Negative Feedback, Inverting & Non-Inverting Amplifier: Complete Guide + Examples

24.2   NEGATIVE FEEDBACK

24.2.1   Feedback in Amplifiers

1.   Definition

Quite often we intentionally send part of the output voltage V_o of an amplifier back to the input of the same amplifier. This action of sending a fraction of the output of an amplifier to the input of the same amplifier is known as feedback.

2.   Feedback signal path

Figure 24.12 The figure shows an op-amp circuit with a feedback network connected between the output and the input to control the amplifier’s behavior.

(a) Feedback is achieved by connecting a feedback network between the input and the output. One of the several ways of doing this is shown in Figure 24.12.

(b) The circuit above shows the ‘voltage feedback’ method. In this method, the feedback network extracts a fraction of the output voltage V_o. The network then introduces this fraction of the output voltage into the input section of the amplifier.

(c) The voltage introduced into the input section is βV_o where β indicates the fraction of the original output voltage V_o fed to the input.

(d) Because there is a circuit linking the input to the output, the amplifier is no longer in the open-loop mode. It is now in the closed-loop mode.

24.2.2   one Example of Feedback Network

Figure 24.13 (a) The figure shows a feedback network where a fraction of the output voltage (βVo) is fed back to the input through a potential divider formed by resistors. (b) The figure shows the rearranged circuit known as a non-inverting amplifier, where the input signal is applied to the non-inverting terminal and positive feedback is established through the resistor network.

1. Refer to Figure 24.13(a). The fraction βV_o of the output voltage fed into the input section of the amplifier is obtained from the p.d. across the resistor R₂, which forms part of the potential divider circuit across the output.

2. The fraction βV_o of the output voltage is given by

$βV_0 = p.d. \ across \ R_2$

$= \left(\frac{ R_1}{R_1 + R_2} \right)V_0$

$β = \frac{R_1}{R_1 + R_2}$

Let R₁ = 20 kΩ, R₂ = 2.0 kΩ. Then we get

$β = \frac{2}{2 + 20} = 0.09$

3. Figure 24.13(b) shows the same circuit but with the resistors rearranged. This circuit is known as the non-inverting amplifier.

24.2.3   Positive and Negative Feedback

Feedback can be

(a) positive feedback
A fraction of the output voltage is introduced into the input to increase the input voltage. The effective voltage V_d directly across the + input and the − input is now given by

$V_d = V_{in} + βV_0$

(b) negative feedback
A fraction of the output voltage is introduced into the input to decrease the input voltage. In this case,

$V_d = V_{in} − βV_0$

24.2.4   Advantages of Negative Feedback

1. The voltage gain of the amplifier with negative feedback is very stable. It is determined mainly by the parameters of those components used in the feedback network. If we replace a damaged op-amp by a similar one, the gain would more or less remain the same. In other words, the gain of the amplifier is less dependent upon the parameters of the op-amp itself.

2. The voltage gain bandwidth is greatly increased.

3. Noises in signals are much reduced.

4. Distortions of signals are much reduced.

Note: The main disadvantage is that the gain is drastically reduced.

24.3   INVERTING AMPLIFIER

24.3.1   The Inverting Amplifier

Figure 24.14 The figure shows an op-amp connected with external resistors to form an inverting amplifier, where the input signal is applied to the inverting terminal and the output is amplified with a phase reversal.

1.   Circuit of inverting amplifier

Figure 24.14 shows how an op-amp is connected to some external components to become an inverting amplifier. Notice the following:

(a) The + input is grounded. This means that V₁ = 0.
(b) The − input is connected to an input signal V_in via resistor R_i.
(c) The point X in the output and the point Y in the input are connected via resistor R_F. Hence, the path from X to Y through R_F acts as a negative feedback circuit. Part of the output voltage will be fed back to the input.
(d) The amplifier is operating in the closed-loop mode.

2.   The sign of V_o

Because the + input is grounded, V₁ = 0. Hence, we have

$V_d = V_1 − V_2 = −V_2$

(a) If V₂ is positive, V_d and V_o become negative.
(b) If V₂ is negative, V_d and V_o become positive.

