Measure a Wheatstone Bridge Sensor Signal with an ADC

I received a message from one of my readers asking me to help with a Wheatstone bridge circuit. Since my response to him bounced back, and this being an interesting subject, I decided to write this article. Here is what he writes:

I found a circuit to condition the output of the Wheatstone bridge in the National ADC1205 datasheet, page 16. It uses an Op Amp configured as follows: V1 from the bridge thru 10K resistor to (–) input of Op Amp, 1.5Meg feedback resistor and Vout connects to the V- of the 5V ADC. V2 from the bridge connects directly to (+) input of the op amp and the V+ of the ADC.

The bridge V2-V1 is 0 mV to 30 mV. This is both at 5.000 V (0 mV) and V1 = 4.985 V, V2 = 5.015 V (30 mV). Please advise the equations to calculate how this works. Since the ADC is 5 V, I cannot see how the Vout can exceed that voltage. Is it true that Vout = 5.015 V when V1 = V2 = 5.015 V and ADC Out = 0 V?

The National ADC1205 is an obsolete component now, but the advice and application notes are still valid. We can use any ADC if we can correctly adjust Vref and the operating conditions. We can use an Arduino board as well, with its 10-bit ADC to achieve a complete system.

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How to Design a Circuit from its Transfer Function Graph

Sometimes all we know about a circuit is its transfer function graph.   The transfer function might look like the one in Figure 1.  How can we design a circuit so that its input-output behavior will match the graph?

Figure 1

The design starts with the mathematical form of the transfer function.  This is a linear function, with the general form of a first order polynomial function.

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The Virtual Ground

In my articles I talked about the op amp virtual ground and sometimes I wrote a brief explanation of this concept. In this article I will show you why an op amp input can be considered at a zero potential, without being galvanically connected to ground. Let’s take a simple circuit, the inverting amplifier.

inverting_amplifier_1Figure 1

In : How to Derive the Inverting Amplifier Transfer Function I showed the proof of its formula by using the virtual ground. The inverting input is at a zero potential, therefore virtual ground, which is a direct consequence of the feedback provided by R2 and the op amp high gain. Let’s see why.

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Derive the Transfer Function of the Common Collector Amplifier with Thevenin’s Theorem

How to Apply Thevenin’s Theorem for Solving Circuits with Dependent Sources

Besides its use to simplify and calculate currents in electrical circuits, Thevenin’s Theorem is also a great tool that we can use to derive transfer functions. This article will illustrate how to derive the small signal transfer function of the Common-Collector Amplifier with bipolar junction transistors (BJTs).

The circuit is shown in Figure 1. It is also called a repeater, so we expect that the calculated transfer function to be close to unity gain.


Figure 1

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How to Derive the Instrumentation Amplifier Transfer Function

The Instrumentation Amplifier (IA) resembles the differential amplifier, with the main difference that the inputs are buffered by two Op Amps.  Besides that, it is designed for low DC offset, low offset drift with temperature, low input bias currents and high common-mode rejection ratio.  These qualities make the IA very useful in analog circuit design, in precision applications and in sensor signal processing.


Figure 1

Figure 1 shows one of the most common configurations of the instrumentation amplifier.  Its clever design allows U1 and U2 operational amplifiers to share the current through the feedback resistors R5, R6 and RG.  Because of that, one single resistor change, RG, changes the instrumentation amplifier gain, as we will see further.  RG is called the “gain resistor”.  If the amplifier is integrated on a single monolithic chip, RG is usually left outside so that the user can change the gain as he wishes.  One example of such instrumentation amplifier is Texas Instruments’ INA128/INA129.

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