the closed-loop voltage gain of an inverting amplifier equals , closed loop amplifier | formula , gain , bandwidth explanation full derivation step by step.
Closed-Loop Amplifier
The gain of the op-amp can be controlled if feedback is introduced in the circuit. that is an output signal is feedback to the input either directly or via another network. if the signal feedback is of opposite or out phase by 1800 w.r.t the input signal, the feedback is callednegative feedback.
An amplifier with negative feedback has a self-correcting ability of change in output voltage caused by changes in environmental conditions. it is also known as degenerative feedback because it reduces the output voltage and in ter, reduces the voltage gain.
If the signal is feedback in phase with the input signa,the feedback is called positive feedback. in positive feedback, the feedback signal aids the input signal. it is also known as regenerative feedback positive feedback is necessary in oscillator circuits.
The negative feedback stabillzes the gain, increases the bandwidth and changes, the input and output resistances. other benefits are reduced distortion and reduced offiset output voltage. it also reduces the effect of temperature and supply voltage variation on the output of an op-amp.
A closed-loop amplifier
A closed-loop amplifier can be represented by two blocks one for an op-amp and other for a feedback circuits. there are four following ways to connect these blocks. these connections are shown in fig. 34.
These connections are classified according to whether the voltage or current is feedback to the input in series or in parallel:
. Voltage series feedback
. Voltage shunt feedback
. Current series feedback
. Current shunt feedback
In all these circunit of fig. 34, the signal direction is from input to output for op-amp and output to input for feedback circuit . only first two, feedback in circuits are important.
Voltage Series Feedback
It is also called non-inverting voltage feedback circuit. wiyh this type of feedback , the input signal drives the non-inverting input of an amplifier, a fraction of the output voltage is then feedback to the inverting input. The op-amp is represented by its symbol including its large signal voltage gain Ad or A and the feedback circuit is composed of two resistors R1 or Rf as shown in Fig. 35
Open -loop voltage gain Ad = Vo/ Vd
Open- loop voltage gain Acl =Vo /Vin
feedback circuit gain B=Vf /Vo
The different voltage input Vd = Vin – Vf
The feedback voltage always oppose the input voltage . [or is out phase by 1800 w.r.t input voltage], hence the feedback is said to be negative.
The closed-loop voltage gain is given by,
ACL = V0 / VIN
V0 = A (V1 – V2) = A (VIN – VF)
= A (VIN – BV0)
V0 = (1 + AB) = AVIN
V0 / VIN = A / 1 + 48
The product of A and B is called loop gain. the gain loop gain is very large such that AB>> 1
V0 / VIN = A / AB = 1 / B
= R1 + RF / R1 = 1 + RF / R1
This shows that overall voltage gain of the circuit equals the reciprocal of B, the feedback gain. it means that closed-loop gain is no longer dependent no the gain of the op-amp but depends on the feedback of the voltage divider. the feedback gain B can be precisely controlled and it is independent of the amplifier.
Physically, what is happening in the circuit? the gain is approximately constant, even though differential voltage gain may change. suppose, A increases for some reasons (temperature change). then, the ontput voltage will try to increase. this means that more voltage is feedback to the inverting input, causing vd voltage to decrease. this almost completely offset the attempted increases in output voltage.
Similarly, if A decreases, the output voltage decreases. it reduces the feedback voltage VF and hence, VD voltage increases. thus, the output voltage increases almost to same level.
Different input Voltage is ideally Zero
Again considering the voltage equation
V0 = ADVD
VD = V0|AD
Since, AD is very large (ideally infinite)
VD >> 0
and V1 = V2 (ideal)
This says, that the voltage at non-inverting input terminal of an op-amp is approximately equal to that at the inverting input terminal provided that AD is very large. this concept is useful in the analysis of closed-loop op-amp circuits. for example, ideal closed loop voltage gain can be obtained using the results.
V1 = VIN
V2 = VF = R1 / R1 + RF = V0
V1 = V2
VIN = R1 / R1 + RF = V0
V0 = (1 + R1/RF) VIN
Input resistance with feedback
Fig. 36 shows a voltage series feedback with the op-amp equivalent circuit.
