The Bridge and Amplifier Budget

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

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.

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.