feedback amplifier is also called as | feedback amplifier solved problems pdf

feedback amplifier solved problems pdf , feedback amplifier is also called as.
Simplified Common-Emitter Hybrid Model
In most practical cases it is appropriate to obtain approximate values. A_V, A_I etc; rather than calculating exact values. how the circuit can be modified without greatly reducing the accuracy? fig. 37 shows the CE amplifier equivalent circuit in terms of h-parameters. since, 1/h_oe in parallel with R_L is approximately equal to R_L, if 1/h_oe >> R_L then, h_oe may be neglected. under these conditions,
I_C = h_feI_b
h_reV_c = h_reI_cR_L = h_reh_feI_bR_L
Since, h_fe . h_re >> 0.01, this voltage may be neglected in comparison with h_icI_b drop across h_ie provided R_L is not very large. if load resistance R_L is small than h_oe and h_re can be neglected.
A_I = – h_fe / 1 + h_oeR_L ~ – h_fe
R_i = h_ie
A_V = A_IR_L / R_I = – h_feR_l / h_ie
Output impedance seems to be infinite. when V_S = 0 and an external voltage is applied at the output we find I_B = 0, I_C = 0, True value depends upon R_S and lies between 40 k and 80 k.
On the same lines, the calculations for CC and CB can be done.
CE Amplifier with an Emitter Resistor
The voltage gain of a CE stage depends upon h_fe. this transistor parameter depends upon temperature, aging and the operating point. moreover, h_fe may very widely from device to device, even for same type of transistor. to stabilize voltage gain A_V of each stage, it should be independent of h_fe. A simple and effective way is to connect an emitter resistor R_e as shown in fig. 38. the resistor provides negative feedback and provide stabilization.
An approximate analysis of the circuit can be made using the simplified model.
Current gain     A_i = I_L / I_b = – I_C / I_B = – h_feI_B / I_B = – h_fe
It is unaffected by the addition of R_C .
Input resistance is given by
R_i = V_i / I_B
= h_ieI_B + (1 + h_fe) I_BR_E / I_B
= h_ie = (1 + h_ie) R_E
The input resistance increases by (1+ h_fe) R_e
A_V = A_IR_L / R_I = -h_feR_L / h_ie + (1 + h_fe)R_e
Clearly, the addition of R_E reduces the voltage gain. if (1 + h_fe) R_e  >> h_ie and h_fe >> 1
A_V = -h_feR_L / (1 + h_fe) R_e ~ – R_L / R_E
Subject to above approximation A_V is completely stable. the output resistance is infinite for the approximate model.
Feedback Amplifier
Feedback is defined as the process whereby a portion of the output signal is fed to the input signal in order to form a part of the system output control. this action tends to make the system self regulating. feedback is used to make the operating point of a transistor insensitive to both manufacturing variations in B as well as temperature.
Negative Feeback
Negative feedback or degenerative feedback in which the signal feedback from output to input is 180⁰ out of phase with the applied excitation.
Negative Feedback has many advantages

1. In the control of impedance levels.
2. Bandwidth improvement.
3. In rendering the circuit performance relatively insensitive to manufacturing as well as to environmental changes.
4. It increases bandwidth and input impedance and lowers the output impedance.

