RC Time Constant (tau), capacitor charging voltage & -3 dB cutoff frequency

63.2% at 1 tau, 99.3% at 5 tau, and -3 dB low-pass corner frequency

Pairing a 10,000 Ω resistor (A2) with a 100 µF (0.0001 F, B2) capacitor yields an RC Time Constant tau = R × C = 1.00 second. When connected to a Vs = 5.0 V step supply (C2), capacitor voltage rises exponentially via Vc(t) = Vs × (1 - EXP(-t / (R × C))): reaching 63.21% (3.1606 V) at t = 1 tau and 99.33% (4.9663 V—practically fully charged) at t = 5 tau!

In the frequency domain, that exact same RC network forms a 1st-order low-pass or high-pass filter whose -3 dB half-power cutoff (corner) frequency is fc = 1 / (2 × PI() × R × C) = 0.1592 Hz!

Assumes an ideal step source, an initially uncharged capacitor, positive resistance and capacitance, and an unloaded first-order network. The charging equation describes capacitor voltage. Filter output across the capacitor is low-pass; output across the resistor is high-pass. Real source/load impedances, leakage and component tolerances alter the response.

Capacitor charging voltage Vc(t) and -3 dB filter cutoff frequency fc

Enter R in Ohms and C in Farads (1 µF = 1E-6 F, 1 nF = 1E-9 F) so R × C is directly in seconds and fc in Hertz.

=IF(OR(COUNT(A2:B2)<>2,A2<=0,B2<=0,NOT(ISNUMBER(D2)),D2<0,NOT(ISNUMBER(C2))),"Invalid RC inputs",ROUND(C2 * (1 - EXP(-D2 / (A2 * B2))), 4))

Passive RC filter -3 dB cutoff frequency fc = 1 / (2 * PI * R * C)

At frequency fc, output amplitude is 1/SQRT(2) = 70.71% (-3.01 dB) of input amplitude.

=IF(OR(COUNT(A2:B2)<>2,A2<=0,B2<=0),"Invalid RC inputs",ROUND(1 / (2 * PI() * A2 * B2), 4))

Further reading

• Microsoft Support — EXP function
• Microsoft Support — PI function