The step response of a **RC (Resistor-Capacitor)** circuit refers to how the voltage across the capacitor changes when a step input voltage (a sudden change from 0 to a fixed value, such as a DC voltage) is applied to the circuit.
### Types of RC Circuits:
1. **Series RC Circuit**: A resistor and capacitor are connected in series with a voltage source.
2. **Parallel RC Circuit**: A resistor and capacitor are connected in parallel to a voltage source.
Here, we focus on the **series RC circuit**, as it's more commonly analyzed in step response scenarios.
### Step Response of a Series RC Circuit
When a step voltage \( V_{\text{in}}(t) \) is applied to a series RC circuit at \( t = 0 \), the capacitor voltage \( V_C(t) \) changes according to the charging or discharging behavior of the capacitor.
#### 1. **Charging the Capacitor** (Applying a Step Input):
Consider a step input voltage \( V_{\text{in}}(t) = V_0 u(t) \), where \( u(t) \) is the unit step function. For \( t \geq 0 \), the input suddenly jumps from 0 to \( V_0 \).
- **Governing Equation**: The voltage across the capacitor, \( V_C(t) \), and the current through the resistor \( R \) are governed by Kirchhoff’s Voltage Law (KVL):
\[
V_{\text{in}}(t) = V_R(t) + V_C(t)
\]
where \( V_R(t) = i(t)R \) and \( i(t) = C \frac{dV_C(t)}{dt} \).
Substituting into KVL:
\[
V_0 = RC \frac{dV_C(t)}{dt} + V_C(t)
\]
This is a first-order linear differential equation.
- **Solution**:
Solving this equation for \( V_C(t) \):
\[
V_C(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right)
\]
where \( RC \) is the time constant \( \tau \) of the circuit. The time constant represents the time it takes for the voltage to rise to approximately 63% of its final value.
- **Interpretation**:
- At \( t = 0 \), the capacitor voltage \( V_C(0) = 0 \) (initially uncharged).
- As \( t \rightarrow \infty \), the capacitor voltage asymptotically approaches the input voltage, i.e., \( V_C(\infty) = V_0 \).
#### 2. **Discharging the Capacitor** (Removing the Step Input):
If the capacitor is charged initially to \( V_0 \) and then the input is removed (or set to zero), the capacitor discharges through the resistor.
- **Governing Equation**:
\[
0 = RC \frac{dV_C(t)}{dt} + V_C(t)
\]
- **Solution**:
Solving this differential equation, we get:
\[
V_C(t) = V_0 e^{-\frac{t}{RC}}
\]
This describes how the voltage across the capacitor decreases exponentially with time as it discharges.
- **Interpretation**:
- At \( t = 0 \), \( V_C(0) = V_0 \).
- As \( t \rightarrow \infty \), \( V_C(\infty) = 0 \).
### Key Concepts:
- **Time Constant (\( \tau = RC \))**: It determines the rate at which the capacitor charges or discharges. After \( t = \tau \), the voltage has reached about 63% of its final value (for charging) or has decreased to about 37% of its initial value (for discharging).
- **Steady-State and Transient Response**:
- The **transient response** refers to the period during which the voltage is changing, i.e., the exponential growth or decay.
- The **steady-state response** is the final value of the capacitor voltage after a long period, which is either \( V_0 \) (charging) or 0 (discharging).
### Practical Examples:
1. **Charging**: Suppose \( V_{\text{in}}(t) = 10 \, \text{V} \), \( R = 1 \, \text{k}\Omega \), and \( C = 1 \, \mu F \). The time constant is:
\[
\tau = RC = 1 \times 10^3 \times 1 \times 10^{-6} = 1 \, \text{ms}
\]
So, after 1 ms, the capacitor voltage will be approximately 63% of 10V, i.e., 6.3V. After 5 ms (5τ), it will be almost fully charged.
2. **Discharging**: If the capacitor starts with 10V, the voltage will decrease to 3.7V after 1 ms and will approach 0V after a longer time.
### Conclusion:
The step response of an RC circuit shows how the capacitor voltage reacts over time to a sudden change in input voltage. It involves exponential charging and discharging behavior, governed by the circuit's time constant \( RC \).