# Wheatstone bridge current consumption

How would you calculate the total current consumption of a Wheatstone bridge made of four 350 Ω resistors, with a supply voltage of 1.8 V? And what is the theory behind this "resistor combination" calculation?

I don't know whether I can look upon a branch of the bridge as one resistor with a value of the sum of these two, and then calculate the total resistance as two parallel resistors.

• Where are you stuck? Aug 27, 2022 at 14:22
• Please edit your question to include a schematic, so we can be sure that you mean what we think you mean. Do you know Ohm's law, and how to calculate the resistance of a parallel/series combination of resistors? Aug 27, 2022 at 14:28
• I don't know whether I can assume branch of bridge as one resistor with value of sum of these two and then calculate total resistance as two paralel resistors. Aug 27, 2022 at 14:28
• There is nothing special about a Wheatstone bridge. It’s just four resistors, two series branches in parallel. The substitute resistance for the bridge is simply 350 ohms. So the whole bridge uses as much power as a single resistor would, but distributes this power across four resistors so they run cooler. Aug 27, 2022 at 16:54

what is the theory behind this "resistor combination" calculation?

## The Wheatstone bridge

Your circuit of {R+R}||{R+R} applies R=350 load across the differential input (2R||2R=R ) and thus biases the common mode voltage to V+/2 which is often ideal (but not always) for most differential amplifiers.

Normally in a full bridge, an unknown sensor has resistance at some known condition that matches the R values chosen in the bridge for greatest sensitivity near zero and linearity for changes. When power is a major concern, a constant current of lower constant voltages may be used.

At the same time it applies a differential load across the power supply of 2R||2R=R. So power is trivial to calculate. $$\P=V^2/R\$$

To minimize load regulation errors your differential amplifier, if used should use R values > 100 x R value or an INstrument Amp (INA) and can be 1Gohm for each input.

If you want to calculate the supply current, you must:

1. calculate the effective resistance between the positive and the negative supply. The rules for parallel and series resistors are useful.
2. use Ohm's law to calculate the current.

Typically there is minimal load on the sense pins of a wheatstone bridge. If your questions states nothing else, it is fair to assume that the schematic is just as drawn by you, with nothing else drawing current at the sense nodes.

Well, we have the following circuit:

simulate this circuit – Schematic created using CircuitLab

The input resistance can be found:

$$\text{R}_\text{in}=\frac{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4\right)}{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}\tag1$$

Now, the input current is given by:

$$\text{I}_\text{in}=\frac{\text{V}_\text{in}}{\text{R}_\text{in}}=\text{V}_\text{in}\cdot\frac{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4\right)}\tag2$$

So, the input power is given by:

$$\text{P}_\text{in}=\text{V}_\text{in}\text{I}_\text{in}=\text{V}_\text{in}^2\cdot\frac{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4\right)}\tag3$$

When all the resistors are the same value, we get:

$$\text{P}_\text{in}=\text{V}_\text{in}^2\cdot\frac{1}{\text{R}}\tag4$$

So, the total energy consumption of this circuit is given by:

$$\text{E}_\text{in}\left(t\right)=\int\limits_0^t\text{P}_\text{in}\space\text{d}\tau=\text{V}_\text{in}^2\cdot\frac{t}{\text{R}}\tag5$$