In this circuit, RIN is the input resistance (open-loop) of the op-amp and RIF is the input resistance of the feedback amplifier. the input resistance with feedback isdefined as
RIF = VIN / IIN
Vin = Vd + Vf
= Vd + BV0
= Vd + BAVd
= Vd (1 + AB)
= (1 + AB) Iin Rin
Vin / Iin = Rin (1 + AB) = Rif
Since, AB is much larger than 1, which means that Rif is much larger than Ri Thus, Rif approaches infinity and therefore, this amplifier approximates an ideal voltage amplifier.
Output resistance with feedback
Output resistance is the resistance determined looking back into the feedback amplifier from the output terminal. to find output resistance with feedback RF input VIN is reduced to zero, an external voltage V0 is applied as shown in fig. 37.
The output resistance (Rof) is defined as
Rof = Vin / Iin
Ia = la + Ib
Since, {(R1||R2) + RF] >> R0
I0 = I0
Applying KVL output loop
V0 = R0Ia + AVd
= ROIO + AVd
VD = V1 – V2 = 0 – BVO
IO = VO – AVD / RO = VO – ABVO / RO
Rof = VO / IO = RO 1 + AB
This shows that the output resistance of the voltage series feedback amplifier is (1/1 + AB) times the output resistance RO of the op-amp. it is very small because (1 + AB) is very large. it approaches to zero for an ideal voltage amplifier.
Reduced non-linear distortion
The final stage of an op-amp has non-linea distortion when the signal swings over most of the AC load line. large swings in current cause the r’e of a transistor to change during the cycle. in other words, the open-loop gain varies throughout the cycle of when a large signal is being applied. it is the changing voltage gain that is a source of the non-linear distortion.
Non-inverting voltage feedback reduces non-linear distortion because the feedback stabilizes the closed-loop voltage gain, making it almost independent of the changes in open-loop voltage gain. As long as loop gain is much greater than 1, the output voltage equals 1/B times the input voltage. this implies that output will be a more faithful reproduction of the input.
Consider, under large signal conditions, the open-loop op-amp circuit produces a distortion voltage, designated Vdist. it can be represented by connecting a source Vdist in series with AVd. without negative feedback , all the distortion voltage Vdist appears at the output. but with negative feedback, a fraction of Vdist is feedback to inverting input. this is amplified and arrives at the output with inverted phase almost completely canceling the original distortion produced by the output stage.
V0 = AVD + Vdist
= A (VIN – BVout) + Vdist
V0 = A / 1 + AB = Vin + 1 / 1 + AB = Vdist
The first term is the amplified output voltage. the second term in the distortion that appears at the final output. the distortion voltage is very much reduced because AB > 1.
Bandwidth with feedback
The bandwidth of an amplifier is defined as the band of frequencies for which the gain remains constant fig. 38, shows the open-loop gain vs frequency curve is 741c op-amp. from this curve for a gain of 2 x 105 the bandwidth is approximately 5 Hz. on the other hand, the bandwidth is approximately 1 MHz, when the gain is unity.
The frequeney at which gain equals 1 is known as the unity gain banwidth. it is the maxiimum frequency the op-amp can be used for.
Furthermore, the gain bandwidth product obtained from the open-loop gain vs frequency curve is equal to the unity gain bandwidth of the op-amp.
Since, the gain bandwidth product is constant obviously the higher the gain the smaller the bandwidth and vice-versa. if negative feedback is used gain decreases from A to A(1 + AB). Therefore, the closed loop bandwidth with feedback
= (1 + AB) x (BW without feedback)
ff = fo (1 + AB)
Output offset voltage
In an op-amp even if the input voltage is zero, an output voltage can exist. there are three cause of this unwanted offset voltage.
- Input offset voltage
- Input bias voltage
- Input offset current
Fig. 39 shows a feedback amplifier with an output offset voltage source in series with the open-loop output AVd. the actual output offest voltage with negative feedback is smaller. the reasoning is similar to that given for distortion. some of the output offset voltage is feedback to the inverting input. after amplification an output of phase voltage arrives at the output cancelling most of the original output offset voltage.
Total output offset voltage with feedback = VOUT / 1 + AB
When loop gain AB is much greater than 1. the closed-loop output offset voltage is much smaller than the open-loop output offest voltage.
