inverting amplifier configuration (A-level only)

In your electronics studies so far, you have seen the "ideal" properties of an operational amplifier (op-amp). Now, it is time to put those properties to work.

What you'll learn in this topic:

How to draw and build an inverting amplifier circuit.
The concept of a "virtual earth" and why it exists.
How to derive the voltage gain formula step-by-step (a common exam question!).
How to calculate output voltages and choose appropriate resistor values.

The Inverting Amplifier Circuit

An operational amplifier on its own is usually too sensitive to be useful—its open-loop gain is so huge that even the tiniest input voltage causes the output to saturate. To control the amplifier, we use a technique called negative feedback, where we connect part of the output back to the inverting ( $-$ ) input.

In an inverting amplifier, we apply our input signal $VinV_{\text{in}}$ to the inverting terminal via an input resistor, $RinR_{\text{in}}$ . We then connect a feedback resistor, $R_f$ , from the output back to the inverting input. Finally, we connect the non-inverting ( $+$ ) terminal directly to ground (0 volts).

Inverting Amplifier Circuit

Because the signal goes into the inverting ( $-$ ) input, the output voltage $VoutV_{\text{out}}$ will be of the opposite sign to $VinV_{\text{in}}$ . If you put in a positive voltage, you get a negative voltage out. If you put in an AC sine wave, the output will be completely out of phase (shifted by $180∘180^\circ$ ).

The Magic of the "Virtual Earth"

Before we can calculate the gain of this circuit, we need to understand a very clever concept called the virtual earth.

Let's look at the non-inverting ( $+$ ) terminal. It is connected directly to ground, so its voltage $V_+$ is exactly 0 volts.

Now, recall two crucial properties of an ideal op-amp:

Infinite open-loop gain ( $\to \infty$ ): The op-amp multiplies the potential difference between its inputs by a huge number to create the output voltage: $Vout=A(V+−V−)V_{\text{out}} = A(V_+ - V_-)$ .
The output voltage is finite: The output is limited by the power supply (usually about $V\pm 15\text{ V}$ ).

If the output is a sensible, finite number, and the gain $A$ is effectively infinite, then the potential difference between the inputs $V_+ - V_-)$ must be practically zero.

Since $V_+$ is exactly 0 volts, $V_-$ is forced to also be extremely close to 0 volts.

Definition

Virtual Earth

A node in a circuit that is at a potential of 0 volts (relative to ground), but is not physically connected to the ground connection. In an inverting amplifier, the inverting input acts as a virtual earth because the op-amp's infinite open-loop gain forces the potential difference between the input terminals to be zero.

Common Mistake

Is it actually connected to ground?

A frequent mistake is thinking the inverting input is wired to ground. It isn't! If it were wired to ground, current from the input signal would just flow harmlessly into the earth and bypass the op-amp entirely. It just behaves as if it is at 0 volts.

Deriving the Gain Formula (Virtual-Earth Analysis)

AQA loves asking you to derive the formula for the voltage gain of an inverting amplifier. You must use the "virtual earth" concept and the properties of an ideal op-amp to prove it.

Here is the current flow in our circuit:

Current flow in an inverting amplifier

Let's break down the derivation step-by-step:

Calculate the current from the input: A current $IinI_{\text{in}}$ flows from the signal source $VinV_{\text{in}}$ towards the inverting input. Because the inverting input is a virtual earth (at 0 volts), the potential difference across the input resistor is simply $Vin−0V_{\text{in}} - 0$ . Therefore, by Ohm's law:

I_{\text{in}} = \frac{V_{\text{in}}}{R_{\text{in}}}

Where does the current go? An ideal op-amp has infinite input resistance. This means absolutely zero current can flow into the op-amp's inverting terminal itself.
Calculate the current in the feedback loop: Because no current can enter the op-amp, Kirchhoff's First Law tells us that 100% of the current $IinI_{\text{in}}$ must travel up and through the feedback resistor $R_f$ . We will call this feedback current $I_f$ . Therefore:

I_{\text{in}} = I_f

Express the feedback current in terms of voltages: The current $I_f$ flows from the virtual earth (0 volts) to the output ( $VoutV_{\text{out}}$ ). The potential difference across the feedback resistor is $V_{\text{out}})$ . Therefore:

I_f = \frac{0 - V_{\text{out}}}{R_f} = -\frac{V_{\text{out}}}{R_f}

Equate and rearrange: Substitute our voltage expressions into the current equality $Iin=IfI_{\text{in}} = I_f$ :

\frac{V_{\text{in}}}{R_{\text{in}}} = -\frac{V_{\text{out}}}{R_f}

Multiply both sides by $R_f$ and divide by $VinV_{\text{in}}$ to find the voltage gain ( $Vout/VinV_{\text{out}} / V_{\text{in}}$ ):

\frac{V_{\text{out}}}{V_{\text{in}}} = -\frac{R_f}{R_{\text{in}}}

Key Idea

Voltage Gain of an Inverting Amplifier

The closed-loop voltage gain, often written as $G$ or $A_v$ , is given by:

\frac{V_{\text{out}}}{V_{\text{in}}} = -\frac{R_f}{R_{\text{in}}}

The minus sign proves that the signal is inverted.
The gain depends entirely on the ratio of the external resistors, not on the op-amp itself! This makes the circuit incredibly stable and predictable.

Calculations in Practice

When tackling calculations, you simply use the formula. However, always remember to check if your calculated output voltage exceeds the op-amp's power supply rails. If it does, the amplifier will saturate (clip the signal at the maximum supply voltage).

Example

Calculating output voltage

An inverting amplifier is built using an input resistor of $\text{ k}\Omega$ and a feedback resistor of $\text{ k}\Omega$ . The op-amp is powered by a $V\pm 15 \text{ V}$ supply. Calculate the output voltage when the input voltage is $\text{ V}$ .

Identify the given values: $kΩR_{\text{in}} = 10 \text{ k}\Omega$ , $kΩR_f = 47 \text{ k}\Omega$ , $VV_{\text{in}} = 1.2 \text{ V}$ .
State the gain formula:

\frac{V_{\text{out}}}{V_{\text{in}}} = -\frac{R_f}{R_{\text{in}}}

Calculate the voltage gain:

\text{Gain} = -\frac{47 \times 10^3}{10 \times 10^3} = -4.7

Calculate the output voltage:

V_{\text{out}} = \text{Gain} \times V_{\text{in}}

V_{\text{out}} = -4.7 \times 1.2 \text{ V}

V_{\text{out}} = -5.64 \text{ V}

Check for saturation: The magnitude of $\text{ V}$ is comfortably less than the $\text{ V}$ supply, so the amplifier will not saturate. The final answer is $\text{ V}$ (to 2 sig figs).

Tip

Handling prefixes

Because the gain formula is a ratio, you do not always need to convert $kΩ\text{k}\Omega$ to $Ω\Omega$ as long as both resistors are in the same unit. For instance, $kΩ47\text{ k}\Omega / 10\text{ k}\Omega$ gives the exact same result as $Ω47000\text{ }\Omega / 10000\text{ }\Omega$ . Just be careful not to mix $kΩ\text{k}\Omega$ and $MΩ\text{M}\Omega$ !

Sometimes you are asked to work backwards to choose a resistor to achieve a specific gain.

Example

Designing a specific gain

An engineer needs an inverting amplifier with a voltage gain of $- 12$ . They have a $\text{ k}\Omega$ resistor available to use as the feedback resistor. What value should they choose for the input resistor?

State the given values: Gain $= - 12$ , $kΩR_f = 120 \text{ k}\Omega$ .
Write out the gain formula:

\text{Gain} = -\frac{R_f}{R_{\text{in}}}

Substitute the known values:

-\frac{120 \text{ k}\Omega}{R_{\text{in}}}

Rearrange to solve for $RinR_{\text{in}}$ :

R_{\text{in}} = \frac{-120 \text{ k}\Omega}{-12}

R_{\text{in}} = 10 \text{ k}\Omega

Exam technique

In the exam

Learn the derivation off by heart: This is a very common 3 or 4-mark question. You must explicitly state why current does not enter the op-amp ("infinite input resistance") and why the inverting input is at 0V ("infinite open-loop gain and grounded non-inverting input").
Watch the negative sign: If the question asks for the "voltage gain", include the minus sign. If it asks for the "magnitude of the voltage gain", leave the minus sign off. If it asks for $VoutV_{\text{out}}$ , make sure it has the opposite sign to $VinV_{\text{in}}$ .
Always do a sanity check for saturation: If your calculation suggests $VV_{\text{out}} = 25\text{ V}$ but the question states the supply rails are $V\pm 12\text{ V}$ , your final answer for the output voltage must be $V+12\text{ V}$ or $V-12\text{ V}$ (saturated), not $V25\text{ V}$ .

