## Design a Bipolar to Unipolar Converter to Drive an ADC

Most Analog to Digital Converters have a unipolar input that can be a problem when designing bipolar circuits.  Some common ADC input voltage ranges are 0 to 2.5 V, or 0 to 5 V.  However, the analog circuit that drives the ADC can have voltage swings of, –1 V to +1 V, –2 V to +2 V , –5 V to +5 V, and so on.  Bringing the ADC input below ground is a big No-No, because the current from input will flow through the chip substrate creating irreversible changes in the ADC and damage it.  So, how do we connect a bipolar front end circuit with a unipolar ADC?  Enters the bipolar to unipolar converter.  Let’s design one.

The converter can be designed with a summing amplifier, as in Figure 1.  How do we calculate the resistors?

Figure 1

## Differential Output Circuit

One of my readers asked me to explain how I designed a circuit I posted in a forum, as a solution to one of the member’s question.  The problem was about designing a circuit with 3 input signals, VA, VB and VCM.  The circuit had to output the sum and difference between VCM and the average of VA and VB as in the following expressions:

 (1)

The solution I posted is the circuit in Figure 1.

Figure 1

What is the easiest way to design this circuit?

## 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.

>>>  <<<

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.

## Converting a Differential Amplifier into a Summing Amplifier

Is there any link between a differential amplifier and a summing amplifier? Yes, it is.  They can be easily converted one into the other one.  While this article shows the conversion, the main purpose is to demonstrate how the same circuit can be viewed as a differential amplifier or as a summing amplifier, depending on the voltage levels in its inputs.

The differential amplifier is shown in Figure 1.  It is mostly used to measure the voltage difference between its inputs.  In addition, it can be used to achieve a linear function Vout(Vin) (Vout of Vin) where Vout is the voltage at the amplifier output, and Vin is the input voltage.  In this case, any one of the inputs V1 or V2 can play the role of Vin.  For details, read my articles Solving the Differential Amplifier Part 1, Part 2, and Part 3.

Figure 1

What is interesting about the differential amplifier, which I call a universal amplifier, is that it can be easily converted into a summing amplifier.  In fact, they are one and the same circuit.  In Figure 2 I drew the summing amplifier, where I noted the voltage levels at its inputs with Vs1 and Vs2.  If you would like to read more about the summing amplifier go to this link: Solving the Summing Amplifier.

Figure 2

The differential amplifier can be seen as a summing amplifier, if we look at it differently.  If, in Figure 1, we take R2 and move it on the left side of the OpAmp, as in Figure 2, without affecting its connections, the differential amplifier looks like a summing amplifier.  The only difference is that, the differential amplifier sums V1 with a zero voltage, because R2 is connected to ground.  Also, R3 is connected to V2 instead of being connected to ground.

Therefore, to convert the differential amplifier into a summing amplifier, all we have to do is to adjust the voltage levels as follows:

1. Disconnect R2 from ground and connect it to Vs2,
2. Connect R3 to ground.

Of course, these steps may be done backwards, if we need to convert a summing amplifier into a differential amplifier.

What about the transfer function of the summing amplifier?  Can we easily derive it as well from a differential amplifier transfer function?  Of course we can, and here is how.

The differential amplifier transfer function is shown in (1).

 (1)

As I already said, the differential amplifier is a summing amplifier between V1 and zero volts, and with V2 = 0.  Let’s rewrite the transfer function (1) as follows:

 (2)

Why did I add 0V at V1 times the resistor ratios?  Because, as I explained in The Differential Amplifier Transfer Function, any voltage on the non-inverting input multiplies the R4, R3 ratio plus one.

Now, if we connect R1 to Vs1 and connect R2 to Vs2, the transfer function becomes

 (3)

where by k I noted the resistor ratio for Vs2.  Using the Superposition Theorem, we realize that Vs2 “sees” a voltage divider made by R1 and R2, with the ratio of

 (4)

For details on how to determine this ratio, read my article How to Derive the Summing Amplifier Transfer Function.

Replacing k in equation (3), we arrive at the summing amplifier transfer function as shown in equation (5).

 (5)

Q. E. D.

## Solving the Summing Amplifier

### How to Design a Summing Amplifier Based on the Input and Output Voltage Level Requirements

In allaboutcircuits.com forum a member asked how can he drive a MOSFET that needs a voltage range of 4V to 5V with a DAC with the output range of 0V to 5V?

Initially I thought he should use a differential amplifier.  However, based on the articles I published, MasteringElectronicsDesign.com: Solving the Differential Amplifier – Part 1, Part 2 and Part3 the solution based on a differential amplifier would require a negative voltage level in the input.   Although V1 can be the input from 0V to 5V, V2 has to be negative, so that the output shifts to positive values.

Then I thought of the Summing Amplifier, or the Non-Inverting Summing Amplifier, which is shown in Figure 1.  It is called a summing amplifier, because two signals are summed in one of the amplifier inputs.  In this case, V1 and V2 are summed in the non-inverting input.

Figure 1

The summing of V1 and V2 is not direct.  Resistors R1 and R2 make a weighted sum and this is what makes this amplifier very useful.  As in the case with the differential amplifier, one can use this circuit to achieve any linear function.  This article shows you how to design a summing amplifier based on the input and output requirements.  You can also solve your amplifier with the calculator I posted here:  MasteringElectronicsDesign.com: Summing Amplifier Calculator.

The transfer function of the summing amplifier is as follows.

 (1)

You can find its demonstration in this article, MasteringElectronicsDesign.com: How to Derive the Summing Amplifier Transfer Function.

Let’s write down what we know:

If Vin1 = 0V then Vout1 = 4V and
If Vin2 = 5V then Vout2 = 5V,

where by Vin1 and Vin2 I noted the input range limits, and by Vout1 and Vout2 I noted the output range limits.

Let’s choose one of the summing amplifier inputs to be Vin, say V1.

Because we have two instances that we know, Vin1 and Vin2 and the corresponding outputs, Vout1 and Vout2, let’s rewrite equation (1) using these two instances.

 (2)

This is a linear system of two equations with a lot of unknowns: R1, R2, R3, R4 and V2.  However, we can simplify our life by grouping the resistors in ratios.  The equations can be rewritten like this,

 (3)

where by k1 and k2 I noted:

 (4)

Now we are left with three unknowns, k1, k2, V2.  I can simply consider V2 as a known value, because I can connect to R2 any voltage I want or, more conveniently, a voltage that I already have in the circuit.  I will target for V2 = 5V, since there is already a DAC in this circuit with an output range of 0 to 5V.  So I can assume there is a 5V reference in this circuit.

If k1 and k2 are the remaining unknowns, then (3) is a system of two equations and two unknowns as in (5), which can be easily solved.

 (5)

It can be easily seen that the second equation becomes

 (6)

so k2 has to be zero.

The result is k1 = 1/4 and k2 = 0.

This result shows that we do not need the resistors R3 and R4. Also, the ratio between R2 and R1 is 1/4.  We can choose R2 = 1 kohm and a standard value for R1 = 4.02 kohm with a 1% tolerance.

The final circuit is shown in Figure 2.

Figure 2

Since k2 is zero, R3 is zero, configuring U1 as a repeater for the summed voltage in the non inverting input.

'