Positive Feedback
Positive feedback or regenerative feedback in which the overall gain of the amplifier is increased. positive feedback is useful in oscillators and while establishing the two stable states of flip-flop.
Conceptual Development through block Diagrams
Sampling Network
The function of the sampling network is to provide a measure of the output signal i.e., a signal that is proportional to the output. two sampling networks are showe in fig. 39. in 39(a), the output voltage is sampled by connecting the output port to the feedback network in parallel with the load. this configuration is called shunt connection. in fig. 39 (b), the output current is sampled and the output port of the feedback network is connected in series with the load. this is a series connection.
Comparison or Summing Network
The two very common networks used for the summing of input and feedback signals are displayed.
The circuit shown in fig 40(a) is a series connection and it is used to compare the signal voltage V_S and feedback signal V_F. the amplifier input signal V_i is proportional to the voltage difference (V_S – V_f ) that results from the comparison. A differential amplifier is uesd for comparison as its output voltage is proportional to the difference between the signals at the two inputs.
A shunt connection is shown in fig. 40(b) in which the source current I_S and feedback current I_f are compared. the amplifier input current I_i is proportional to the difference I_S – I_F.
Basic Ampliffier
The basic amplifier is one of the important parts of the feedback amplifier. the circuit amplifies, the difference signal that results from comparison and this process is responsibe for DC. Sensitivity and control of the output in a feedback system.
Calculations of Open-loop Gain, Closed-loop Gain and Feedback Factors
The general block diagram of an ideal feedback amplifier indicating basic amplifier, feeddback network, external load and corresponding signals is shown in fig. 41.
The ideal feedback amplifier can have any of the four configuration as listed in table.
The input sigual X_S’ the output signal X_o the feedback signal X_f and the difference signal X_i each represent either a voltage or a current. these signals and the transfer ratios A and B are summarized in table for different feedback topologives.
For a positive feedback, we get
X_i = X_s + X_f  …………………………….(1)
The signal X_i , representing the output of the summing network is the amplifier input X_I . If the feedback signal X_f is 180⁰ out of phase with the input X_S – as is true in negative feedback system, then X_I is a difference. signal. therefore, X_I decreases as| X_f| increases. the reverse transmission of the feedback network B is defined by
B = X_F / X_{o            ……………………………………} (2)
The transfer fusction B is a real number, but in general. it is a function of frequency. the gain of the basic amplifier is defined as
A = X_o / X_i     ………………………(3)
Now, from eq. (1), we get
X_I = X_S + X_F
Substituting the value of X_F from eq. (2) as X_F = BX_O in eq. (1) we get
X_I = X_S + X_F = X_S + BX_o ………………………(3a)
From eq. (3), we get
X_o = AX_i …………………………(3b)
Substituting the value of X_I From eq. (3a) we get
X_O = AX_I = A(X_S + BX_O) = AX_S + ABX_O
Or                   X_O (1 – AB) = AX_S
Or                     X_O / X_S = A / 1 – AB …………………..(3c)
The feedback gain A_F is obtained from eq. (3a) as
A_F = X_O / XS  = A / 1 – AB   ……………………(4)
The gian in eq. (3) represents the transfer function without feedback. if B = 0; eliminating the feedback signal, no feedback exists and eq. (4) reduces to eq. (3). frequency A is referred to as the open-loop gain and is designated by A_OL. when B=0, a feedback-loop exists and A_F is often called the closed-loop gain.
If |A_f|<|A|, the feedback is negative. if |A_F|>A, the feedback is positive. in case of a negative feedback , |1-AB|>1. the reason behind this being the barkhausen criterion.
Similarly for negative feedback,
X_I = X_S – X_F
Calculating in the same manner as done for positive feedback, we can represent the feedback gain as
A_F = A / 1 + AB
For negative feedback, |1 + AB| > 1 so, A_F < A i.e., A_F decreases. therefore, the general equation of feedback can be written as
A_F = A / 1 + AB
Loop Gain or Return Ratio
The signal X_I in fig 4. is multiplied by gain A when passing through the amplifier and by B in transmission through the feedback network. such a path takes us from the amplifier input around the loop consisting of the amplifier and the feedback network. the product AB, is called the loop gain or return ratio T. eq. (4) can be written in terms of A_OL and T as
A_F = A / 1 – AB  = A_OL / 1 + T
For negative feedback – AB = T > 0
We can give a physical interpretation for the return ratio by considering the input signal X_S = 0, and keeping the path between X_I and X_I open . if a signal X_I is now applied to amplifier input the
 
T = – AB – X_I / X_I / x_s = 0
The retum ratio is the regative of the ratio of the feedback signal to the amplifier input. often the quantity F = 1 – AB = 1 + T is referred to as the return difference. if negative feedback is considered then, both F and t are greater then zero.
Topologies of the Feedback Amplifier
There are four basic amplifier types. each of these is being approximated by the characteristics of an ideal controlled source. the four feedback topologies are as follows;

1. Voltage-series or series-shunt feedback
2. Current-series or series-series feedback
3. Current-shunt or shunt-series feedback
4. Voltage-shunt or shunt-shunt feedback

