Three resistor voltage divider

Question

I'm looking at a 'range expander' for a data acquisition system. It basically functions as a voltage divider. I don't understand why there's effectively two voltage dividers per line in the resistor network. It almost looks like an unbalanced Wheatstone bridge.

Is there some advantage over a simple 2-resistor divider?

Here's the schematic:

Here's the product in question.

EDIT: How is this functionally different than the following circuit?

Answer 1

Your link provides more detail, which helps. So I'd like to drag into the discussion the fuller image from your link. But I'd also like to add a link to a PDF where that image is just one page of many. Here's the image:

And here is the User's Manual PDF link.

To better understand the table provided in the above image, I'll introduce a small program I wrote some time ago. Each of the SLICE inputs, when "measuring large differential voltages," are now subjected to the following 3-resistor divider model:

(Ignore "Button4." I was playing with it some years ago and didn't remove it.)

As the above shows, you have to supply 7 of the 9 options, leaving the remaining two for calculation. It's vital that you understand this fact. There must be two variables left for computation. You cannot over-specify by requiring 8 and calculating the remaining 1. That does not work. You cannot under-specify, either. That also doesn't work. It is 2 and only 2 that must be left to calculation.

In this case, we can use the above table. Let's take the first entry of that table and plug things into the above program. I don't know what type of resistors are used, so for now let's plug in 2% and see where that takes us:

Before drilling in more, let's just look at the left-hand side where the two calculated output values are generated. Here you find that the max is \$4.0938069\:\text{V}\$ and that the min is \$450.81967\:\text{mV}\$. So we find \$\frac{4.0938069\:\text{V}-450.81967\:\text{mV}}{20\:\text{V}-\left(-20\:\text{V}\right)}=0.0910746807\:\frac{\text{V}}{\text{V}}=91.0746807\:\frac{\text{mV}}{\text{V}}\$. Note that this is an exact match for the added table entry in the above SLICE diagram.

Now, you can also see what the arrangement is doing for you. It is taking a wide range and compressing it into a smaller one.

There is a horrible consequence, though. In using resistor dividers like this, which have initial accuracy tolerances to them (not to mention long term drift, temperature, etc), the right-hand side of the report tells us that when using 2% resistors the maximim has a spread of over \$147\:\text{mV}\$ and that the minimum has a spread of over \$167\:\text{mV}\$.

(In computing the report, my program takes into account every possible variation and calculates the entire multi-dimensional hyper-rectangle and reports the worst-case span.)

That situation is the same for both sides of the differential input. While this may mean the differential could be twice as bad, it also means the probabilities will favor a more consistent differential spread, too. Either way, though, that whole situation may not be acceptable and you may need to use the better resistors possible to any given situation, as their tolerances make a significant difference (unless you also intend on crafted calibration steps for your situation.)

Answer 2

If you look carefully, each input consists of a circuit like this:

^{simulate this circuit – Schematic created using CircuitLab}

On the right is the reduced version, after converting the components in the blue box to their Thevenin equivalent. From there it's easier to derive the relationship between the input signal \$V_{IN}\$ and output \$V_{OUT}\$:

$$ \begin{aligned} V_{OUT} &= (V_{IN}-2.5V)\frac{5k\Omega}{R+5k\Omega} + 2.5V \\ \\ &= V_{IN} \frac{5k\Omega}{R+5k\Omega} + 2.5V(1-\frac{5k\Omega}{R+5k\Omega}) \end{aligned} $$

It's clear that this configuration performs two functions:

• Scales down the input by a factor of \$ \frac{5k\Omega}{R+5k\Omega} \$

• Adds an offset of \$ 2.5V \times (1-\frac{5k\Omega}{R+5k\Omega}) \$

Both the offset and attenuation help bring the input signal to within the acceptable input range of the differential amplifier that follows. If, for example, that amplifier (without any input conditioning) accepts input voltages from \$V_{MIN}\$ to \$V_{MAX}\$, then with this network in place the new range can be calculated by setting \$V_{OUT}\$ to those values, and finding the corresponding \$V_{IN}\$:

$$ \begin{aligned} V_{IN} &= (V_{OUT} - 2.5)\frac{R+5k\Omega}{5k\Omega} + 2.5 \\ \\ V_{IN(MIN)} &= (V_{MIN} - 2.5)\frac{R+5k\Omega}{5k\Omega} + 2.5 \\ \\ V_{IN(MAX)} &= (V_{MAX} - 2.5)\frac{R+5k\Omega}{5k\Omega} + 2.5 \\ \\ \end{aligned} $$

The difference between these two values will be the maximum allowable peak-to-peak amplitude of the input signal, with the extender network in place:

$$ \begin{aligned} V_{IN(PP)} &= V_{IN(MAX)} - V_{IN(MIN)} \\ \\ &= (V_{MAX} - 2.5)\frac{R+5k\Omega}{5k\Omega} + 2.5 - (V_{MIN} - 2.5)\frac{R+5k\Omega}{5k\Omega} + 2.5 \\ \\ &= (V_{MAX} - V_{MIN}) \frac{R+5k\Omega}{5k\Omega} \end{aligned} $$

What this means is that where originally (with no extender) the maximum input amplitude was \$ V_{MAX} - V_{MIN} \$ peak to peak, now you can supply input voltages a factor of \$\frac{R+5k\Omega}{5k\Omega}\$ greater than that.

Real numbers might be more meaningful. For \$R=10k\Omega\$, \$V_{MIN} = 0V\$ and \$V_{MAX} = +5V\$:

$$ V_{IN(PP)} = (V_{MAX} - V_{MIN}) \frac{R+5k\Omega}{5k\Omega} = 15V_{PP}$$

The input must lie between:

$$ \begin{aligned} V_{IN(MIN)} &= (V_{MIN} - 2.5)\frac{R+5k\Omega}{5k\Omega} + 2.5 = -5V\\ \\ V_{IN(MAX)} &= (V_{MAX} - 2.5)\frac{R+5k\Omega}{5k\Omega} + 2.5 = +10V\\ \\ \end{aligned} $$

Here's a graph of what the input \$V_{IN}\$ (blue) might look like, and what the differential amplifier would see at one of its own inputs, \$V_{OUT}\$ (orange):

The differential amplifier has two such inputs, of course. With the same extender network at each input, what we are doing here is conditioning them both individually, but equally. As long as both inputs stay within \$-5V < V < +10V\$, then the amplifier will never see a voltage at either of its inputs outside of \$0V < V < +5V\$, and we never violate its own input constraints.

Answer 3

At first yes using 3 resistors is important to this circuit. To understand why you should use this circuit you should know first that your op amp have range of input voltage any voltage more than that range can damage the input or make the op amp can't know how much is that voltage.

To understand why to use two resistor divider instead of one like what you use in your circuit is your circuit make the input voltage float.

The input voltage has no connection between it and voltage rails only it connected to op amp inputs and some op amps have JFET input (no current go to input) your op amp will damaged in second so you need to have a resistor between every input and one of voltage rail.

The reason why you should use 3 resistor not 2 is to make voltage stay in input voltage range as you can. If you use 2 resistors devider at every input when you put some voltage at input one input will get positive voltage to ground and the other input will get negative voltage below ground wich will make op amp input range problem as I mentioned earlier.

The 3 resistors voltage devider adds some offset voltage to two inputs will get 2.5 voltage and if you put 2 two volt so when there's no voltage at input of circuit both input of op amp for example you will have on input at 3 volt and the author one at 2 volt.