closed loop amplifier | formula , gain , bandwidth , the closed-loop voltage gain of an inverting amplifier equals

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 A_d or A and the feedback circuit is composed of two resistors R₁ or R_f as shown in Fig. 35
Open -loop voltage gain A_d = V_o/ V_d
Open- loop voltage gain A_cl =V_o /V_in
feedback circuit gain B=Vf /V_o
The different voltage input V_d  = V_in – V_f
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,
A_CL = V₀ / V_IN
V₀ = A (V₁ – V₂) = A (V_IN – V_F)
= A (V_IN – BV₀)
V₀ = (1 + AB) = AV_IN
V₀ / V_IN = 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  / V_IN = A / AB = 1 / B
= R₁ + R_F / R₁ = 1 + R_F / 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 v_d 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 V_F and hence, V_D voltage increases. thus, the output voltage increases almost to same level.
Different input Voltage is ideally Zero
Again considering the voltage equation
V₀ = A_DV_D
V_D = V₀|A_D
Since, A_D is very large (ideally infinite)
V_D >> 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 A_D 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 = V_IN
V₂ = V_F = R₁ / R₁ + R_F  = V₀
V₁ = V₂
V_IN = R₁ / R₁ + R_F = V₀
V₀ = (1 + R₁/R_F) V_IN
Input resistance with feedback
Fig. 36 shows a voltage series feedback with the op-amp equivalent circuit.
In this circuit, R_IN is the input resistance (open-loop) of the op-amp and R_IF is the input resistance of the feedback amplifier. the input resistance with feedback isdefined as
R_IF = V_IN / I_IN
V_in = V_d + V_f
= V_d + BV₀
=  V_d + BAV_d
= V_d (1 + AB)
= (1 + AB) I_in R_in
V_in / I_in = R_in (1 + AB) = R_if
Since, AB is much larger than 1, which means that R_if is much larger than R_i Thus, R_if 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 R_F input V_IN is reduced to zero, an external voltage V₀ is applied as shown in fig. 37.
The output resistance (R_of) is defined as
R_of = V_in / I_in
I_a = l_a + I_b
Since,              {(R₁||R₂) + R_F] >> R₀
I₀ = I₀
Applying KVL output loop
V₀ = R₀I_a + AV_d
= R_OI_O + AV_d
V_D = V₁ – V₂ = 0 – BV_O
I_O = V_O – AV_D / R_O = V_O – ABV_O / R_O
R_of = V_O / I_O = R_O 1 + AB
This shows that the output resistance of the voltage series feedback amplifier is (1/1 + AB) times the output resistance R_O 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 V_dist. it can be represented by connecting a source V_dist in series with AV_d. without negative feedback , all the distortion voltage V_dist appears at the output. but with negative feedback, a fraction of V_dist 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 = AV_D + V_dist
= A (V_IN – BV_out) + V_dist
V₀ = A / 1 + AB = V_in + 1 / 1 + AB = V_dist
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)
f_f = f_o (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.

