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. AV, AI 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/hoe in parallel with RL is approximately equal to RL, if 1/hoe >> RL then, hoe may be neglected. under these conditions,
IC = hfeIb
hreVc = hreIcRL = hrehfeIbRL
Since, hfe . hre >> 0.01, this voltage may be neglected in comparison with hicIb drop across hie provided RL is not very large. if load resistance RL is small than hoe and hre can be neglected.
AI = – hfe / 1 + hoeRL ~ – hfe
Ri = hie
AV = AIRL / RI = – hfeRl / hie
Output impedance seems to be infinite. when VS = 0 and an external voltage is applied at the output we find IB = 0, IC = 0, True value depends upon RS 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 hfe. this transistor parameter depends upon temperature, aging and the operating point. moreover, hfe may very widely from device to device, even for same type of transistor. to stabilize voltage gain AV of each stage, it should be independent of hfe. A simple and effective way is to connect an emitter resistor Re 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 Ai = IL / Ib = – IC / IB = – hfeIB / IB = – hfe
It is unaffected by the addition of RC .
Input resistance is given by
Ri = Vi / IB
= hieIB + (1 + hfe) IBRE / IB
= hie = (1 + hie) RE
The input resistance increases by (1+ hfe) Re
AV = AIRL / RI = -hfeRL / hie + (1 + hfe)Re
Clearly, the addition of RE reduces the voltage gain. if (1 + hfe) Re >> hie and hfe >> 1
AV = -hfeRL / (1 + hfe) Re ~ – RL / RE
Subject to above approximation AV is completely stable. the output resistance is infinite for the approximate model.
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 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
- 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 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
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 VS and feedback signal VF. the amplifier input signal Vi is proportional to the voltage difference (VS – Vf ) 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 IS and feedback current If are compared. the amplifier input current Ii is proportional to the difference IS – IF.
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 XS’ the output signal Xo the feedback signal Xf and the difference signal Xi 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
Xi = Xs + Xf …………………………….(1)
The signal Xi , representing the output of the summing network is the amplifier input XI . If the feedback signal Xf is 1800 out of phase with the input XS – as is true in negative feedback system, then XI is a difference. signal. therefore, XI decreases as| Xf| increases. the reverse transmission of the feedback network B is defined by
B = XF / Xo …………………………………… (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 = Xo / Xi ………………………(3)
Now, from eq. (1), we get
XI = XS + XF
Substituting the value of XF from eq. (2) as XF = BXO in eq. (1) we get
XI = XS + XF = XS + BXo ………………………(3a)
From eq. (3), we get
Xo = AXi …………………………(3b)
Substituting the value of XI From eq. (3a) we get
XO = AXI = A(XS + BXO) = AXS + ABXO
Or XO (1 – AB) = AXS
Or XO / XS = A / 1 – AB …………………..(3c)
The feedback gain AF is obtained from eq. (3a) as
AF = XO / 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 AOL. when B=0, a feedback-loop exists and AF is often called the closed-loop gain.
If |Af|<|A|, the feedback is negative. if |AF|>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,
XI = XS – XF
Calculating in the same manner as done for positive feedback, we can represent the feedback gain as
AF = A / 1 + AB
For negative feedback, |1 + AB| > 1 so, AF < A i.e., AF decreases. therefore, the general equation of feedback can be written as
AF = A / 1 + AB
Loop Gain or Return Ratio
The signal XI 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 AOL and T as
AF = A / 1 – AB = AOL / 1 + T
For negative feedback – AB = T > 0
We can give a physical interpretation for the return ratio by considering the input signal XS = 0, and keeping the path between XI and XI open . if a signal XI is now applied to amplifier input the
T = – AB – XI / XI / xs = 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 VI of the basic amplifier is the algebraic sum of input signal VI and the feedback siganal BVO , where Vo 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 IO which is proportional to the input voltage VS, the feedback signal is the voltage VF, which is added to VS 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 Io which is proportional to the input current II this makes it a current is Io = IL
Voltage-Shunt or Shunt-Shunt Feedback
The voltage-shunt or shunt-shunt feedback amplifier is illustrated in fig. 45.