Voltagen follower
The lowest gain that can be obtained from a non-inverting amplifier with feedback is 1. when the non-inverting amplifier given unity gain. it is called voltage follower because the output voltage is equal to the input voltage and is phase with the input voltage. in other words the voltage output follows the input voltage.
To obtain voltage follower, R1 is open circuited and RF is shorted in a negative feedback amplifier of fig. 39. the resultant circuit is shown in fig. 40
VOUT = AVD = A (V1 – V2)
V1 = Vin
V2 = VOUT
V1 = V2 if A >> 1
VOUT = VIN
The gain of the feedback circuit (B) is 1. therefore,
Af = 1/B = 1
Voltage Shunt Feedback
Fig. 41 shows the voltage shunt feedback amplifier using op-amp.
The input voltage drives the inverting terminal and the amplified as well as inverted output signal is also applied to the inverting input via the feedback resistor Rf. this arrangement forms a negative feedback because any increase in the output signal results in a feedback signal into the inverting input signal causing a decrease in the output signal. the non-inverting terminal is grounded. resistor R1 is connected in series with the source.
The closed-loop voltage gain can be obtained by writing kirchhoff’s current equation at the input node V2.
I1 = If + IB
The closed-loop voltage. since R1 is very large, the input current IB is negligibly small.
I1 = If
Vin – V2 /R1 = V2 – VO / RF
And (V1 – V2) = VO/A
V2 = – VO/A bacause V1 = 0)
VIN + VO/A/R1 = (-VO /A) – VO/RF
AF = VO/VIN = RFA/R1 + RF + AR1
Since, A is very very high therefore, AR1 >> (R1 + RF)
AF = – RF /R
= – 1 / B
Since, B = (R1/RF)
The negative sign in equation indicates that the input and output signals are out of phaes by 1800. therefore, it is called inverting amplifier. the gain can be selected by selecting RF and R1 (even < 1).
Inverting input at virtual ground
In the fig. 42 shown earlier, the non-inverting terminal is grounded and the input signal is applied to the inverting terminal via resistor R1. the difference input voltage VD is ideally zero. (Va = VO/A) is the voltage at the inverting terminals (V2) is approimatrly equal to that of the non-inverting terminal (V1) In other words, the inverting terminal voltage (V1) is approximately at ground potential. therefore. it is said to be at virtual ground.
Iin = If
VIN – V2 /R1 = V2 – VO / RF
V1 = V2 = o V
VIN / R1 = -Vo / RF
VO = – RF / R1 = VIN
Input resistance with feedback
To find the input resistance, miller equivalent of the feedback resister RF is obtained i.e., RF is splitted into its two miller components as shows in fig. 43, therefore input resistance with feedback RIF is
RIF = R1 + (RIF / 1 + A)
Since, RI and A are very large. therefore,
(RF / 1 + A) ||RI ~ 0
RIF = RI
Output resistance with feedback
The output resistance with feedback ROF is the resistance measured at the output terminal of the feedback amplifier. the output resistance can be obtained using thevenin’sequivalent circuit, shown in fig. 44.
IO = Ia + Ib
Since, RO is very small as compared to RF + (R1 ||R2)
Therefore, IO = Ia
Vo = ROIO + AVD
Vd = VI – V2 = 0 -BVO
IO = VO – AVD / RO
= VO – ABVO / RO
ROF = VO /IO = RO / 1 + AB
B = R1 / RF
Similarly, the bandwidth incresasses by (1 + AB) and tatal output offset voltage reduces by (1 + AB).
Example 6. An inverting amplifier shown in figure with RI = 10 and R2 = 1 M is driven by source VI = 0.1 V.Find the closed-loop gain A, the percentage division of A from the ideal value – R2/R1 and the inverting input voltage VN for the cases A = 100 V/V. 105 and 105 V/V.
Sol. We have, AF = -RFA / (R1 + RF + AR1)
When A = 103.
AF = – 10 x 1 x 106 / (10 x 103) + (1 x 106) + (103 x 105)
= -109 / (10 x 104 + 106 + 106)
= -109 / 104 + (H200) = – 497.5
Now, AF(ideal) = -RF / R1 = 103
Deviation = AF (deal) – AF (actual) / AF (ideal) x 100%
= 502.5 x 100% / 1000 = 50.25%
VO = 0.1 x 497.5 = 49.75 V
Or 10 – 100V1 = V1 – 49.75
V1 = 59.75 / 100 V
= 0.5975 V = 0.6 V = 0.6 V
Example 7. Find VN, V1 and VO for the circuit shown in fig.
