Useful Operational Amplifier Formulas and Configurations

Non-inverting Amplifier

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

Voltage Follower

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.

How to Derive the Transfer Function of the Inverting Summing Amplifier

The inverting summing amplifier does exactly what its name says: adds the input signals and inverts the result.  This amplifier presents a major advantage versus the non-inverting summing amplifier.  The input signals are added with their own gain.  The disadvantage is the inversion of the sum, which might not be desirable in some cases.

Figure 1

Figure 1 shows the non-inverting summing amplifier with two inputs.  Its transfer function is shown in equation (1).

 (1)

As you can see, this is a simple function. Each signal is added with its own gain created by the feedback resistor, Rf, and the corresponding resistor for that signal.  But, why is that?  Why is this transfer function a lot simpler than the non-inverting summing amplifier?  How can we derive this function?  What is the transfer function of the inverting summing amplifier with 3, 4, or n inputs?  This article answers all these questions.

The Transfer Function of the Non-Inverting Summing Amplifier with “N” Input Signals

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?

How to Derive the Summing Amplifier Transfer Function

The summing amplifier, or the non-inverting summing amplifier, is an analog processing circuit with the transfer function (the summing amplifier formula as some say) shown in the following equation.

 (1)

The first term of the product is the actual summing, while the second term is a gain due to the R3 and R4 resistors.  I prefer this type of summing amplifier as shown in Figure 1, because it is more flexible and allows us to achieve any linear function we want.

Figure 1

Some authors prefer the following schematic,

Figure 2

with the transfer function

 (2)

One can see that the summing amplifier in Figure 2 is a subset of my preferred schematic in Figure 1.  In Figure 2, R4 is zero, while R3 is infinity (open connection).  It performs the analog summation between V1 and V2, with a gain of 1.  Therefore, the amplifier in Figure 1 gives us more choices when designing a function with this circuit.  If the gain is not needed, this should come up from calculations, as in this article Solving the Summing Amplifier.

If you followed this website, by now you probably figured that I am not a promoter of learning formulas by heart.  I like to derive the transfer function if I need it. So, how do we prove this formula?

We will use the Superposition Theorem, which says that, the effect of all the sources in a circuit is equal with the sum of the effects of each source taken separately in the same circuit.  Therefore, if we take out one source, V2, and replace it with a wire, we then can find the voltage in each node and the current in each branch of this circuit due to the remaining source V1.  Then we do the same with V1 and then sum up the currents on each branch and the voltage levels on each node.  We are only interested in Vout, so this should be simple.

We will first make V2 = 0V, by connecting R2 to ground, as in Figure 3.

Figure 3

The Op Amp is considered an ideal component, so that the input bias currents are negligible.  If the current in the non-inverting input is zero, R1 and R2 make a voltage divider for V1.  The non-inverting input voltage V1n, can be written as

 (3)

and, based on the non-inverting amplifier transfer function, Vout1 is

 (4)

By replacing V1n in (4), the output voltage is

 (5)

In the second part of my demonstration, based on the Superposition Theorem, R2 is connected back to V2 and V1 = 0, by connecting R1 to ground.  Following the same train of thought Vout2 can be written as

 (6)

Now we have to add Vout1 to Vout2 to complete the third step of the Superposition Theorem.  After factorizing the gain component 1+R4/R3, the summing amplifier transfer function becomes the mathematical relation shown in (7).

 (7)

Q.E.D.

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This formula shows that this sum is a weighted sum between V1 and V2.  This is better than a direct sum V1 plus V2, because, again, brings flexibility in design.  Together with the differential amplifier, this circuit brings another treat in the art of electronics design.

How to Derive the Differential Amplifier Transfer Function

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.

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).

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).

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.)

Solving the Differential Amplifier – Part 1

Design a Differential Amplifier Based on the Input and Output Voltage Level Requirements

The differential amplifier, also known as the difference amplifier, is a universal linear processing circuit in the analog domain.  Why?  Because you can achieve any linear transfer function with it.  It can be reduced to a simple inverter, a voltage follower or a gain circuit.  It can also be transformed in a summing amplifier.

Figure 1

So, what is the common usage of the differential amplifier in Figure 1?  When the resistor ratios are equal

the amplifier transfer function is

and the circuit amplifies the difference between the input signals.

However, there are times when the electronics designer is faced with the following design requirements:  Given an input range of, say, -0.5V to 5.5V, the output has to swing between, say -1.25V and +2.365V.  It is clear that this requires an amplifier with a certain gain and an offset different than zero.  How can we design the differential amplifier to achieve such a function?