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 180^{0} 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 R_{1 }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 180^{0} w.r.t input voltage], hence the feedback is said to be negative.

The closed-loop voltage gain is given by,

A_{CL} = V_{0} / V_{IN}

V_{0} = A (V_{1} – V_{2}) = A (V_{IN} – V_{F})

= A (V_{IN} – BV_{0})

V_{0} = (1 + AB) = AV_{IN}

V_{0} / 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

V_{0 } / V_{IN} = A / AB = 1 / B

= R_{1} + R_{F} / R_{1} = 1 + R_{F} / R_{1}

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

V_{0} = A_{D}V_{D}

V_{D} = V_{0}|A_{D}

Since, A_{D} is very large (ideally infinite)

V_{D} >> 0

and V_{1} = V_{2} (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.

V_{1} = V_{IN}

V_{2} = V_{F} = R_{1} / R_{1} + R_{F} = V_{0}

V_{1} = V_{2}

V_{IN} = R_{1 }/ R_{1} + R_{F} = V_{0}

V_{0} = (1 + R_{1}/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} + BV_{0}

= 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 V_{0} 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, {(R_{1}||R_{2}) + R_{F}] >> R_{0}

I_{0} = I_{0}

Applying KVL output loop

V_{0} = R_{0}I_{a} + AV_{d}

= R_{O}I_{O} + AV_{d}

V_{D} = V_{1} – V_{2} = 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.

V_{0} = AV_{D} + V_{dist}

= A (V_{IN} – BV_{out}) + V_{dist}

V_{0} = 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 10^{5} 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.

- 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 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, R_{1} 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 (V_{1} – V_{2})

V_{1} = V_{in}

V_{2} = V_{OUT}

V_{1} = V_{2} 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 R_{1} 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 V_{2}.

I_{1} = I_{f} + I_{B}

The closed-loop voltage. since R_{1} is very large, the input current I_{B} is negligibly small.

I_{1} = I_{f}

V_{in} – V_{2} /R_{1} = V_{2} – V_{O} / R_{F}

And (V_{1 – }V_{2}) = V_{O}/A

V_{2} = – V_{O}/A bacause V_{1} = 0)

V_{IN} + V_{O/}A/R_{1} = (-V_{O} _{/}A) – V_{O}/R_{F}

A_{F} = V_{O}/V_{IN} = R_{F}A/R_{1} + R_{F} + AR_{1}

Since, A is very very high therefore, AR_{1} >> (R_{1} + 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 180^{0}. therefore, it is called inverting amplifier. the gain can be selected by selecting R_{F} and R_{1} (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 R_{1}. the difference input voltage V_{D} is ideally zero. (V_{a} = V_{O}/A) is the voltage at the inverting terminals (V_{2}) is approimatrly equal to that of the non-inverting terminal (V_{1}) In other words, the inverting terminal voltage (V_{1}) is approximately at ground potential. therefore. it is said to be at virtual ground.

I_{in} = I_{f}

V_{IN} – V_{2} /R_{1} = V_{2} – V_{O} / R_{F}

V_{1} = V_{2} = o V

V_{IN} / R_{1} = -V_{o }/ R_{F}

V_{O} = – R_{F} / R_{1} = 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} = R_{1} + (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} + (R_{1} ||R_{2})

Therefore, I_{O} = I_{a}

V_{o} = R_{O}I_{O} + AV_{D}

V_{d} = V_{I} – V_{2} = 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 = R_{1} / 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 R_{2} = 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 – R_{2}/R_{1} and the inverting input voltage V_{N} for the cases A = 100 V/V. 10^{5} and 10^{5} V/V.

**Sol.** We have, A_{F} = -R_{F}A / (R_{1} + R_{F} + AR_{1})

When A = 10^{3}.

A_{F} = – 10 x 1 x 10^{6} / (10 x 10^{3}) + (1 x 10^{6}) + (10^{3} x 10^{5})

= -10^{9} / (10 x 10^{4} + 10^{6} + 10^{6})

= -10^{9} / 10^{4} + (H200) = – 497.5

Now, A_{F}(ideal) = -R_{F} / R_{1} = 10^{3}

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 – 100V_{1} = V_{1} – 49.75

V_{1} = 59.75 / 100 V

= 0.5975 V = 0.6 V = 0.6 V

**Example 7. **Find V_{N}, V_{1} 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} = V_{0}

V_{N} = V_{0}/3

Now, V_{0} – V_{I} = 6 as point A and N are virtuall shorted.

