What is the step response of a series RC circuit?
2 Answers
The step response of a series RC (Resistor-Capacitor) circuit describes how the voltage across the capacitor (or resistor) changes over time when a step input (such as a sudden application of a DC voltage) is applied. This response is an important concept in electrical engineering and signal processing because it helps us understand how RC circuits behave in time-varying situations.
### Step Response of a Series RC Circuit
#### 1. **Understanding the Series RC Circuit**
A series RC circuit consists of a resistor \( R \) and a capacitor \( C \) connected in series with a voltage source. When a step input voltage \( V_{\text{in}}(t) \) is applied to this circuit, we are interested in how the voltage across the capacitor, \( V_C(t) \), changes over time.
#### 2. **Applying Kirchhoff's Voltage Law (KVL)**
Kirchhoff's Voltage Law states that the sum of the potential differences (voltages) around any closed loop in a circuit is zero. For a series RC circuit:
\[
V_{\text{in}}(t) = V_R(t) + V_C(t)
\]
where:
- \( V_{\text{in}}(t) \) is the input step voltage.
- \( V_R(t) = i(t) \cdot R \) is the voltage across the resistor.
- \( V_C(t) = \frac{1}{C} \int i(t) \, dt \) is the voltage across the capacitor.
The current \( i(t) \) in the circuit is the same through both the resistor and the capacitor because they are in series.
#### 3. **Formulating the Differential Equation**
The current \( i(t) \) through the capacitor is related to the rate of change of the voltage across the capacitor:
\[
i(t) = C \frac{dV_C(t)}{dt}
\]
Substituting this into \( V_R(t) = i(t) \cdot R \), we get:
\[
V_R(t) = RC \frac{dV_C(t)}{dt}
\]
Thus, the input voltage equation becomes:
\[
V_{\text{in}}(t) = RC \frac{dV_C(t)}{dt} + V_C(t)
\]
Rearranging this, we get a first-order linear differential equation:
\[
\frac{dV_C(t)}{dt} + \frac{1}{RC} V_C(t) = \frac{V_{\text{in}}(t)}{RC}
\]
#### 4. **Solving the Differential Equation**
For a step input voltage \( V_{\text{in}}(t) = V_0 \) (a constant voltage applied at \( t = 0 \)), the differential equation becomes:
\[
\frac{dV_C(t)}{dt} + \frac{1}{RC} V_C(t) = \frac{V_0}{RC}
\]
To solve this, we can use the standard method for solving linear differential equations:
1. **Find the homogeneous solution** \( V_C(t)_h \) by setting the right-hand side to 0:
\[
\frac{dV_C(t)_h}{dt} + \frac{1}{RC} V_C(t)_h = 0
\]
The solution is:
\[
V_C(t)_h = A e^{-\frac{t}{RC}}
\]
where \( A \) is a constant determined by initial conditions.
2. **Find the particular solution** \( V_C(t)_p \) for the non-homogeneous part. Since the right-hand side is a constant, we assume \( V_C(t)_p = K \). Substituting this into the differential equation gives:
\[
K = V_0
\]
3. **General Solution**: The general solution is the sum of the homogeneous and particular solutions:
\[
V_C(t) = A e^{-\frac{t}{RC}} + V_0
\]
#### 5. **Applying Initial Conditions**
At \( t = 0 \), the capacitor is initially uncharged, so \( V_C(0) = 0 \). Substituting this into the general solution:
\[
0 = A e^0 + V_0 \implies A = -V_0
\]
Thus, the complete solution for the voltage across the capacitor is:
\[
V_C(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right)
\]
#### 6. **Interpreting the Step Response**
- At \( t = 0 \), \( V_C(0) = 0 \). The capacitor starts charging.
- As \( t \to \infty \), \( V_C(t) \to V_0 \). The capacitor voltage asymptotically approaches the input step voltage \( V_0 \).
- The time constant \( \tau = RC \) determines how fast the capacitor charges. When \( t = RC \), the voltage across the capacitor reaches approximately \( 63.2\% \) of \( V_0 \).
The plot of \( V_C(t) \) versus \( t \) is an exponential curve that starts at zero and approaches \( V_0 \) asymptotically.
#### 7. **Current Response in the Circuit**
The current \( i(t) \) in the circuit, which is the same as the current through the resistor, can be found using:
\[
i(t) = C \frac{dV_C(t)}{dt} = \frac{V_0}{R} e^{-\frac{t}{RC}}
\]
This shows that the current starts at \( \frac{V_0}{R} \) when \( t = 0 \) and decays exponentially to zero as the capacitor becomes fully charged.
### Summary
- The voltage across the capacitor in a series RC circuit following a step input is given by:
\[
V_C(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right)
\]
- The current in the circuit decays exponentially:
\[
i(t) = \frac{V_0}{R} e^{-\frac{t}{RC}}
\]
The response characteristics are governed by the time constant \( RC \), which determines how quickly the circuit responds to changes in input.
