## Design a Unipolar to Bipolar Converter the Easy Way with Microsoft Mathematics

Many analog circuits can be calculated with simple algebra. This may involve an equation or a system of equations, but the calculations are quite simple. Take the differential amplifier, as an example. In a previous article, MasteringElectronicsDesign: Design a Differential Amplifier the Easy Way with Mathcad, I showed how to design the differential amplifier by solving a system of two equations with two unknowns using Mathcad. Since then, readers asked me if there is any other substitute for Mathcad that they can use to solve the system of equations. And the answer is, yes, there is one.

Microsoft Mathematics is a free application which is loaded with features. Besides its graphing, math formulas and units converter, it has an equation solver that can easily handle systems of equations. By changing a few values and letting the application calculate the unknowns, a user can tweak his circuit to match the design requirements.

## How to Design a Circuit from its Transfer Function Graph

Sometimes all we know about a circuit is its transfer function graph.   The transfer function might look like the one in Figure 1.  How can we design a circuit so that its input-output behavior will match the graph?

Figure 1

The design starts with the mathematical form of the transfer function.  This is a linear function, with the general form of a first order polynomial function.

## Differential Amplifier Calculator

### Unipolar to Bipolar Converter Example

If you need to design a differential amplifier, here is a handy calculator. All you need to define are the input range, the output range and a choice of voltage reference.

The differential amplifier was explained in different articles on this website. Solving the Differential Amplifier – Part 1, Part 2 and Part 3 shows a numerical example and how to design such an amplifier. Also, the common mode voltage level and the common mode output error were explained in the series of articles The Differential Amplifier Common-Mode Error – Part 1 and Part 2.

Enter the input range, Vin1 to Vin2, the output range, Vout1 to Vout2 and a reference voltage Vref. You need to choose two resistors, R2 and R3. The calculator will compute R1 and R4.

## 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.)

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