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Circuits, electronics and noise
Inverting integrator output under constant input
Output of an ideal inverting integrator, input resistor R and feedback capacitor C, for a constant input Vi.
- Ref
- ELN-28
- Rev
- v1
- Field
- Circuits, electronics and noise
- Result
- Vo(t) [V]
An ideal inverting integrator holds its inverting input at virtual ground, so a constant input Vi gives a constant current Vi/R into the capacitor C. The output voltage is the integral of that current, Vo(t) = Vo(0) − Vi t/(RC). It ramps at −Vi/(RC) volts per second and stops at the op amp’s supply rails. A resistor across C is needed in practice to stop DC drift.
Output of an ideal inverting integrator, input resistor R and feedback capacitor C, for a constant input Vi. It ramps linearly until it reaches the supply limits. A negative input gives a rising output.
Calculator
- Ramp rate, −Vi/(RC)−200 V/s
- Time constant, RC0.001 s
- Input current, Vi/R2×10−5 A
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The output ramps at −200 V/s and reaches −0.5 V after 0.005 s.
- With R C = 0.001 s, an input of 0.2 V takes 0.025 s to move the output by 5 V.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (V) | Local step | Change in result |
|---|---|---|---|---|---|---|
| C Feedback capacitor | 1e-9 to 1e-5 F | −99.5 | 0.49 | 99.99 | +10% | −18.2 % |
| Vi Constant input voltage | −5 to 5 V | −24.5 | 25.5 | 50 | +0.2 V | +200 % |
| t Elapsed time | 1e-4 to 0.1 s | −19.5 | 0.48 | 19.98 | +10% | +20 % |
| Vo(0) Initial output voltage | −5 to 5 V | −6 | 4 | 10 | +0.2 V | −40 % |
| R Input resistor | 1,000 to 1,000,000 Ω | −9.5 | 0.49 | 9.99 | +10% | −18.2 % |
Sweep of Elapsed time
| Elapsed time (s) | Output voltage (V) |
|---|---|
| 1e-4 | 0.48 |
| 2.371e-4 | 0.4526 |
| 5.623e-4 | 0.3875 |
| 0.001334 | 0.2333 |
| 0.003162 | −0.1325 |
| 0.007499 | −0.9998 |
| 0.01778 | −3.057 |
| 0.04217 | −7.934 |
| 0.1 | −19.5 |
Sweep one variable
Variables and how each one contributes
Output voltage
What the equation gives you. Everything else feeds it.
Initial output voltage
The initial output voltage, in V. Adds directly to the output. Raising it by 10 % (to 0.55) moves the result by −10 %, from −0.5 to −0.45 V.
Constant input voltage
The constant input voltage, in V. A larger positive input gives a steeper downward ramp. Raising it by 10 % (to 0.22) moves the result by +20 %, from −0.5 to −0.6 V.
Elapsed time
The elapsed time, in s. A longer time gives a larger change, in direct proportion. Raising it by 10 % (to 0.0055) moves the result by +20 %, from −0.5 to −0.6 V.
Input resistor
The input resistor, in Ω. A larger R reduces the input current and so the ramp rate, in inverse proportion. Raising it by 10 % (to 11,000) moves the result by −18.2 %, from −0.5 to −0.4091 V.
Feedback capacitor
The feedback capacitor, in F. A larger C needs more charge for each volt, so the ramp rate falls in inverse proportion. Raising it by 10 % (to 1.1×10−7) moves the result by −18.2 %, from −0.5 to −0.4091 V.
Worked example with the values above
- Time constant: RC = 10,000 × 1×10−7 = 0.001 s.
- Change in output: −Vi t/(RC) = −0.2 × 0.005 / 0.001 = −1 V.
- Vo = 0.5 + −1 = −0.5 V.
Assumptions and limits
- An ideal op amp with infinite gain and no input current, in its linear range with negative feedback.
- Output limited only by the supply, and signals within the useful bandwidth.
Common mistakes
- Forgetting the minus sign.
- Running it past the supply rails, where the formula no longer holds.
- Leaving off the resistor across the capacitor, so that a DC offset integrates to the rail.
Statistics and uncertainty
Resistor tolerances set the gain accuracy, so use 1 % parts or better for precision, and a matched network for the difference amplifier, where mismatch converts common-mode voltage into error. Real op amps add offset, bias current and finite gain.
Where it is used
- Signal mixing, scaling, integration and current-to-voltage conversion in instrumentation.
- Photodiode, sensor and ramp-generator circuits.
Maths you need
- Virtual ground: the inverting input sits at the non-inverting voltage
- Negative feedback: the output acts to cancel the input difference
- Kirchhoff’s current law: currents at a node add to zero