The Bridge and Amplifier Budget

The Bridge and Amplifier Budget

Ada re-derives this chapter’s own numbers step by step, at full precision

foundations
math-foundations
calculation-audit
sensors
Ada ADA · CALCULATION AUDIT

The Bridge and Amplifier Budget

Four calculations, ~4 minutes

Bridge numbers are small enough that rounding can hide a design mistake. Keep the physics path visible: strain changes resistance, the bridge turns that into millivolts, and the amplifier must make it readable without clipping.

A foil strain gauge under 1000 microstrain with gauge factor 2.0 and 5.0 V excitation produces just 2.5 mV out of the bridge — too small for a 3.3 V ADC, so the chapter amplifies it with an AD620 at gain 100 to 250 mV. The same gauge can see 1500 microstrain in an overload test, where gain 1000 would push the 3.75 mV raw signal to 3.75 V and clip that same 3.3 V ADC. This audit asks the question that overload case invites: does the chapter’s bridge, gain-resistor, and headroom arithmetic actually hold up end to end, digit by digit?

Companion to the chapter Bridge Linearity and Instrumentation — every number here comes from that chapter.

See the relationship before changing it

The figure reads from left to right. The blue card is amplifier gain. The middle card applies this page's rule. The green card is amplified bridge output. Walk the arrows once: set the input, apply the rule, then read the result with its unit.

The retained audit below checks several chapter fixtures. This model keeps those stated values fixed and changes only amplifier gain, so the numeric fixture does not switch without explanation.

Amplifier gain changes amplified bridge output An input card leads through the rule amplified output = 2.5 mV x gain to the amplified bridge output result. INPUT PAGE INPUT APPLY THE RULE predict calculate check units OUTPUT RESULT
Walk the arrows. More gain enlarges the strain signal but can consume ADC headroom during overload.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline is 100 times.

  2. 2

    Name the relationship. amplified output = 2.5 mV x gain

  3. 3

    Substitute with units. 2.5 mV x 100 = 250.0 mV

  4. 4

    Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.

Predict, then change amplifier gain

Try Predict the direction of amplified output = 2.5 mV x gain. Test another amplifier gain, then compare amplified bridge output.

100 times
Chapter baseline
Amplified bridge output

Observe More gain enlarges the strain signal but can consume ADC headroom during overload. Reset amplifier gain to 100 and compare amplified bridge output.

Explain More gain enlarges the strain signal but can consume ADC headroom during overload.

Check yourself

What should you do before trusting a moved-control result?
Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.
What does this small model leave out?
Answer: Only amplifier gain moves here. Field effects named in the technical boundary stay fixed.
Try

Apply GF = 2.0, strain = 0.001, and Vex = 5.0 V, then use Check audit after sizing RG for gain 100.

Observe

Bridge output reads 2.5 mV and RG approximately 499 ohm; the amplified signal becomes 250 mV while rejected common-mode residue stays near 0.1 mV.

Explain

The quarter bridge scales fractional resistance change by 5 V excitation, then gain 100 lifts the differential signal while CMRR suppresses shared noise.

1. Convert strain into bridge voltage

The chapter’s quarter-bridge approximation and foil-gauge example use GF = 2.0, ε = 1000 µε = 0.001, and Vex = 5.0 V.

ΔR/R = GF x ε = 2.0 x 0.001 = 0.002
Vout = Vex x (ΔR/R) / 4 = 5.0 x 0.002 / 4 = 0.0025 V = 2.5 mV
Check Arithmetic Result
Raw sensor movement 2.0 x 0.001 0.002, or 0.2% resistance change
Bridge output 5.0 x 0.002 / 4 0.0025 V = 2.5 mV
Gain 100 output 2.5 mV x 100 250 mV
12-bit, 3.3 V ADC scale 3.3 V / 4096 = 0.8057 mV/count; 250 / 0.8057 about 310 counts, matching the chapter’s readout estimate

2. Check the gain resistor

The AD620 relation in the chapter is G = 1 + 49.4 kΩ / RG. Rearranging for G = 100 gives:

RG = 49.4 kΩ / (100 - 1) = 49,400 Ω / 99 = 498.99 Ω ≈ 499 Ω

3. Test overload headroom before committing

The same formula with 1500 µε = 0.0015 gives ΔR/R = 2.0 x 0.0015 = 0.003, so 5.0 x 0.003 / 4 = 0.00375 V = 3.75 mV. Gain 100 outputs 375 mV; gain 1000 outputs 3.75 V, which exceeds a 3.3 V ADC range.

4. Translate common-mode rejection into a voltage check. A 100 dB common-mode rejection ratio is a voltage factor of 10^(100/20) = 100,000. So the chapter’s 100 mV shared motor noise becomes 100 mV / 100,000 = 1 µV equivalent input. After gain 100, that is 0.1 mV at the output, while the wanted 2 mV differential signal becomes 200 mV.

Every number above is taken from the chapter’s own bridge and amplifier examples and re-derived step by step.

Technical boundaries: The small-signal bridge and ideal CMRR equations omit resistor tolerance, bridge nonlinearity, amplifier offset and noise, finite bandwidth, supply headroom, temperature drift, and input common-mode limits.

Ready: work the ledger before checking it.