RC circuit calculator
Cutoff frequency and time constant of a resistor with a capacitor.
Fill in the fields and the result will appear here automatically.
Connect the time constant with cutoff frequency for one ideal RC stage. Resistance is entered in Ω and capacitance in nF. The displayed 5τ reaches 99.3262% of final voltage when charging from zero, not complete charge. An ideal source and no output loading are assumed; source and load resistances can alter the response.
How it works
Formula and logic
τ=R·C, fc=1/(2πRC), C=CnF·10⁻⁹ F. Charging from zero: Uc(t)=U∞(1−e^(−t/τ)); discharge: Uc(t)=U0·e^(−t/τ). R, C>0. An unloaded first-order filter has transfer magnitude 1/√2 at fc, or−3.0103 dB.
Example
R=10000 Ω and C=100 nF give τ=0.001 s, fc=159.15 Hz and 5τ=0.005 s. 1000 Ω and 1000 nF give the same times and cutoff. A 1% residual needs−τ·ln(0.01)=4.60517τ; complete charge has no finite arrival time.
Fields and units
- Resistance R — Ω
- Capacitance C — nF
How to use
- — Enter Ω: 10 kΩ means 10000 Ω.
- — Enter nF: 0.1 µF means 100 nF.
- — Check that source and attached load do not add significant resistance.
- — Output across C gives low-pass and across R high-pass; the tool does not select topology or plot the response.
Method and limitations
- Calculation method
- Formula and logic
- Data or methodology source
- OpenStax: exponential RC charge and discharge
- Limitation
- Ideal unloaded first-order RC stage; 5τ means 99.3262%, not complete charge.
FAQ
Why do different R and C pairs give the same frequency?
Because only the product R·C enters the formula. 10 kΩ with 100 nF and 1 kΩ with 1000 nF are the same product, hence the same cutoff frequency and the same time constant.
How does cutoff differ from the edge of the passband?
Cutoff is not a sharp boundary. Transfer magnitude is 1/√2 of passband level, or−3.0103 dB. Half-power wording needs an equal-resistance comparison. 20 dB/decade is an asymptotic slope, not an exact drop from any starting frequency.
What does the settling time show?
One τ reaches 63.2121%; 5τ reaches 99.3262%, leaving 0.6738%. Required settling time follows the allowed error: t=−τlnδ. Five τ is a chosen reference, not complete charge.
Does this work for a second-order filter?
The model has one passive RC pole. A second-order response requires its own topology and parameters; directly cascaded passive stages also load each other.