Voltage-series or Series-shunt Feedback
The configuration of voltage-series or series-shunt feedback circuit is illustrates in fig. 42.
The input voltage V_I of the basic amplifier is the algebraic sum of input signal V_I and the feedback siganal BV_O , where V_o is the output voltage.
Current-Series or Series-Series Feedback
The current-series or series-series feedback topology is illustrated in fig. 43.
As mentioned in table, the trans conductance feedback amplifier provides an output current I_O which is proportional to the input voltage V_S, the feedback signal is the voltage V_F, which is added to V_S at the input of input of the basic amplifier.
Current-Shunt or Shunt-Series Feedback
The current-shunt or shunt-series feedback amplifier, as shown in fig. 44 supplies an output current I_o which is proportional to the input current I_I this makes it a  current is I_o = I_L
Voltage-Shunt or Shunt-Shunt Feedback
The voltage-shunt or shunt-shunt feedback amplifier is illustrated in fig. 45.
This provides an output voltage V_o in proportion to the input current I_s the input current I_i of the basic amplifier is the algebraic sum of I_s and the feedback current I_f
Effect of Feedback On Gain, Input and Output Impedances
Feedback is applied with the objective of improving the performance of an amplifier. the operation of an amplifier is regulated by controlling the gain and impedance.
Effect of Feedback On Input Impedance
Voltage-series feedback
Fig. 46 shows the equivalent thevenin’s model of the voltage-series amplifier of fig 45 in the circuit A_v represents open-circuit voltage gain taking Z_s into account. from fig. 46 the input impedance with the feedback is
Z_if = V_s / I_i ……………………….(6)
and                             V_s = I_iZ_i +V_f = I_iZ_I + BV_o ………………………..(7)
Using voltage-divider rule, we get
V_O = A_VV_IZ_L / Z_O + Z_L = A_VI_IZ_L = A_VI_IZ_L …………………….(8)
where,                  I_I = V_I / Z_O + Z_L
Now,                   V_O = A_VI_IZ_L = A_VV_I
Or                            A_V = V_O /I_I
From fig. 46, the input impedance without feedback is
Z_I = V_I / I_I …………………(9)
Now                   Z_IF = V_S / I_I = V_I (1 + A_VB) / I_I
From eqs. (6) and (7), we have
Z_IF = V_S / I_I = Z_I (1 + BA_V)
Thus, the input-impedance is increased.
Although A_V represents the open-circuit voltage gain without feedback, eq.(6) indicates that A_V is the voltage gain without feedback taking the load Z_L into account.
Current-series feedback
In a similar manner as for voltage series, for current series feedback as shown in fig. 43. we obtain
Z_IF = Z_I (1 + BY_M) ……………………..(11a)
where, Y_M is the short- circuit transadmitance without feedback considering the load impedance and is given by
Y_M = I_O / V_I = Y_M Z_O / Z_O + Z_L ……………………(11b)
From eq. (11a) it is clear that for series mixing Z_IF > Z_I.
Current-shunt feedback
Fig. 47 shows the current-shunt feedback in which the amplifier is replaced by its norton equivalent circuit. if A_I is the short-circuit current gain then from fig. 47.
I_S = I_I + I_F = I_I + BI_O ………………………..(12)
and               I_O = A_I f_I / Z_O + Z_L = A_II_I ……………………..(13)
Where,                  A_I = I_O / I_I = A_IZ_O / Z_O + Z_L …………………………(14)
From eqs. (12) and (13) , we have
I_S = I_I (1 + BA_I) ……………………….(15)
From figure
Z_IF = V_I / I_S and Z_I = V_I / I_I
Using eq. (15) we obtain
Z_IF = V_I / I_I (1 + BA_I) = Z_I / 1 + BA_I ……………….(16)
Voltage-Shunt Feedback
For voltage-shunt feedback, proceeding in a similar way as we have dove in the previous section S, we obtain
Z_IF = Z_I / 1 + BZ_M …………………(17a)
where, Z_M is the transimpedance without feedback considering the load and is give by
Z_M = V_O / I_I = Z_MZ_L / Z_O + Z_L ………………………(17b)
where, Z_M is the open-circuit transimpedance without feedback.
From eq. (17b) it is clear that for shunt comparison Z_IF < Z_I
Effect of Feedback on Output Impedance
Voltage-series feedback
To find the output resistance with feedback Z_OF looking into output terminals with Z_L disconnected, external signals must be removed (V_S = 0 or I_S = 0 ) let Z_L = ₀₀, impress a voltage V across the output terminals which delivers current I.
Therefore,                      Z_OF = V / I ………………..(18)
Replacing V_O by V in fig. 47, we get
I = V – A_VV_I / Z_O = V + BA_V V/ Z_O ………………………(19)
with V_S = 0,               V_I = V_F = – BV
Hence,                          Z_OF = V / I = Z_O / 1 + BA_V …………………..(20)
The output resistance with feedback Z’_OF, Which includes Z_L, is given by Z_OF in parallel with Z_L. SO,
Z’_OF = Z_OFZ_L / Z_OF + Z_L
= Z_OZ_L / Z_O + Z_L +BA_VZ_L
= Z_OZ_L /(Z_O + Z_L) / 1 + BA_VZ_L/(Z_O + Z_L) ………………..(21)
It should be noted that Z’_O = Z_O||Z_L is the output resistance without feedback.
Using eq. (9) in eq. (21). we obtain
Z’_OF = Z’_O / 1 + BA_V ………………….(22)