Amplify a small sensor signal

Use a non-inverting amplifier to scale a 50 mV sensor signal.

Physics Phoebe, your practice guide

Physics Phoebe
Predict the reading, then compare it with the measurement.

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Version tested: Prepared circuit opened in the browser on 2026-09-06; no tool version number exposed. Date: 2026-09-06.

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Steps

Screens captured against Falstad CircuitJS Prepared circuit opened in the browser on 2026-09-06; no tool version number exposed on 2026-09-06; the tool may have moved on — the text steps are the contract.

  1. Step 1

    Do
    In the circuit canvas, inspect the 50 mV source, op-amp, and feedback resistor labels.
    You will see
    Source panel: "Vd = 50 mV". Feedback resistor panels: "R = 90 kΩ"; "R = 10 kΩ". Op-amp panel: "Vout = 499.95 mV"; "range = -5 V to 5 V". Output wire panel: "V = 499.95 mV".
    Why it matters
    Expected: Feedback is 90 kΩ over 10 kΩ, with a 50 mV input. Meaning: This is a non-inverting amplifier with a finite 5 V output rail.

  2. Step 2

    Do
    In the circuit canvas, use the feedback resistor labels to calculate the gain 1 + 90000 / 10000.
    You will see
    Source panel: "Vd = 50 mV". Feedback resistor panels: "R = 90 kΩ"; "R = 10 kΩ". Op-amp panel: "Vout = 499.95 mV"; "range = -5 V to 5 V". Output wire panel: "V = 499.95 mV".
    Why it matters
    Expected: 1 + 90/10 = gain 10. Meaning: The feedback divider sets the ideal closed-loop gain.

  3. Step 3

    Do
    In the circuit canvas, point at the output wire and read its measurement panel.
    You will see
    Source panel: "Vd = 50 mV". Feedback resistor panels: "R = 90 kΩ"; "R = 10 kΩ". Op-amp panel: "Vout = 499.95 mV"; "range = -5 V to 5 V". Output wire panel: "V = 499.95 mV".
    Why it matters
    Expected: 50 mV × 10 = 500 mV ideal; the model reports 499.95 mV. Meaning: The small difference reflects the simulated op-amp model.

  4. Step 4

    Do
    In the source’s right-click menu, choose Edit and set Voltage to 0.1 V.
    You will see
    Source panel: "Vd = 100 mV". Feedback resistor panels: "R = 90 kΩ"; "R = 10 kΩ". Op-amp panel: "Vout = 999.9 mV"; "range = -5 V to 5 V". Output wire panel: "V = 999.9 mV".
    Why it matters
    Expected: 100 mV × 10 = 1 V ideal; the model reports 999.9 mV. Meaning: The amplifier remains in its linear range for this input.

  5. Step 5

    Do
    In the source’s right-click menu, choose Edit and set Voltage to 0.6 V.
    You will see
    Source panel: "Vd = 600 mV". Feedback resistor panels: "R = 90 kΩ"; "R = 10 kΩ". Op-amp panel: "Vout = 5 V"; "range = -5 V to 5 V". Output wire panel: "V = 5 V".
    Why it matters
    Expected: 600 mV × 10 = 6 V ideal, above the +5 V output rail. Meaning: The observed 5 V output is clipping, not a gain of 8.33.

  6. Step 6

    Do
    In the source’s right-click menu, choose Edit and restore Voltage to 0.05 V.
    You will see
    Source panel: "Vd = 50 mV". Feedback resistor panels: "R = 90 kΩ"; "R = 10 kΩ". Op-amp panel: "Vout = 499.95 mV"; "range = -5 V to 5 V". Output wire panel: "V = 499.95 mV".
    Why it matters
    Expected: Restoring 50 mV should restore approximately 500 mV output. Meaning: The response is reversible while the circuit is back in its linear range.

Chapter checks

These questions refer to the chapter’s examples. Use the return links to review their answers.

  1. A strain gauge Wheatstone bridge is balanced at zero load with all four 350ohm resistors. Under maximum load, one gauge changes by 0.1% (to 350.35ohm). With 5V excitation, what differential output voltage do you expect?

    Return to the chapter’s knowledge check

  2. A pressure sensor outputs 0-50mV for 0-100 PSI range. You want to interface it with a 3.3V, 12-bit ADC. What amplifier gain provides the best resolution while staying within the ADC input range?

    Return to the chapter’s knowledge check

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