Self review

Check yourself

Can you draw the schematic for an inverting amplifier from memory, making sure the input signal goes to the correct terminal?
What two ideal properties of an op-amp are used to prove that $Iin=IfI_{\text{in}} = I_f$ and that the inverting input is a virtual earth?
If $R_f$ is $kΩ100\text{ k}\Omega$ and $RinR_{\text{in}}$ is $kΩ20\text{ k}\Omega$ , and the input is $V-2\text{ V}$ , what will the output voltage be?

In your electronics studies so far, you have seen the "ideal" properties of an operational amplifier (op-amp). Now, it is time to put those properties to work.

What you'll learn in this topic:

How to draw and build an inverting amplifier circuit.
The concept of a "virtual earth" and why it exists.
How to derive the voltage gain formula step-by-step (a common exam question!).
How to calculate output voltages and choose appropriate resistor values.

The Inverting Amplifier Circuit

Inverting Amplifier Circuit

The Magic of the "Virtual Earth"

Before we can calculate the gain of this circuit, we need to understand a very clever concept called the virtual earth.

Let's look at the non-inverting ( $+$ ) terminal. It is connected directly to ground, so its voltage $V_+$ is exactly 0 volts.

Now, recall two crucial properties of an ideal op-amp:

Infinite open-loop gain ( $\to \infty$ ): The op-amp multiplies the potential difference between its inputs by a huge number to create the output voltage: $Vout=A(V+−V−)V_{\text{out}} = A(V_+ - V_-)$ .
The output voltage is finite: The output is limited by the power supply (usually about $V\pm 15\text{ V}$ ).

If the output is a sensible, finite number, and the gain $A$ is effectively infinite, then the potential difference between the inputs $V_+ - V_-)$ must be practically zero.

Since $V_+$ is exactly 0 volts, $V_-$ is forced to also be extremely close to 0 volts.

Definition

Virtual Earth

Common Mistake

Is it actually connected to ground?

Deriving the Gain Formula (Virtual-Earth Analysis)

AQA loves asking you to derive the formula for the voltage gain of an inverting amplifier. You must use the "virtual earth" concept and the properties of an ideal op-amp to prove it.

Here is the current flow in our circuit:

Current flow in an inverting amplifier

Let's break down the derivation step-by-step:

Calculate the current from the input: A current $IinI_{\text{in}}$ flows from the signal source $VinV_{\text{in}}$ towards the inverting input. Because the inverting input is a virtual earth (at 0 volts), the potential difference across the input resistor is simply $Vin−0V_{\text{in}} - 0$ . Therefore, by Ohm's law:

I_{\text{in}} = \frac{V_{\text{in}}}{R_{\text{in}}}

Where does the current go? An ideal op-amp has infinite input resistance. This means absolutely zero current can flow into the op-amp's inverting terminal itself.
Calculate the current in the feedback loop: Because no current can enter the op-amp, Kirchhoff's First Law tells us that 100% of the current $IinI_{\text{in}}$ must travel up and through the feedback resistor $R_f$ . We will call this feedback current $I_f$ . Therefore:

I_{\text{in}} = I_f

Express the feedback current in terms of voltages: The current $I_f$ flows from the virtual earth (0 volts) to the output ( $VoutV_{\text{out}}$ ). The potential difference across the feedback resistor is $V_{\text{out}})$ . Therefore:

I_f = \frac{0 - V_{\text{out}}}{R_f} = -\frac{V_{\text{out}}}{R_f}

Equate and rearrange: Substitute our voltage expressions into the current equality $Iin=IfI_{\text{in}} = I_f$ :

\frac{V_{\text{in}}}{R_{\text{in}}} = -\frac{V_{\text{out}}}{R_f}

Multiply both sides by $R_f$ and divide by $VinV_{\text{in}}$ to find the voltage gain ( $Vout/VinV_{\text{out}} / V_{\text{in}}$ ):

\frac{V_{\text{out}}}{V_{\text{in}}} = -\frac{R_f}{R_{\text{in}}}

Key Idea

Voltage Gain of an Inverting Amplifier

The closed-loop voltage gain, often written as $G$ or $A_v$ , is given by:

\frac{V_{\text{out}}}{V_{\text{in}}} = -\frac{R_f}{R_{\text{in}}}

The minus sign proves that the signal is inverted.
The gain depends entirely on the ratio of the external resistors, not on the op-amp itself! This makes the circuit incredibly stable and predictable.