Sol. Applying KCL at N,
VN / 10 + VN – VO / 20 = 0
Or 2vn + VN = V0
VN = V0/3
Now, V0 – VI = 6 as point A and N are virtuall shorted.
Therefore, V0 – VN = 6
V0 = VN + 6
V0 = V0 /3 + 6
Or 2vo/3 = 6 V0 = 9V
Therefore, VN = VI = 3 V
Example 8. Find a relationship between V0 and V1 through V6 in the circuit.
Sol. Let’s consider of V1 (single)by shorting the others. i.e,. the circuit then looks like as shown in figure below.
The current flowing through the rsistor R into the circuit.
i.e., I = VI – V1 / R
V1~ 0.1 = V1/R
The current when passes through R, output an operational value of
R/R V1 = – V …………….(1)
The net output V’ = – (V1 + V3 + V5) ………………(2)
Let as now consider the case of v2 with other inputs shorted, circuit looks like as shown in figure below.
Now, V0 is given by
VN (1 + R / R3) = V0 – VN (4)
VN = V2 / R + R/3 x R/3 = V2 /4
V0 = VN(4) – V2 /4 x 4 = V2
Same thing to V4 and V6,
Net output voltage V” = V2 + V4 + V6
From eqs. (2) and (3),
V’ + V” (V2 + V4 + V6) – (V1 + V3 + V5)
So, V0 = V2 + V4 + V6 – V1 – V3 – V5
Differential Amplifier
The basic differential amplifier is shown in fig. 45.
Since, there are two inputs superposition theorem can be used to find the output voltage. when VB = 0, then the circuit becomes inverting amplifier, hence the output due to Va only is
VO(a) = – (RF / R1) Va
Similarly, when Va = 0, the configuration is a inverting amplifier having a voltage a voltage divided network at the non-inverting input.
V1 = R3 / R2 + R3 = VB
The output due to VB only is
VO(B) = (1 + RF/R1) (R3/R2 + R3) VB
= (R1 + RF / R1) (R3 /R2 + R3) VB
If R1 = R2 and = R3, then
V0(B) = RF / R1 = VB
Therefore, the total output voltage VO is given by
V0 = V0 (A) = V0(B)
V0 = RF/R1 (-Va + Vb)
Example 9. Find VOUT and IOUT for the circuit shown in figure below. the input voltage is simusoidal with amplitude of 0.5 v.
Sol. we begin by writing the KCL equation at both the positive and negative terminals of the op-amp.
For the negative terminal,
V – VOUT / 9.8 x 104 + V – – 0/7000 = 0
Threrfore, 15 v = vout
For the positive teminal,
V+ – VIN / 104 + V+ – 0 / 2 x 104 = 0
This yields two equations in three unknowns VOUT, V+ and V– The third equation is the relationship between V+ and V– for the ideal op-amp
V+ = V–
Solving these equaions we find
VOUT = 10 VIN = 5 sin 00t V
Since, 2 k resistor forms the load of the op-amp, then the current IOUT is given by IOUT = VOUT/ROUT x 2.5 sin 00t mA.
Example 10. For the different amplifiers shown in figure below verify that
V0 = – (1 – 2R2 / R3) RF /R1(VX – VY)
Sol. Since, the differential input voltage of op-amp is negligible, therefore,
V1 = VX
And V2 = VY
The input impedance of op-amp is very large and therefore, the input current of op-amp is negligible.
Thus Va – V1 / R2 = V1 – V2 / R3 ……………..(1)
And V1 – V2 / R3 = V2 – VB / R2 ……………(2)
From eq. (1),
VA – VX / R2 = VX – VY / R3
Or VA = R2 / R3(VX – VY) + VX
From eq. (2)
VX – VY / R3 = VY – VB / R2
or VB = VY – R2 / R3 (VX – VY)
The op-amp is working as differential ampifier, therefore,
V0 = RF / R1 (VB – VA)
= RF / R1 [VY – VX – 2 R2 /R3(VX – VY)]
V0 = – (1 + 2R2/ R3) RF / R1 (VX – VY)