This circuit provides a voltage change as the resistance changes, allowing a microcontroller or other device to measure determine the resistance of the unknown element by reading the voltage through an ADC. In other words, if you know the values of three of the resistors, you can calculate the resistance of an unknown fourth resistor simply by measuring the voltage across the bridge. The simplest type of Wheatstone bridge balances two legs of a bridge circuit, one leg of which includes the unknown component. What is a Wheatstone Bridge?Ī Wheatstone bridge uses two balanced legs in a bridge circuit (i.e., two voltage dividers) to provide a link between the voltage across the bridge circuit and some unknown resistance of one resistor in the Wheatstone bridge. Note: There is some math in this article to help you understand how to determine the unknown resistance, but it is very simple! It might look like a bunch of formulas, but don’t tune out, as it’s the same formula broken down in different ways with the hope of making it easier to explain. The simplicity and effectiveness of a Wheatstone bridge makes it an incredibly powerful circuit, even if it does have a relatively niche application. Virtually every digital scale uses a Wheatstone bridge-based load cell, for example. Wheatstone bridges are one of those circuits that you might not feel like you’ve come across before, but you probably have inside some device or sensor you are using. While in some devices, there is an integrated circuit providing amplification of minute voltage changes, it’s also common to have direct access to the bridge, such as in a strain gauge/load cell. You’ll find Wheatstone bridge circuits in all kinds of compression and tension-based devices, such as air and fluid pressure sensors, strain gauges and more. Many sensor types use a Wheatstone bridge internally, as the resistance measured in the circuit can be linked back to some other phenomenon that causes the resistance of the sensor to change. In this article, we’ll take a look at Wheatstone bridge circuits, how they work and how we can effectively use them with modern electronics. Despite the simplicity of a Wheatstone bridge, it can be a challenge to make use of one effectively. So (150k / 50k) + 1 = 4.If you need to measure resistance precisely, a Wheatstone bridge is a simple circuit that provides a way to do so by taking a voltage measurement. Since R1c and R1d are in parallel, we get 50k. In the simulation you can see the opamp +IN swings from 0V to 1.25V, so it needs a gain of 4 to output 0V to 5V. Here is an example of how it should look (the a and b suffixes are just to keep SPICE happy): It needs (R1 + R2) for the feedback resistor. The summing amp you show has a problem though, the inverting gain resistors shown will not correct for the divider. The above non-inverting circuit is a bit like your summing amplifier. Here's a non-inverting method for reference also:Īnd the simulation (the "to_adc" is the output voltage): Note the input impedance is defined by R3, so you may need to increase this (and R2 by the same) or buffer if the source is high impedance. You will need a rail to rail output opamp. This can be replaced by a dedicated voltage reference if desired. The resistive divider supplies 1.25V to the non-inverting input. You'll need an RRIO (Rail-to-Rail I/O) opamp if you want to power if from a single 5 V supply. To make that 5 V out we have to amplify by 1.5, so R3 must be twice R4. If we have 2.5 V on the Vin input and with R2 = 2 \$\times\$R1 we get 3.33 V on the non-inverting input of the opamp. Next we have to find the amplification, which is determined by R3 and R4: So R1 and R2 form a voltage divider, and R2 should be twice R1 to get the 0 V. If we have -2.5 V on the Vin input the non-inverting input should be zero if you want 0 V out, regardless of the values of R3 and R4. You would think that we simply have to add 2.5 V, but do you have that? I'm assuming you have 5 V, so let's use that and see where it gets us. This is a non-inverting summing amplifier. But it's clear that this won't work here: a resistor adder always attenuates the signal, and we need a \$\times\$1 amplification. First thing to try is a simple resistor adder, without opamp.
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