the steps that we took whereas the video beneath it will provide you the idea of the CE-amplifier that we had designed in this post.

Here is the video which will give you a more clear idea of how an amplifier works at different frequencies.

CONCLUSION :-
So we had designed our Ce Amplifier properly since it is giving the desired gain what we wanted and it was exactly 180 degrees out of phase to the input signal.

If you are loving my content then please do comment below this will motivate me thankyou.