Calculations in Practice

Example

Calculating output voltage

Identify the given values: $kΩR_{\text{in}} = 10 \text{ k}\Omega$ , $kΩR_f = 47 \text{ k}\Omega$ , $VV_{\text{in}} = 1.2 \text{ V}$ .
State the gain formula:

\frac{V_{\text{out}}}{V_{\text{in}}} = -\frac{R_f}{R_{\text{in}}}

Calculate the voltage gain:

\text{Gain} = -\frac{47 \times 10^3}{10 \times 10^3} = -4.7

Calculate the output voltage:

V_{\text{out}} = \text{Gain} \times V_{\text{in}}

V_{\text{out}} = -4.7 \times 1.2 \text{ V}

V_{\text{out}} = -5.64 \text{ V}

Check for saturation: The magnitude of $\text{ V}$ is comfortably less than the $\text{ V}$ supply, so the amplifier will not saturate. The final answer is $\text{ V}$ (to 2 sig figs).

Tip

Handling prefixes

Sometimes you are asked to work backwards to choose a resistor to achieve a specific gain.

Example

Designing a specific gain

State the given values: Gain $= - 12$ , $kΩR_f = 120 \text{ k}\Omega$ .
Write out the gain formula:

\text{Gain} = -\frac{R_f}{R_{\text{in}}}

Substitute the known values:

-\frac{120 \text{ k}\Omega}{R_{\text{in}}}

Rearrange to solve for $RinR_{\text{in}}$ :

R_{\text{in}} = \frac{-120 \text{ k}\Omega}{-12}

R_{\text{in}} = 10 \text{ k}\Omega

Exam technique

In the exam

Learn the derivation off by heart: This is a very common 3 or 4-mark question. You must explicitly state why current does not enter the op-amp ("infinite input resistance") and why the inverting input is at 0V ("infinite open-loop gain and grounded non-inverting input").
Watch the negative sign: If the question asks for the "voltage gain", include the minus sign. If it asks for the "magnitude of the voltage gain", leave the minus sign off. If it asks for $VoutV_{\text{out}}$ , make sure it has the opposite sign to $VinV_{\text{in}}$ .
Always do a sanity check for saturation: If your calculation suggests $VV_{\text{out}} = 25\text{ V}$ but the question states the supply rails are $V\pm 12\text{ V}$ , your final answer for the output voltage must be $V+12\text{ V}$ or $V-12\text{ V}$ (saturated), not $V25\text{ V}$ .

Self review

Check yourself

Can you draw the schematic for an inverting amplifier from memory, making sure the input signal goes to the correct terminal?
What two ideal properties of an op-amp are used to prove that $Iin=IfI_{\text{in}} = I_f$ and that the inverting input is a virtual earth?
If $R_f$ is $kΩ100\text{ k}\Omega$ and $RinR_{\text{in}}$ is $kΩ20\text{ k}\Omega$ , and the input is $V-2\text{ V}$ , what will the output voltage be?

The Inverting Amplifier Circuit

The Magic of the "Virtual Earth"

Deriving the Gain Formula (Virtual-Earth Analysis)

Calculations in Practice

Review inverting amplifier configuration (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

inverting amplifier configuration (A-level only)

The Inverting Amplifier Circuit

The Magic of the "Virtual Earth"

Deriving the Gain Formula (Virtual-Earth Analysis)

Calculations in Practice

Review inverting amplifier configuration (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

inverting amplifier configuration (A-level only)

The Inverting Amplifier Circuit

The Magic of the "Virtual Earth"

Deriving the Gain Formula (Virtual-Earth Analysis)

Calculations in Practice

Review inverting amplifier configuration (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

inverting amplifier configuration (A-level only) Revision Guide

inverting amplifier configuration (A-level only)

The Inverting Amplifier Circuit

The Magic of the "Virtual Earth"

Deriving the Gain Formula (Virtual-Earth Analysis)

Calculations in Practice

Review inverting amplifier configuration (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

inverting amplifier configuration (A-level only) Revision Guide