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_{fe}I_{b}

h_{re}V_{c} = h_{re}I_{c}R_{L} = h_{re}h_{fe}I_{b}R_{L}

Since, h_{fe} . h_{re} >> 0.01, this voltage may be neglected in comparison with h_{ic}I_{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_{oe}R_{L} ~ – h_{fe}

R_{i} = h_{ie}

A_{V} = A_{I}R_{L} / R_{I} = – h_{fe}R_{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_{fe}I_{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_{ie}I_{B} + (1 + h_{fe}) I_{B}R_{E} / I_{B}

= h_{ie} = (1 + h_{ie}) R_{E}

The input resistance increases by (1+ h_{fe}) R_{e}

A_{V} = A_{I}R_{L} / R_{I} = -h_{fe}R_{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_{fe}R_{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 180^{0} out of phase with the applied excitation.

**Negative Feedback has many advantages**

- In the control of impedance levels.
- Bandwidth improvement.
- In rendering the circuit performance relatively insensitive to manufacturing as well as to environmental changes.
- 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 180^{0} 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;

- Voltage-series or series-shunt feedback
- Current-series or series-series feedback
- Current-shunt or shunt-series feedback
- 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_{i}Z_{i} +V_{f} = I_{i}Z_{I} + BV_{o} ………………………..(7)

Using voltage-divider rule, we get

V_{O} = A_{V}V_{I}Z_{L} / Z_{O} + Z_{L} = A_{V}I_{I}Z_{L} = A_{V}I_{I}Z_{L} …………………….(8)

where, I_{I} = V_{I} / Z_{O} + Z_{L}

Now, V_{O} = A_{V}I_{I}Z_{L} = A_{V}V_{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_{V}B) / 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_{I}I_{I} ……………………..(13)

Where, A_{I} = I_{O} / I_{I} = A_{I}Z_{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_{M}Z_{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_{V}V_{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_{OF}Z_{L} / Z_{OF} + Z_{L}

= Z_{O}Z_{L} / Z_{O} + Z_{L} +BA_{V}Z_{L}

= Z_{O}Z_{L} /(Z_{O} + Z_{L}) / 1 + BA_{V}Z_{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)