# Circuit VR: The Wheatstone Bridge Analog Computer We’re always a fan of something so simple that can actually be so complicated. For example, what do you think goes into an analog computer? Of course, a “real” analog computer has opamps that can do logarithms, square roots, multiplication and division. But will it surprise you that you can make an analog device like a chip base using a Wheatstone bridge – they are basically two voltage dividers. You don’t even need any active hardware at all. It’s an old idea that has been popping up in e-magazines every now and then. I’ll show you how it works and simulate the device so you don’t have to build it unless you just want to.

The voltage divider is one of the easiest circuits in the world to analyze. Consider the presence of two resistors, Ra and Rb, respectively. The voltage comes in at the top of Ra and grounds at the bottom of Rb. The node connecting Ra and Rb – let’s call it Z – is what we will consider the output.

Suppose we have a battery powered by a voltage of 10 volts and an ideal voltmeter that does not load the circuit connected to Z. According to Kirchhoff’s current law, we know that the current through Ra and Rb must be the same. After all, there’s nowhere else to go. We also know that the voltage drop across Ra plus the voltage drop across Rb must equal 10 V. Kirchoff, energy conservation, whatever you want to call it. Let’s call these quantities I, Va and Vb.

There are many ways to go from here, but let’s accept that the current through series resistors will be as if it were a single resistor of equal value. That is, a 1 KΩ and 2 KΩ resistor in series will draw current as a 3 KΩ resistor. This means that Ohm’s law tells us:

`I = 10/(Ra+Rb)`

Now you can solve for each voltage drop:

```Va = I Ra

Vb = I Rb```

In fact, our voltmeter will measure at Z Vb because it is grounded.

## hair big deal

Of course, you probably know about voltage dividers. But we were talking about Wheatstone Bridges. The truth is that these two are just dividing the voltage in parallel and you are measuring the voltage between the two outputs (call them Z1 and Z2). You often see this circle drawn like a diamond, but don’t let that fool you. There are still only two voltage dividers.

Without using any math, you can see that if the voltage dividers are the same, then Z1 and Z2 will be the same, so no current will flow because the voltage between the two points is zero. What happens when the septum is different? There will be more voltage on one Z point than on the other.

Historically, this was used to measure resistance. You can use two identical resistors on part of the bridge, that have an unknown resistance in one of the remaining legs and a variable resistor with a dial calibrated to read the ohms. You turn the dial until the counter reads zero and reads the resistance value from the dial. If the power source is alternating current, you can also measure the reaction using a similar electrical circuit.

## But the slide rule?

So how do you go from a piece of antique test equipment to a slide base? Let’s change the bridge so that the left rail has two resistors Ra and Rb while the other one has Rc and Rd. We can look at algebra:

```Z1=V (Rb/(Ra + Rb))

Z2=V (Rd/(Rc + Rd))```

We want Z1 to equal Z2 so:

`V (Rb/(Ra + Rb)) = V (Rd/(Rc + Rd))`

We can divide both sides by V and get rid of this term:

`(Rb/(Ra + Rb)) = (Rd/(Rc + Rd))`

So to balance the bridge we need:

```(Ra + Rb)/Rb  = (Rc + Rd)/Rd    reciprocal both sides

(Ra Rd + Rb Rd) = (Rc Rb + Rb Rd)    multiply both sides by Rb Rd

Ra Rd = Rc Rb      subtract Rb Rd from both sides

Ra = (Rb Rc)/Rd    Solve for Ra```

As a simple thought experiment, imagine that path = 1. If you set Rb and Rc, you can set Ra to balance and the value of Ra will be the answer. Or you can set Rb to 1 and enter the numbers in Rc and Rd. Once you balance Ra, you will know the result of the division.

But in practice, you may want to measure the result, especially for division. For example, if Rb = 1, Rc = 2, and Rd = 1000, you will need to set A to .002 ohms which is very difficult to do. In this case, though, you can set Rb to a scale factor. If it is, say, 10 k, then Ra can be set to 20 ohms.

## simulation

You can crack a few potentiometers and do just that. We like to suggest linear unless you are very useful in making logarithmic disks. But since this is Circuit VR, we prefer a simulation. Falstad fits the bill, but any simulator is well suited for the task.

There are two switches in the simulation. The upper ‘C’ switch allows you to toggle in the upper resistor or range resistor 10x, 100X, or 1000X for C. The lower ‘D’ switch allows you to select the 1 ohm resistor or variable resistor for D. The center shows bridge balance and will read 0A when you are in equilibrium .

Speaking of variable resistors, I have placed sliders for each of the resistors on the right sidebar of the simulator. However, using them often puts values ​​like 10.002K which reads 10K on the screen and is a source of error. Of course, you will have the same problem with real pots, so this is probably a good simulation. However, it is better to double-click on the resistor you want to change and directly enter its value. Obviously, the three fixed C resistors or the fixed D resistor should not be changed.