This provides an output voltage Vo in proportion to the input current Is the input current Ii of the basic amplifier is the algebraic sum of Is and the feedback current If
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
Fig. 46 shows the equivalent thevenin’s model of the voltage-series amplifier of fig 45 in the circuit Av represents open-circuit voltage gain taking Zs into account. from fig. 46 the input impedance with the feedback is
Zif = Vs / Ii ……………………….(6)
and Vs = IiZi +Vf = IiZI + BVo ………………………..(7)
Using voltage-divider rule, we get
VO = AVVIZL / ZO + ZL = AVIIZL = AVIIZL …………………….(8)
where, II = VI / ZO + ZL
Now, VO = AVIIZL = AVVI
Or AV = VO /II
From fig. 46, the input impedance without feedback is
ZI = VI / II …………………(9)
Now ZIF = VS / II = VI (1 + AVB) / II
From eqs. (6) and (7), we have
ZIF = VS / II = ZI (1 + BAV)
Thus, the input-impedance is increased.
Although AV represents the open-circuit voltage gain without feedback, eq.(6) indicates that AV is the voltage gain without feedback taking the load ZL into account.
In a similar manner as for voltage series, for current series feedback as shown in fig. 43. we obtain
ZIF = ZI (1 + BYM) ……………………..(11a)
where, YM is the short- circuit transadmitance without feedback considering the load impedance and is given by
YM = IO / VI = YM ZO / ZO + ZL ……………………(11b)
From eq. (11a) it is clear that for series mixing ZIF > ZI.
Fig. 47 shows the current-shunt feedback in which the amplifier is replaced by its norton equivalent circuit. if AI is the short-circuit current gain then from fig. 47.
IS = II + IF = II + BIO ………………………..(12)
and IO = AI fI / ZO + ZL = AIII ……………………..(13)
Where, AI = IO / II = AIZO / ZO + ZL …………………………(14)
From eqs. (12) and (13) , we have
IS = II (1 + BAI) ……………………….(15)
ZIF = VI / IS and ZI = VI / II
Using eq. (15) we obtain
ZIF = VI / II (1 + BAI) = ZI / 1 + BAI ……………….(16)
For voltage-shunt feedback, proceeding in a similar way as we have dove in the previous section S, we obtain
ZIF = ZI / 1 + BZM …………………(17a)
where, ZM is the transimpedance without feedback considering the load and is give by
ZM = VO / II = ZMZL / ZO + ZL ………………………(17b)
where, ZM is the open-circuit transimpedance without feedback.
From eq. (17b) it is clear that for shunt comparison ZIF < ZI
Effect of Feedback on Output Impedance
To find the output resistance with feedback ZOF looking into output terminals with ZL disconnected, external signals must be removed (VS = 0 or IS = 0 ) let ZL = 00, impress a voltage V across the output terminals which delivers current I.
Therefore, ZOF = V / I ………………..(18)
Replacing VO by V in fig. 47, we get
I = V – AVVI / ZO = V + BAV V/ ZO ………………………(19)
with VS = 0, VI = VF = – BV
Hence, ZOF = V / I = ZO / 1 + BAV …………………..(20)
The output resistance with feedback Z’OF, Which includes ZL, is given by ZOF in parallel with ZL. SO,
Z’OF = ZOFZL / ZOF + ZL
= ZOZL / ZO + ZL +BAVZL
= ZOZL /(ZO + ZL) / 1 + BAVZL/(ZO + ZL) ………………..(21)
It should be noted that Z’O = ZO||ZL is the output resistance without feedback.
Using eq. (9) in eq. (21). we obtain
Z’OF = Z’O / 1 + BAV ………………….(22)