The polarity of the output voltage V_o is always the reverse of the polarity of V₂ at the − input and also the signal V_in. Because of this, the amplifier is known as the inverting amplifier. It means that V_d and V_o are out of phase by 180°.

3.   Virtual ground

The voltage difference V_d between the + input and the − input is nearly equal to zero, i.e.,

$V_d = V_1 − V_2 ≈ 0$

Because the + input is grounded, we have V₁ = 0. This means that V₂ ≈ 0. Hence, the potential at point Y at the − input is nearly equal to zero. Because of this, point Y is referred to as the virtual ground.

24.3.2   Voltage Gain of Inverting Amplifier

1.   Definition of gain

The closed-loop voltage gain A_CL of the inverting amplifier is defined as the ratio

$A_{CL} = \frac{V_0}{V_{in}}$

where V_o and V_in are the output voltage and the input signal voltage respectively.

Note: (a) If A_CL is constant then

$V_0 ∝ V_{in}$

(b) This linear relationship is true only if the output voltage does not get saturated.

2.   A_CL in terms of R_i and R_F

(a) Refer to the amplifier circuit shown in Figure 24.14. Current flows in the direction as shown.
(b) Move from P to Y through R_i, then to the − input, to ground and back to P through V_in. Using Kirchhoff’s voltage law, we get

$V_{in}− I_iR_i + V_d = 0$

Assume that the amplifier is ideal. We get

$V_d = 0$     (ideal)

Then,

$I_iR_i = V_{in}$

$I_i = \frac{V_{in}}{R_i}$

(c) Move from the ground to the + input, to the − input, to Y, to X through R_F and back to ground through R_L. Using Kirchhoff’s voltage law, we get

$V_d − I_FR_F + V_0 = 0$

$I_FR_F = V_0$     ($V_d = 0$)

$I_F = \frac{V_0}{R_F}$

(d) Apply Kirchhoff’s current law to the junction Y.

$I_i = I_F + I_2$

But,

$I_2 = 0$     (ideal)

Hence,

$I_i = I_F$

(e) We have

$I_i = \frac{V_{in}}{R_i}$

and

$I_F = \frac{V_0}{R_F}$

Hence,

$\frac{V_0}{V_{in}}= \frac{R_F}{R_i}$

According to definition

$A_{CL} = \frac{V_0}{V_{in}}$

or,

$V_0 = A_{CL}V_{in}$

But the sign of V_o is always the reverse of the sign of V_in. Hence, we have

$+V_0 = A_{CL}(−V_{in})$

$A_{CL} = \frac{− R_F}{R_i}$

This is the voltage gain of an ideal inverting amplifier.

3.   Stability of A_CL

Notice that A_CL is determined only by external components R_i and R_F. Hence, the instability of A_OL does not affect A_CL. This means that the voltage gain of an op-amp which has a negative feedback is very stable.

EXAMPLE   24.4

Design a basic amplifier with a voltage gain of 25 and an input resistance of 2.5 kΩ. The signal voltage applied to the inputs which can cause the output voltage to go into saturation is about ±0.3 V. The signal voltage and the output voltage are to be out of phase by 180°.

Answer

(a) Since the signal voltage and the output voltage are to be out of phase by 180°, we must construct an inverting op-amp. We must connect the + input of the op-amp to the ground.

(b) Connect one end of the resistor with resistance R_i = 2.5 kΩ to the − input.

(c) Connect one end of resistor R_F to the output and the other end to the − input. The value of R_F is determined as follows:

$A_{CL}= \frac{− R_F}{R_i}$

$−25 = \frac{− R_F}{2.5 \  kΩ}$

$R_F = 63 \ kΩ$

(d) The voltages ±V_s supplied to the op-amp is determined as follows:

$A_{CL} = \frac{V_0}{V_{in}}$

V_o becomes V_sat at about V_in = ±0.3 V. Hence we have

$25 = \frac{V_{sat}}{±0.3 \ V}$

$V_{sat} = ±7.5 \ V$

Assume that

$V_{sat} ≅ 0.9 \ V_s$

$V_s = 8.3 \ V$

Hence, we use voltage supplies of ±9 V.

