**Figure 1**

Widely used in Analog Design, the inverting amplifier in Figure 1 has a simple transfer function.

What is the proof of this function?

**Figure 1**

Widely used in Analog Design, the inverting amplifier in Figure 1 has a simple transfer function.

What is the proof of this function?

The Instrumentation Amplifier (IA) resembles the differential amplifier, with the main difference that the inputs are buffered by two Op Amps. Besides that, it is designed for low DC offset, low offset drift with temperature, low input bias currents and high common-mode rejection ratio. These qualities make the IA very useful in analog circuit design, in precision applications and in sensor signal processing.

**Figure 1**

Figure 1 shows one of the most common configurations of the instrumentation amplifier. Its clever design allows U1 and U2 operational amplifiers to share the current through the feedback resistors R5, R6 and RG. Because of that, one single resistor change, RG, changes the instrumentation amplifier gain, as we will see further. RG is called the “gain resistor”. If the amplifier is integrated on a single monolithic chip, RG is usually left outside so that the user can change the gain as he wishes. One example of such instrumentation amplifier is Texas Instruments’ INA128/INA129.

One of the most common amplifiers in Analog Design is the non-inverting amplifier.

**Figure 1**

Its transfer function is

_{} |
(1) |

How do you derive this function?

My friends advised me that it would be helpful to have on this site the most common operational amplifier configurations and transfer functions or formulas. So, here they are. This article is not just a simple collection of circuits and formulas. It also has links to the transfer function proof for these circuits so I hope it will be very helpful. Make sure you post a comment and let me know how I can improve this page. This article will be updated, so do check it often.

**Note**: The proof of this transfer function can be found here: How to Derive the Non-Inverting Amplifier Transfer Function.

**Note**: This configuration can be considered a subset of the Non-inverting Amplifier. When Rf2 is zero and Rf1 is infinity, the Non-inverting Amplifier becomes a voltage follower. When a resistor has an infinity value, in practice it means it is disconnected.

In a previous article, How to Derive the Summing Amplifier Transfer Function, I deduced the formula for the non-inverting summing amplifier with two signals in its input. But what if we have 3, 4 or an ** n** number of signals? Can we add them all with one amplifier?

Theoretically, yes. Practically, it is a different story. There is a practical limit on how many signals can be summed up with one amplifier. When the number of input signals grows, each signal component in the sum decreases in value. By the end of this article you will understand why.

**Figure 1**

We already saw that, for a summing amplifier with two input signals (Figure 1), the transfer function is

_{} |
(1) |

If we need to add 3 signals, the circuit schematic looks like the one in Figure 2. What is the transfer function of this summing amplifier with 3 inputs?

The transfer function of the differential amplifier, also known as difference amplifier, can be found in articles, websites, formula tables, but where is it coming from? Why is the differential amplifier transfer function as in the following mathematical relation?

(1) |

where the resistors are those shown in Figure 1.

First, an important remark: This formula applies only for an ideal operational amplifier. This means that the amplifier has a large gain, so large that it can be considered infinity, and the input offset sufficiently small, so that it can be considered zero. Also, the input bias currents are sufficiently small so that they can be considered zero. I was once asked “but what is sufficiently small?” A voltage or current in electronics is considered sufficiently small, when its numerical value is 1/100 or less versus the dominant voltages or currents in the circuit. For example, if the input voltage levels, in the circuit in Figure 1, are around a few volts, and the operational amplifier input offset is millivolts or sub-millivolts, then we can neglect the input offset and consider it zero.

Having said that, do we need to know this formula by heart? Of course not. All we need to know is how to derive it. This is not difficult at all.

The transfer function can be derived with the help of the Superposition Theorem. This theorem says that the effect of all sources in a linear circuit is the algebraic sum of all of the effects of each source taken separately, in the same circuit. In other words (back at Figure 1), if we remove V1, and replace it with a short circuit to ground and calculate the output voltage, and then we do the same with V2, the output voltage of the differential amplifier is the sum of both output voltages as they were calculated with each source separately.

Let’s first remove V1. R1 cannot be left unconnected, because in the initial circuit there was current flowing through it. Therefore, let’s ground R1 (see Figure 2).

We can see that our amplifier becomes an inverter, which has its non-inverting input connected to ground through R1 and R2. Vout2 is given in equation (2).

_{} |
(2) |

Read MasteringElectronicsDesign.com: How to Derive the Inverting Amplifier Transfer Function for a proof of this function.

Now let’s remove V2 and ground R3 (see Figure 3).

This is a non-inverting amplifier. For an ideal operational amplifier, Vout1 is a function of V, which is the voltage referred to ground at the non-inverting input of the operational amplifier.

_{} |
(3) |

The resistors R1 and R2 are an attenuator for V1, so that V can be determined as in the following relation.

_{} |
(4) |

By replacing V in equation (3), Vout1 becomes:

_{} |
(5) |

Now that we have Vout1 and Vout2, and using the Superposition Theorem, Vout is the algebraic sum of Vout1 and Vout2,

_{} |
(6) |

which is the differential amplifier transfer function. (**Q.E.D.**)

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