Analog Electronic Circuits | Analog Circuits Basics

Analog circuits: An analog circuit is a circuit with continuous, variable signal (i.e. an analog signal) as resist or opposed to a digital circuit where a signal must be one of two discrete levels. Analog circuits within electrical equipment can convey/ carries information through changes in the current, voltage, (or) frequency.

ANALYSIS OF AMPLIFIERS:

In analog circuits we have two types. Those are

1.      DC analysis

2.    AC analysis

Ac analysis is again sub-divided into two types. Those are

1.      Low frequency analysis

2.    High frequency analysis

While performing these analog circuit analysis problems we have to remember these following important points.

DC analysis:

1.   find out V_CE, I_C. because these are operating points of a transistor in a DC analysis.

2.  In DC analysis we make all  ac input signals to zero.

3.   All external capacitors are open circuited in DC analysis.

4.  Amplifies operates in active region

5.   Apply KVL, KCL to obtain the operating points of amplifier.

AC analysis:

In AC analysis is required to obtain gain, input resistance, output resistance and frequency of operation.

1.   in Ac analysis we make all DC sources to zero.

2.  All external capacitors are short-circuited.

3.   in low frequency analysis transistor is replaced by its low frequency small signal model.

4.  in high frequency analysis transistor is replaced by its high frequency small signal model.

5.   After these following steps now we will apply to the KCL, KVL to obtain gain, input output resistance and frequency response of amplifiers.

Before learning AC analysis of transistor now we will see basics of hybrid parameters and its diagrams.

Hybrid parameter two-port network:

With the help of hybrid parameters we will find input, output voltages & currents using these two equations.

          

                                         

                                   

V₁ = h₁₁I₁ + h₁₂V₂

  I₂= h₂₁I1 + h₂₂V₂

h₁₁= V₁/I₁   (at V₂ =0)     this is input resistance with o/p short-circuit

h₁₂= V₁/V₂ (at I₁=0)    reverse voltage gain with i/p open-circuit

h₂₁ = I₂/I₁  (at V₂=0)   forward current gain with o/p short-circuit

h_{22 ­}=  I₁/V₂ (at I₁=0)    output conductance with i/p open-circuit

Transistor Low Frequency Hybrid model:

Transistor can be considered as two port network, one is input port and one is output port. Input is applied to the input port and output is taken from output port of hybrid networks.

     The transistor configurations and their hybrid model shown in below.

 Euqtion for the CE hybrid model is

V_be = h_iei_b + h_reV_ce

I_c = h_fei_b + h_oeV_ce

                                                                                 

Euqtion for the CC hybrid model is

V_bc = h_ici_b + h_rcV_ec

I_e = h_fci_b + h_ocV_ec

                                  

                                       

Euqtion for the CB hybrid model is

V_eb = h_ibi_e + h_rbV_cb

I_c = h_fbi_e + h_obV_cb

Analysis of two port network

Here we dicuss the hybrid model of two port network which is connected to source resistance R_s and source voltage V_s at input port and output port is connected to load resistor R_L as shown in below fig.
                                            

                                  

Current gain

A_I = I₀/I₁

From the output loop we can see that I0 = -I2

Applying KVL in the output loop

(-1/h22) (i2-h21I1) + I0R_L=0

Replacing –I2=I0

(I0/h22) + I1(h21/h22)+I0R_L=0

A_I=I0/I1 = -h21/(1+h22R_L)

Input resistance:

R_i = Vi/Ii
From input loop we will get

V1 - h11I1 -h12V2=0 -----------------(1)

V2=I0R_L

Substituting I0=A_II1

V2=A_II1R_L

Therefore the equation (1) of input loop is

V1 - h11I1 -h12 A_II1R_L =0

R_i = h11 + h12A_IR_L

Voltage gain:

Av=v2/v1

= A_II1R_L/ R_iI1

Output resistance
                                              

  Applying kvl at this output loop is

Vg=(1/h22)(Ig-h21I1)

From the input loop

-(Rs + h11)I1 – h12V2=0

   
In input loop we substitute V2=Vg

-(Rs + h11)I1 – h12Vg=0

(Rs + h11)I1 = - h12Vg

I1 = -( h12Vg/Rs+h11)

Then I1 substitute in the output loop quation

Vg=(1/h22)(Ig-h21(-( h12Vg/Rs+h11)))

Vg=(1/h22)(Ig + h21h12Vg / Rs+h11)

h22Vg= Ig + h21h12Vg / Rs+h11

Ig= h22Vg – ( h21h12Vg / Rs+h11)

Y0 = Ig/Vg = h22-(h12h21/Rs+h11)

R0= 1/Y0