EXAMPLE   24.5

Refer to Figure 24.14. Given R_i = 3 kΩ, R_F = 30 kΩ, R_L = 25 kΩ, V_in = 0.50 V (rms), V_s = ±15 V. Determine

(a) the closed-loop voltage gain of the inverting amplifier
(b) the output voltage
(c) current I_i
(d) current I_F
(e) current I_L through R_L
(f) current I which flows through the amplifier.
State one assumption about the output voltage.

Answer

(a) $A_{CL}= \frac{R_F}{R_i}$

$= \frac{− 100 \ kΩ}{10 \ kΩ} = −10$

(b) $V_0 = A_{CL}V_{in}$

$= (−10)(+0.5)$

$= −5.0 \ V$ (rms)

(c) $I_i = \frac{V_{in}}{R_i}$

$= \frac{0.5 \ V}{10 \ kΩ} = 0.05 \ mA$

(d) $I_F = I_i$

$= 0.05 \ mA$

(e) $I_L = \frac{V_0}{R_L}$

$= \frac{5.0 \ V}{22 \ kΩ} = 0.20 \ mA$

(f) $I = I_F + I_L$

$= 0.05 + 0.20$

$= 0.25 \ mA$

Assumption: We assume that the output voltage does not go into saturation so that we can apply the linear relationship

$V_0 ∝ V_{in}$

EXAMPLE   24.6

Figure 24.15 (a) Inverting amplifier circuit with 20 kΩ feedback resistor, (b) Alternating input voltage waveform showing signal variation with time.

(a) Refer to Figure 24.15(a). Assume that the op-amp is ideal.
(i) Does the op-amp act as an inverting or non-inverting amplifier?
(ii) Determine the voltage gain of this amplifier.
(iii) Determine the current which flows through the 20 kΩ resistor at the instant when the input signal voltage is +40 mV.

(b) An alternating voltage whose waveform is as shown in Figure 24.15(b) is applied to the − input.
Estimate the maximum signal voltage which would get the output voltage into saturation.
Sketch a graph to show how
(i) the output voltage varies with time.
(ii) the output voltage varies with the input signal voltage.

Answer

(a) (i) It acts as an inverting amplifier.
(ii) Given R_i = 2.0 kΩ and R_F = 20 kΩ.

$A_{CL} = \frac{− R_F}{R_i}$

$= \frac{− 20}{2} = −10$

(iii)

$I_i = \frac{V_{in}}{R_{in}}$

$= \frac{40 \ mV}{2 \ kΩ} = 20 \ μA$

$I_F = I_i = 20 \ μA$

(b) (i) To plot the graph, we need the following:

1. $V_{sat} = ±0.9V_s$

     $= ±0.9(15) = ±14 \ V$

2. $V_{sat} = A_{CL}V_{in}$

     $±14 = (10)V_{in}$

     $V_{in} = ±1.4 \ V$

Hence, beyond V_in = ±1.4 V, V_o will go into saturation.

3. When the signal voltage is a positive maximum, the output voltage becomes

$V_0 = A_{CL}V_{in}$

$= (−10)(+0.05 V) = −0.50 \ V$

Figure 24.16 The figure shows how the output voltage of the op-amp varies with time, illustrating the waveform of the amplified signal.

Figure 24.16 shows how the output voltage changes with time.

(ii)

Figure 24.17 shows how V_o varies with V_in. Notice the following:

(a) The maximum value of V_o is ±0.50 V whereas V_sat = ±14 V. Hence, the amplifier is functioning within the linear region.
(b) When V_o is positive, we have V_in negative, and vice versa. This means that the output voltage and the input signal are out of phase by 180°. The amplifier is an inverting one.

Figure 24.17 The figure shows the relationship between output voltage (Vo) and input voltage (Vin), illustrating how the output varies proportionally with the input within the operating range.

24.4   NON-INVERTING AMPLIFIER

24.4.1   The Non-inverting Amplifier

Figure 24.18 The figure shows an op-amp connected as a non-inverting amplifier, where the input signal is applied to the non-inverting terminal and the output is amplified without phase reversal.

