The Core Formula for Discharging Capacitor Circuits
When you short a charged capacitor through a resistor, the voltage doesn't drop linearly; it decays exponentially. The foundational formula for discharging capacitor circuits dictates that the voltage across the capacitor at any given time is a function of the initial voltage, the resistance, and the capacitance. On the bench, this is the equation you use to size bleeder resistors, calculate debounce delays, and ensure high-voltage power supplies don't shock you after being unplugged.
The primary voltage decay equation is:
V(t) = V0 × e−t / (R × C)
The corresponding current decay equation (which starts at maximum and drops to zero) is:
I(t) = (V0 / R) × e−t / (R × C)
| Symbol | Definition | Standard Unit |
|---|---|---|
| V(t) | Voltage across the capacitor at time t | Volts (V) |
| V0 | Initial voltage at t = 0 (fully charged state) | Volts (V) |
| I(t) | Current flowing through the resistor at time t | Amperes (A) |
| t | Elapsed time since discharge began | Seconds (s) |
| R | Resistance of the discharge path | Ohms (Ω) |
| C | Capacitance | Farads (F) |
| e | Euler's number (mathematical constant ≈ 2.71828) | Dimensionless |
| τ (tau) | Time constant (R × C), time to reach 36.8% of V0 | Seconds (s) |
Rearranged Forms: Solving for Time, Resistance, and Capacitance
In practical design, you rarely just want to find the voltage at a random time. Usually, you have a target voltage (like a safe-to-touch threshold of 50V) and need to find the required time, or you have a fixed time constraint and need to calculate the exact resistor value. By applying the natural logarithm (ln) to both sides, we isolate the variables in the exponent.
- Solving for time (t):
t = -R × C × ln( V(t) / V0 ) - Solving for resistance (R):
R = -t / [ C × ln( V(t) / V0 ) ] - Solving for capacitance (C):
C = -t / [ R × ln( V(t) / V0 ) ] - Solving for initial voltage (V0):
V0 = V(t) / e-t / (R × C)
Assumptions, Boundaries, and Unit Traps
The exponential decay formula is an idealized model. Before you plug numbers into your calculator, you need to understand where this model holds up and where it breaks down on a real workbench.
When the Formula Applies
This formula assumes a lumped-element DC transient circuit. It applies when a capacitor with a known initial charge is connected to a purely resistive load, with no external voltage source actively driving the circuit. It assumes the resistor and capacitor are ideal, meaning the resistance is constant regardless of voltage or temperature, and the capacitance doesn't vary with applied voltage (a caveat for Class 2 ceramic capacitors like X7R or Y5V, which lose significant capacitance at high DC bias).
The Unit Mistakes That Break Your Math
Realistic Answer Magnitudes
If your calculator spits out a time constant of 4,000 seconds for a small PCB bleeder circuit, you've missed a decimal. In typical electronics, RC time constants range from microseconds (high-speed signal filtering) to tens of seconds (power supply bleeder networks). Bleeder resistors for mains-derived DC buses usually fall in the 100kΩ to 2MΩ range, yielding discharge times of 5 to 60 seconds.
Worked Examples with Strict Unit Tracking
Let's run two scenarios with explicit intermediate steps to prove the math.
Problem 1: Calculating Discharge Time to a Safe Threshold
Scenario: You have a 400V DC bus with a 220μF filter capacitor. You install a 100kΩ bleeder resistor. How long does it take for the voltage to drop to a safe 50V?
- Convert units: C = 220 × 10-6 F (0.00022 F); R = 100 × 103 Ω (100,000 Ω).
- Calculate Time Constant (τ): τ = R × C = 100,000 × 0.00022 = 22 seconds.
- Apply rearranged formula for t:
t = -22 × ln(50 / 400) - Solve the ratio and natural log:
50 / 400 = 0.125
ln(0.125) ≈ -2.0794 - Final multiplication:
t = -22 × -2.0794 = 45.75 seconds.
Result: It takes roughly 46 seconds for the bus to reach 50V. (Note: 5 time constants, or 110 seconds, would be required to discharge to <1% of the original voltage, or ~2.7V).
Problem 2: Sizing a Resistor for a Specific Discharge Window
Scenario: A 12V audio circuit uses a 4700μF decoupling capacitor. To prevent a loud 'pop' from the speaker on power-down, the voltage must drop below 1V within 2 seconds. What resistance is required?
- Convert units: C = 4700 × 10-6 F = 0.0047 F; t = 2 s; V0 = 12V; V(t) = 1V.
- Apply rearranged formula for R:
R = -2 / [ 0.0047 × ln(1 / 12) ] - Solve the natural log:
ln(0.08333) ≈ -2.4849 - Multiply denominator:
0.0047 × -2.4849 = -0.011679 - Final division:
R = -2 / -0.011679 = 171.24 Ω.
Result: You need a resistor of 171Ω or lower. A standard 150Ω resistor would be the correct E12 series pick to guarantee the 2-second window.
Decision Path: Sizing a Bleeder Resistor for a Power Supply
When designing a mains-powered supply, you can't just guess the bleeder resistor. It must discharge the capacitors quickly enough to meet safety standards (like IEC 60950 or IEC 62368-1, which often mandate dropping to 60V or 37% of peak within 1-2 seconds for accessible parts), but it must also not dissipate so much continuous heat that it burns up or wastes standby power.
| Design Step | Action / Calculation |
|---|---|
| 1. Define Parameters | 120V AC mains rectified to ≈170V DC peak. Filter cap C = 100μF (0.0001 F). Target V(t) = 50V in t = 5 seconds. |
| 2. Calculate Max R | R = -5 / [ 0.0001 × ln(50/170) ] = -5 / [ 0.0001 × -1.2237 ] = 40,859 Ω |
| 3. Calculate Steady-State Power | P = V2 / R = 1702 / 40859 = 28900 / 40859 = 0.707 Watts |
| 4. Apply Derating Rule | Resistors should run at ≤50% of rated power for longevity and to prevent PCB scorching. Required wattage = 0.707W × 2 = 1.41W minimum. |
| 5. Select Standard Values | Next standard E24 resistor value above 40.8kΩ is 47kΩ (which slightly increases discharge time but lowers steady-state power to 0.61W). Next standard wattage is 2W. |
Real-World Parasitics: When the Ideal Formula Fails
The exponential formula assumes ideal components. On the bench, parasitics alter the discharge curve, particularly in the first and last few milliseconds.
Equivalent Series Resistance (ESR): Every real capacitor has internal resistance. If your external bleeder resistor is very low (e.g., 10Ω for fast active discharge), the capacitor's ESR (which might be 0.5Ω for a large electrolytic) becomes a significant percentage of the total R. The actual discharge will be slightly faster than the ideal formula predicts because the true R is Rexternal + ESR. According to Georgia State University HyperPhysics, while ESR dominates high-frequency ripple heating, its impact on DC discharge time constants is usually negligible unless you are dealing with ultra-low resistance discharge paths.
Dielectric Absorption (Memory Effect): This is the most dangerous parasitic for high-voltage work. In electrolytic and film capacitors, the dielectric material physically polarizes under high voltage. When you discharge the cap to 0V using a resistor, the surface charge bleeds off, but the deep bulk charge remains trapped. If you remove the bleeder resistor, the capacitor will 'rebound' or recover a voltage—sometimes tens of volts—over the next few minutes. This is why high-voltage safety protocols require a permanent, hard-wired bleeder resistor, or a grounding stick applied multiple times before servicing, rather than relying on a single temporary discharge cycle.
