Library / Circuits, electronics and noise / ELN-29
Circuits, electronics and noise
Inverting differentiator output for a ramp input
Output of an ideal inverting differentiator for an input that changes at a steady rate.
- Ref
- ELN-29
- Rev
- v1
- Field
- Circuits, electronics and noise
- Result
- Vo [V]
For a ramp input the current through the input capacitor is C·dVi/dt, and all of it flows through Rf. The output is Vo = −Rf C dVi/dt, proportional to the slope of the input and inverted. A rising input gives a negative output. A real differentiator needs a small series resistor with C and a small capacitor across Rf, or it amplifies high-frequency noise and oscillates.
Output of an ideal inverting differentiator for an input that changes at a steady rate. It is proportional to the slope of the input, and negative for a rising input. Real circuits add a resistor and capacitor to limit noise gain.
Calculator
- Time constant, Rf C1×10−4 s
- Capacitor current, C dVi/dt2×10−5 A
- Gain at 1 kHz, Rf C 2π f0.6283 V/V
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.
- A ramp of 2,000 V/s through Rf C = 1×10−4 s gives −0.2 V at the output.
- The gain rises with frequency, 0.628 at 1 kHz, which is why noise is amplified more than the signal.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (V) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Ci Input capacitor | 1e-10 to 1e-6 F | −20 | −0.002 | 20 | +10% | +10 % |
| Rf Feedback resistor | 1,000 to 1,000,000 Ω | −20 | −0.02 | 19.98 | +10% | +10 % |
| dVi/dt Input slope | −10,000 to 10,000 V/s | −1 | 1 | 2 | +400 V/s | +20 % |
Sweep of Input slope
| Input slope (V/s) | Output voltage (V) |
|---|---|
| −10,000 | 1 |
| −7,500 | 0.75 |
| −5,000 | 0.5 |
| −2,500 | 0.25 |
| 0 | 0 |
| 2,500 | −0.25 |
| 5,000 | −0.5 |
| 7,500 | −0.75 |
| 10,000 | −1 |
Sweep one variable
Variables and how each one contributes
Output voltage
What the equation gives you. Everything else feeds it.
Feedback resistor
The feedback resistor, in Ω. Scales the output size in direct proportion. Raising it by 10 % (to 11,000) moves the result by +10 %, from −0.2 to −0.22 V.
Input capacitor
The input capacitor, in F. A larger C carries more current for the same slope, so the output size rises in direct proportion. Raising it by 10 % (to 1.1×10−8) moves the result by +10 %, from −0.2 to −0.22 V.
Input slope
The slope of the input, dVi/dt, in V/s. A steeper input raises the output size in direct proportion; the sign of the output is opposite to that of the slope. Raising it by 10 % (to 2,200) moves the result by +10 %, from −0.2 to −0.22 V.
Worked example with the values above
- Time constant: Rf C = 10,000 × 1×10−8 = 1×10−4 s.
- Slope: dVi/dt = 2,000 V/s.
- Vo = −1×10−4 × 2,000 = −0.2 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 of the inverting stage.
- Leaving out the small series resistor, so that the circuit is unstable.
- Using the slope in volts per millisecond with SI components.
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