Capacitor Charge and Discharge Curves

Understanding how capacitors charge and discharge is fundamental to modern electronics, timing circuits, and even cardiac defibrillators. The characteristic curve of a capacitor's voltage over time isn't linear; it's exponential. Mastering this behavior and the mathematics behind it allows you to predict circuit performance, design timing delays, and measure unknown component values with precision.

The Physical Basis of Exponential Change

A capacitor is a device that stores electrical energy in an electric field between two conductive plates. When connected to a voltage source through a resistor, charge doesn't flow onto the plates instantly. The resistor limits the current, causing the capacitor to charge gradually. Conversely, when the charged capacitor is connected to just a resistor, it discharges its stored energy over time.

The rate at which this happens depends on two factors: the resistance $R$ in the circuit (in ohms, $\Omega$) and the capacitance $C$ of the capacitor (in farads, F). Their product, $RC$, is a crucial parameter called the time constant, denoted by $\tau$ (tau). The time constant $\tau = RC$ has units of seconds and represents a specific, meaningful point on the charge or discharge curve. It is the time taken for the voltage across the capacitor to fall to approximately 37% of its initial value during discharge, or rise to about 63% of the supply voltage during charging.

The Exponential Decay and Growth Equations

The voltage $V$ across a capacitor and the charge $Q$ stored on it change according to precise exponential laws. For a capacitor discharging through a resistor, starting from an initial voltage $V_0$ or charge $Q_0$, the equations are: $$V = V_0 e^{-t/RC} \quad \text{and} \quad Q = Q_0 e^{-t/RC}$$ where $e$ is the base of the natural logarithm (approximately 2.718), and $t$ is time. The current $I$ during discharge also follows an exponential decay: $I = I_0 e^{-t/RC}$, where $I_0 = V_0 / R$.

For a capacitor charging from zero through a resistor towards a supply voltage $V_s$, the voltage equation is: $$V = V_s (1 - e^{-t/RC})$$ The charge and current follow similar charging patterns. Notice how the discharging formula involves decay from an initial value, while the charging formula describes growth towards a maximum value.

Example Calculation (Discharge): A $100 \mu\text{F}$ capacitor is charged to 12 V and then discharged through a $10 \text{k}\Omega$ resistor. What is the voltage across the capacitor after 2 seconds?

1. Calculate the time constant: $\tau = RC = (10 \times 10^3) \times (100 \times 10^{-6}) = 1 \text{ second}$.
2. Apply the discharge equation: $V = V_0 e^{-t/RC} = 12 \times e^{-2/1} = 12 \times e^{-2}$.
3. Compute: $e^{-2} \approx 0.1353$, so $V \approx 12 \times 0.1353 = 1.62 \text{ V}$.

Linearization Using Natural Logarithms

Plotting voltage against time for a discharge produces a curve that asymptotically approaches zero. It can be difficult to accurately determine the time constant $\tau$ from this curve. The solution is to linearize the data by exploiting the properties of logarithms.

Starting with the discharge equation $V = V_0 e^{-t/RC}$, taking the natural logarithm (ln) of both sides yields: $$\ln(V) = \ln(V_0) - \frac{t}{RC}$$ This is now in the linear form $y = mx + c$, where:

• $y = \ln(V)$
• $x = t$
• Gradient $m = -\frac{1}{RC}$
• $y$-intercept $c = \ln(V_0)$

Therefore, a plot of $\ln(V)$ against time $t$ for a discharging capacitor will yield a straight line with a negative gradient. The magnitude of the gradient is equal to $1/RC$. This is a powerful experimental method: $|\text{gradient}| = 1 / \tau$.

Step-by-step analysis: If you collect data of voltage $V$ at various times $t$ during discharge:

1. Calculate $\ln(V)$ for each data point.
2. Plot $\ln(V)$ on the y-axis against $t$ on the x-axis.
3. Draw a line of best fit.
4. Calculate the gradient of this line (which will be negative).
5. The time constant is $\tau = \frac{1}{|\text{gradient}|}$.
6. If $R$ is known, capacitance can be found from $C = \tau / R$.

Designing Experiments to Measure Capacitance

You can use the discharge curve method to determine an unknown capacitance. The required apparatus includes a capacitor, a known resistor, a power supply, a voltmeter (preferably a data logger for rapid measurements), and a switch.

Procedure:

1. Set up a series circuit with the power supply, the capacitor $C$ (unknown), the known resistor $R$, and a switch.
2. Close the switch to charge the capacitor fully to the supply voltage $V_s$. Record $V_s$.
3. Open the switch to disconnect the power supply and simultaneously start timing. The capacitor now discharges through the known resistor $R$.
4. Record the voltage $V$ across the capacitor at regular time intervals (e.g., every 5 or 10 seconds) until the voltage is very small.
5. Process the data as described in the logarithmic analysis section. Plot $\ln(V)$ vs. $t$.
6. Determine the gradient of the line of best fit. Calculate $\tau = 1 / |\text{gradient}|$.
7. Finally, calculate the capacitance using $C = \tau / R$.

This experiment highlights the practical application of the theory and reinforces the relationship between the graphical gradient and the circuit's physical components.

Common Pitfalls

1. Misidentifying the Time Constant: A common mistake is to think the time constant $\tau$ is the time for the capacitor to fully discharge. In reality, after one time constant ($t=\tau$), the voltage is $V_0 e^{-1} \approx 0.37V_0$ (37% remaining). It takes about $5\tau$ for the voltage to fall below 1% of $V_0$. Confusing $\tau$ with the "total discharge time" will lead to significant errors in calculations.

1. Incorrect Logarithmic Plots: When creating a $\ln(V)$ vs. $t$ plot, students sometimes plot $\log_{10}(V)$ instead of $\ln(V)$. While this will still give a straight line, the gradient will be different: $m_{\log10} = -\frac{1}{2.303RC}$. Forgetting to take the logarithm at all and trying to fit a line to a curved $V$ vs. $t$ plot is another frequent error. Always remember the transformation step: $y$-axis = $\ln(V)$.

1. Confusing Charge and Discharge Equations: Applying the decay equation $V = V_0 e^{-t/RC}$ to a charging scenario is incorrect. For charging, you must use the growth equation $V = V_s (1 - e^{-t/RC})$. A good check: at time $t=0$, a discharging capacitor starts at $V_0$, while a charging capacitor starts at 0. Mentally test the initial and final conditions for your scenario before selecting an equation.

1. Ignoring the Gradient Sign in $ln(V)$ Analysis: The gradient of the $\ln(V)$ vs. $t$ line is negative because voltage decreases with time. When calculating $\tau = 1/RC$, you use the magnitude (absolute value) of this gradient. Using the negative value directly will give a negative time constant, which is physically meaningless. Always state the gradient as a negative number but use its positive magnitude in the $\tau$ calculation.

Summary

• The charging and discharging of a capacitor through a resistor follow exponential laws, governed by the time constant $\tau = RC$.
• For discharge, voltage decays as $V = V_0 e^{-t/RC}$; for charge, it grows as $V = V_s (1 - e^{-t/RC})$.
• Plotting $\ln(V)$ against time $t$ for a discharge linearizes the exponential relationship, where the gradient of the line equals $-1/RC$.
• This logarithmic graph provides an accurate method to determine the time constant $\tau$ and, if $R$ is known, the capacitance $C$.
• In experimental design, meticulous measurement of voltage during discharge, followed by logarithmic analysis, allows for the precise determination of unknown capacitance values.