1. Input offset voltage
2. Input bias voltage
3. Input offset current

Fig. 39 shows a feedback amplifier with an output offset voltage source in series with the open-loop output AV_d. 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 = V_OUT / 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 R_F is shorted in a negative feedback amplifier of fig. 39. the resultant circuit is shown in fig. 40
V_OUT = AV_D = A (V₁ – V₂)
V₁ = V_in
V₂ = V_OUT
V₁ = V₂ if A >> 1
V_OUT = V_IN
The gain of the feedback circuit (B) is 1. therefore,
A_f = 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 R_f. 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 = I_f + I_B
The closed-loop voltage. since R₁ is very large, the input current I_B is negligibly small.
I₁ = I_f
V_in – V₂ /R₁ = V₂ – V_O / R_F
And          (V_{1 –} V₂) = V_O/A
V₂ = – V_O/A                bacause V₁ = 0)
V_IN + V_O/A/R₁ = (-V_{O /}A) – V_O/R_F
A_F = V_O/V_IN = R_FA/R₁ + R_F + AR₁
Since, A is very very high therefore, AR₁ >> (R₁ + R_F)
A_F = – R_F /R
= – 1 / B
Since,                         B = (R_1/R_F)
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 R_F 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 V_D is ideally zero. (V_a = V_O/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.
I_in = I_f
V_IN – V₂ /R₁ = V₂ – V_O / R_F
V₁ = V₂ = o V
V_IN / R₁ = -V_o / R_F
V_O = – R_F / R₁  = V_IN
Input resistance with feedback
To find the input resistance, miller equivalent of the feedback resister R_F is obtained i.e., R_F is splitted into its two miller components as shows in fig. 43, therefore input resistance with feedback R_IF is
R_IF = R₁ + (R_IF / 1 + A)
Since, R_I and A are very large. therefore,
(R_F / 1 + A) ||R_I ~ 0
R_IF = R_I
Output resistance with feedback
The output resistance with feedback R_OF 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.
I_O = I_a + I_b
Since, R_O is very small as compared to R_F + (R₁ ||R₂)
Therefore, I_O = I_a
V_o = R_OI_O + AV_D
V_d = V_I – V₂ = 0 -BV_O
I_O = V_O – AV_D / R_O
= V_O – ABV_O / R_O
R_OF = V_O /I_O = R_O / 1 + AB
B = R₁ / R_F
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 R_I  = 10 and R₂ = 1 M is driven by source V_I = 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 V_N for the cases A = 100 V/V. 10⁵ and 10⁵ V/V.
Sol.  We have, A_F = -R_FA / (R₁ + R_F + AR₁)
When                A = 10³.
A_F = – 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,             A_F(ideal) = -R_F / R₁ = 10³
Deviation    = A_F (deal) – A_F (actual) / A_F (ideal) x 100%
= 502.5 x 100% / 1000 = 50.25%
V_O = 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 V_N, V₁ and V_O for the circuit shown in fig.
Sol. Applying KCL at N,
V_N / 10 + V_N – V_O / 20 = 0
Or                      2v_n + V_N = V₀
V_N = V₀/3
Now, V₀ – V_I = 6 as point A and N are virtuall shorted.
Therefore,                 V₀ – V_N = 6
V₀ = V_N + 6
V₀ = V₀ /3 + 6
Or                         2v_o/3 = 6         V₀ = 9V
Therefore,            V_N = V_I = 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 = V_I – 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
V_N (1 + R / R₃) = V₀ – V_N (4)
V_N = V₂ / R + R/3 x R/3 = V₂ /4
V₀ = V_N(4) – V₂ /4 x 4 = V₂
Same thing to V₄ and V_6,
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 V_B = 0, then the circuit becomes inverting amplifier, hence the output due to V_a only is
V_O(a)  = – (R_F / R₁) V_a
Similarly, when V_a = 0, the configuration is a inverting amplifier having a voltage a voltage divided network at the non-inverting input.
V₁ = R₃ / R₂ + R₃ = V_B
The output due to V_B only is
V_O(B) = (1 + R_F/R₁) (R₃/R₂ + R₃) V_B
= (R₁ + R_F / R₁) (R₃ /R₂ + R₃) V_B
If R₁ = R₂ and = R₃, then
V_0(B) = R_F / R₁ = V_B
Therefore, the total output voltage V_O is given by
V₀ = V_{0 (A)}  = V_0(B)
V₀ = R_F/R₁ (-V_a + V_b)
Example 9. Find V_OUT and I_OUT 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 – V_OUT / 9.8 x 10⁴ + V _– – 0/7000 = 0
Threrfore,                       15 v = v_out
For the positive teminal,
V₊ – V_IN / 10⁴ + V₊ – 0 / 2 x 10⁴ = 0
This yields two equations in three unknowns V_OUT, 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
V_OUT = 10 V_IN = 5 sin ₀₀t V
Since, 2 k resistor forms the load of the op-amp, then  the current I_OUT is given by I_OUT = V_OUT/R_OUT x 2.5 sin ₀₀t  mA.
Example 10. For the different amplifiers shown in figure below verify that
V₀ = – (1 – 2R₂ / R₃) R_F /R₁(V_X – V_Y)
Sol. Since, the differential input voltage of op-amp is negligible, therefore,
V₁ = V_X
And                                V₂ = V_Y
The input impedance of op-amp is very large and therefore, the input current of op-amp is negligible.
Thus                      V_a – V₁ / R₂ = V₁ – V₂ / R₃ ……………..(1)
And                        V₁ – V₂ / R₃ = V₂ – V_B / R₂ ……………(2)
From eq. (1),
V_A – V_X / R₂ = V_X – V_Y / R₃
Or                            V_A = R₂ / R₃(V_X – V_Y) + V_X
From eq. (2)
V_X – V_Y / R₃ = V_Y – V_B / R₂
or                       V_B = V_Y – R₂ / R₃ (V_X – V_Y)
The op-amp is working as differential ampifier, therefore,
V₀ = R_F / R₁ (V_B – V_A)
= R_F / R₁ [V_Y – V_X – 2 R₂ /R₃(V_X – V_Y)]
V₀ = – (1 + 2R₂/ R₃) R_F / R₁ (V_X – V_Y)