Therefore, V_{0} – V_{N} = 6

V_{0} = V_{N} + 6

V_{0} = V_{0} /3 + 6

Or 2v_{o}/3 = 6 V_{0} = 9V

Therefore, V_{N} = V_{I} = 3 V

**Example 8. **Find a relationship between V_{0} and V_{1} through V_{6} in the circuit.

**Sol.** Let’s consider of V_{1} (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} – V_{1} / R

V_{1}~ 0.1 = V_{1}/R

The current when passes through R, output an operational value of

R/R V_{1} = – V …………….(1)

The net output V’ = – (V_{1} + V_{3} + V_{5}) ………………(2)

Let as now consider the case of v_{2} with other inputs shorted, circuit looks like as shown in figure below.

Now, V_{0} is given by

V_{N} (1 + R / R_{3}) = V_{0} – V_{N} (4)

V_{N} = V_{2} / R + R/3 x R/3 = V_{2} /4

V_{0} = V_{N}(4) – V_{2} /4 x 4 = V_{2}

Same thing to V_{4} and V_{6,}

Net output voltage V” = V_{2} + V_{4} + V_{6}

From eqs. (2) and (3),

V’ + V” (V_{2} + V_{4} + V_{6}) – (V_{1} + V_{3} + V_{5})

So, V_{0} = V_{2} + V_{4} + V_{6} – V_{1} – V_{3} – V_{5}

**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} / R_{1}) 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.

V_{1} = R_{3} / R_{2} + R_{3} = V_{B}

The output due to V_{B} only is

V_{O(B)} = (1 + R_{F}/R_{1}) (R_{3}/R_{2} + R_{3}) V_{B}

= (R_{1} + R_{F} / R_{1}) (R_{3} /R_{2} + R_{3}) V_{B}

If R_{1} = R_{2} and = R_{3}, then

V_{0(B)} = R_{F} / R_{1} = V_{B}

Therefore, the total output voltage V_{O} is given by

V_{0} = V_{0} _{(A) }= V_{0(B)}

V_{0} = R_{F}/R_{1} (-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 10^{4} + V _{–} – 0/7000 = 0

Threrfore, 15 v = v_{out}

For the positive teminal,

V_{+} – V_{IN} / 10^{4} + V_{+} – 0 / 2 x 10^{4} = 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 _{00}t 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 _{00}t mA.

**Example 10. **For the different amplifiers shown in figure below verify that

V_{0} = – (1 – 2R_{2} / R_{3}) R_{F} /R_{1}(V_{X} – V_{Y})

**Sol.** Since, the differential input voltage of op-amp is negligible, therefore,

V_{1} = V_{X}

And V_{2} = V_{Y}

The input impedance of op-amp is very large and therefore, the input current of op-amp is negligible.

Thus V_{a} – V_{1} / R_{2} = V_{1} – V_{2} / R_{3} ……………..(1)

And V_{1} – V_{2} / R_{3} = V_{2} – V_{B} / R_{2} ……………(2)

From eq. (1),

V_{A} – V_{X} / R_{2} = V_{X} – V_{Y} / R_{3}

Or V_{A} = R_{2} / R_{3}(V_{X} – V_{Y}) + V_{X}

From eq. (2)

V_{X} – V_{Y} / R_{3} = V_{Y} – V_{B} / R_{2}

or V_{B} = V_{Y} – R_{2} / R_{3} (V_{X} – V_{Y})

The op-amp is working as differential ampifier, therefore,

V_{0} = R_{F} / R_{1} (V_{B} – V_{A})

= R_{F} / R_{1} [V_{Y} – V_{X} – 2 R_{2} /R_{3}(V_{X} – V_{Y})]

V_{0} = – (1 + 2R_{2}/ R_{3}) R_{F} / R_{1} (V_{X} – V_{Y})