1.   Circuit of non-inverting amplifier

Figure 24.18 shows how an op-amp is connected to become a non-inverting amplifier. Notice the following:

(a) The input voltage V_in is now connected to the + input.
(b) The path from X to Y through R_F acts as a path for negative feedback.
(c) The amplifier functions in the closed-loop mode.

2.   Sign of V_o

The sign of V_o is determined by the sign of V_in. Note that

(a) if V_in is positive, V_o becomes positive.
(b) if V_in is negative, V_o becomes negative.

Because of this, the amplifier is known as a non-inverting amplifier. This means that V_d and V_o are in phase with each other.

24.4.2   Voltage Gain of Non-inverting Amplifier

1.   Definition of voltage gain

The closed-loop voltage gain A_CL for this amplifier is defined as the ratio

$A_{CL}= \frac{V_0}{V_{in}}$

where V_o and V_in are the output voltage and input signal respectively.

2.   A_CL in terms of R and R_F

(a) Refer to Figure 24.18. Current flows in the direction shown.
(b) Travel from ground to the + input through V_in, to the − input, to Y through R and back to ground. Use Kirchhoff’s voltage law. We get

$V_{in} − V_d − V = 0$

Assuming that the amplifier is ideal, then we have

$V_d = 0$

$V_{in} = V$     (ideal)

where

$V = IR$     ..... (1)

(c) To determine V in terms of V_o, use

$V_0 = V + V_F$

$= IR + I_FR_F$

Apply Kirchhoff’s current law to the junction Y.

$I_F = I + I_2$

But,

$I_2 = 0$

Hence,

$I_F = I$     (ideal)

$V_0 = I(R + R_F)$     ..... (2)

(1) ÷ (2),

$\frac{V}{V_0} = \frac{R}{R + R_F}$

$V = \left(\frac{ R}{R + R_F}\right ) V_0$

(d) We have

$V_{in} = V$

Hence

$V_{in} = \left(\frac{ R}{R + R_F}\right ) V_0$

$A_{CL} = \frac{V_0}{V_{in}}$

$A_{CL} = 1 + \frac{R_F}{R}$

3.   Stability of A_CL

A_CL is determined by the external components R and R_F. Hence, A_CL is very stable.

EXAMPLE   24.7

Refer to Figure 24.18. Assume that the op-amp is ideal. Given R = 2 kΩ, R_F = 80 kΩ, V_s = ±15 V, V_in = (0.5 V)sin100πt where time t is in second. Determine V_o.

Answer

$A_{CL} = 1 + \frac{R_F}{R}$

$A_{CL} = 1 + \frac{80 \ kΩ}{2 \ kΩ}=41$ 

$V_0 = A_{CL}V_{in}$

$= (41)(0.5 \ sin \ 100πt) \ V$

$= (20.5 \ sin \ 100πt) \ V$

The maximum voltage is 20.5 V. But the supply voltage is only ±15 V. Hence, the output voltage has reached saturation.

$V_0 = ±0.9V_s$

$= ±0.9 (15 \ V)$

$= ±14 \ V$

EXAMPLE   24.8

Figure 24.19 (a) The figure shows an op-amp circuit used to determine whether it operates as an inverting or non-inverting amplifier and to calculate its voltage gain. (b) The figure shows an alternating input voltage waveform applied to the non-inverting (+) terminal of the op-amp.

(a) Refer to Figure 24.19(a). Assume that the op-amp is ideal.
(i) Does the op-amp act as an inverting or non-inverting amplifier ?
(ii) Determine the voltage gain of this amplifier.

(b) An alternating voltage whose waveform is as shown in Figure 24.19(b) is applied to the + input. Sketch a graph to show how
(i) the output voltage varies with time.
(ii) the output voltage varies with V_in

Answer

(a) (i) It acts as a non-inverting amplifier.
(ii) Given R = 2.0 kΩ and R_F = 20 kΩ.

$A_{CL} = 1 + \frac{R_F}{R}$

$= 1 + \frac{20}{2} = 11$

(b) (i) To plot the graph, we need the following:

1. $V_{sat} = ±0.9V_s$
    
$= ±0.9(13) = ±11 \ V$

2. $V_{sat} = A_{CL}V_{in}$

     $±11 = (11)V_{in}$

     $V_{in} = ±1.0 \ V$

Hence, beyond V_in = ±1.0 V, V_o will go into saturation.