### Step Response of a Series RC Circuit
#### 1. **Understanding the Series RC Circuit**
A series RC circuit consists of a resistor \( R \) and a capacitor \( C \) connected in series with a voltage source. When a step input voltage \( V_{\text{in}}(t) \) is applied to this circuit, we are interested in how the voltage across the capacitor, \( V_C(t) \), changes over time.
#### 2. **Applying Kirchhoff's Voltage Law (KVL)**
Kirchhoff's Voltage Law states that the sum of the potential differences (voltages) around any closed loop in a circuit is zero. For a series RC circuit:
\[
V_{\text{in}}(t) = V_R(t) + V_C(t)
\]
where:
- \( V_{\text{in}}(t) \) is the input step voltage.
- \( V_R(t) = i(t) \cdot R \) is the voltage across the resistor.
- \( V_C(t) = \frac{1}{C} \int i(t) \, dt \) is the voltage across the capacitor.
The current \( i(t) \) in the circuit is the same through both the resistor and the capacitor because they are in series.
#### 3. **Formulating the Differential Equation**
The current \( i(t) \) through the capacitor is related to the rate of change of the voltage across the capacitor:
\[
i(t) = C \frac{dV_C(t)}{dt}
\]
Substituting this into \( V_R(t) = i(t) \cdot R \), we get:
\[
V_R(t) = RC \frac{dV_C(t)}{dt}
\]
Thus, the input voltage equation becomes:
\[
V_{\text{in}}(t) = RC \frac{dV_C(t)}{dt} + V_C(t)
\]
Rearranging this, we get a first-order linear differential equation:
\[
\frac{dV_C(t)}{dt} + \frac{1}{RC} V_C(t) = \frac{V_{\text{in}}(t)}{RC}
\]
#### 4. **Solving the Differential Equation**
For a step input voltage \( V_{\text{in}}(t) = V_0 \) (a constant voltage applied at \( t = 0 \)), the differential equation becomes:
\[
\frac{dV_C(t)}{dt} + \frac{1}{RC} V_C(t) = \frac{V_0}{RC}
\]
To solve this, we can use the standard method for solving linear differential equations:
1. **Find the homogeneous solution** \( V_C(t)_h \) by setting the right-hand side to 0:
\[
\frac{dV_C(t)_h}{dt} + \frac{1}{RC} V_C(t)_h = 0
\]
The solution is:
\[
V_C(t)_h = A e^{-\frac{t}{RC}}
\]
where \( A \) is a constant determined by initial conditions.
2. **Find the particular solution** \( V_C(t)_p \) for the non-homogeneous part. Since the right-hand side is a constant, we assume \( V_C(t)_p = K \). Substituting this into the differential equation gives:
\[
K = V_0
\]
3. **General Solution**: The general solution is the sum of the homogeneous and particular solutions:
\[
V_C(t) = A e^{-\frac{t}{RC}} + V_0
\]
#### 5. **Applying Initial Conditions**
At \( t = 0 \), the capacitor is initially uncharged, so \( V_C(0) = 0 \). Substituting this into the general solution:
\[
0 = A e^0 + V_0 \implies A = -V_0
\]
Thus, the complete solution for the voltage across the capacitor is:
\[
V_C(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right)
\]
#### 6. **Interpreting the Step Response**
- At \( t = 0 \), \( V_C(0) = 0 \). The capacitor starts charging.
- As \( t \to \infty \), \( V_C(t) \to V_0 \). The capacitor voltage asymptotically approaches the input step voltage \( V_0 \).
- The time constant \( \tau = RC \) determines how fast the capacitor charges. When \( t = RC \), the voltage across the capacitor reaches approximately \( 63.2\% \) of \( V_0 \).
The plot of \( V_C(t) \) versus \( t \) is an exponential curve that starts at zero and approaches \( V_0 \) asymptotically.
#### 7. **Current Response in the Circuit**
The current \( i(t) \) in the circuit, which is the same as the current through the resistor, can be found using:
\[
i(t) = C \frac{dV_C(t)}{dt} = \frac{V_0}{R} e^{-\frac{t}{RC}}
\]
This shows that the current starts at \( \frac{V_0}{R} \) when \( t = 0 \) and decays exponentially to zero as the capacitor becomes fully charged.
### Summary
- The voltage across the capacitor in a series RC circuit following a step input is given by:
\[
V_C(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right)
\]
- The current in the circuit decays exponentially:
\[
i(t) = \frac{V_0}{R} e^{-\frac{t}{RC}}
\]
The response characteristics are governed by the time constant \( RC \), which determines how quickly the circuit responds to changes in input.