Figure 24. 20 This figure shows how the output voltage varies with time. The waveform follows the input signal but is amplified and inverted due to the inverting amplifier. When the input signal is too large, the output reaches its maximum limit and becomes clipped (saturated) at the top and bottom.

3. When the signal voltage is a positive maximum, the maximum output voltage becomes

$V_0 = A_{CL}V_{in}$

$= (11)(+0.05 \ V) = +0.55 \ V$

Figure 24.20 shows how the output voltage changes with time.

(ii)

Figure 24.21 This figure shows how the output voltage (Vo) varies with the input voltage (Vin). The graph is linear in the central region, where the output is proportional to the input with a constant gain. However, when the input becomes too large, the output reaches its maximum and minimum limits and becomes saturated, resulting in flat regions at the top and bottom of the graph.

Figure 24.21 shows how V_o varies with V_in. Notice the following:

(a) The maximum value of V_o is ±0.55 V whereas V_sat ≅ ±11 V. Hence, the amplifier is functioning within the linear region.
(b) When V_o is positive, we have V_in also positive. This means that the output voltage and the input signal are in phase. The amplifier is a non-inverting one. The line has a positive gradient.

Real Applications of Amplifiers

Negative Feedback, Inverting & Non-Inverting Amplifier

Application 1

Audio Amplifier System

In audio systems, amplifiers are used to increase weak sound signals from microphones or instruments. Negative feedback is applied to improve sound quality by reducing noise and distortion.

• Stable gain
• Low distortion
• Better sound clarity

Example: Speakers, music systems, studio equipment.

Application 2

Sensor Signal Conditioning

Inverting amplifiers are used to process signals from sensors such as temperature, pressure, and light sensors where signal inversion or scaling is required.

• Signal inversion
• Precise voltage scaling
• Noise reduction using feedback

Example: Industrial sensors, IoT devices, automation systems.

Application 3

Voltage Follower (Buffer Circuit)

Non-inverting amplifiers are used as voltage followers to isolate circuits and prevent loading effects.

• High input impedance
• Low output impedance
• No signal loss

Example: ADC input stages, measurement systems.

Application 4

Comparator Circuit

Op-amps without feedback act as comparators to compare two voltages and produce digital-like output.

• Used in switching circuits
• Decision making systems
• Threshold detection

Example: Battery level detector, zero-crossing detector.

Application 5

Control Systems

Negative feedback is widely used in control systems to maintain stability and accuracy.

• Automatic control
• Error correction
• System stability

Example: Robotics, motor control, temperature control systems.

Conclusion

Negative feedback plays a crucial role in improving the performance of amplifiers by enhancing stability, reducing noise, and minimizing signal distortion. Through the use of feedback networks, amplifiers can operate in a controlled closed-loop system, making them more reliable and predictable.

• Inverting amplifiers produce output signals that are 180° out of phase with the input.
• Non-inverting amplifiers maintain the same phase between input and output signals.
• Voltage gain depends mainly on external resistors, making the system stable.
• Negative feedback improves bandwidth and reduces distortion significantly.

Overall, understanding feedback and amplifier configurations is essential for designing efficient electronic systems in audio, control, and signal processing applications.

Frequently Asked Questions (FAQ)

What is negative feedback in amplifiers? ▼

Negative feedback is the process of feeding a portion of the output signal back to the input to improve stability, reduce distortion, and control gain.

What is the difference between inverting and non-inverting amplifiers? ▼

In an inverting amplifier, the output signal is 180° out of phase with the input, while in a non-inverting amplifier, the output is in phase with the input.

Why is negative feedback important? ▼

Negative feedback improves amplifier performance by increasing bandwidth, reducing noise, and making the gain more stable and predictable.

What is the formula for voltage gain in an inverting amplifier? ▼

The voltage gain is given by A = -Rf / Ri, where Rf is the feedback resistor and Ri is the input resistor.

What is a comparator in op-amp circuits? ▼

A comparator is an op-amp used without feedback to compare two input voltages and produce a high or low output depending on which input is greater.

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