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Exercise E1 Machine-Vision Strobe: Capacitor Charging and Safe Discharge
A high-speed machine-vision inspection system on a production line uses a short high-voltage flash pulse. For this purpose, an energy-storage capacitor is charged by a regulated DC high-voltage source and must be safely discharged before maintenance.
Data: \begin{align*} C &= 1~{\rm \mu F} \\ W &= 0.1~{\rm J} \\ I_{\rm max} &= 100~{\rm mA} \\ R_i &= 10~{\rm M\Omega} \end{align*}
Assume the DC charging source is adjusted to the required final capacitor voltage.
1. What voltage must be present across the capacitor so that it stores the required energy?
2. The charging current must not exceed $100~\rm mA$ at the start of charging. What series charging resistor is required?
3. How long does the charging process take until the capacitor is considered practically fully charged?
In engineering practice, a capacitor is usually considered fully charged after about $5\tau$.
\begin{align*} t_{\rm charge} &\approx 5\tau \approx 5 \cdot 4.47~{\rm ms} \approx 22.4~{\rm ms} \end{align*}
Mathematically, the charging time to exactly $100\%$ is infinite.
4. Sketch the capacitor voltage and the voltage across the charging resistor during charging.
with
\begin{align*} U_0 &= 447.2~{\rm V} \\ RC &= 4.47~{\rm ms} \end{align*}
Initial and final values:
\begin{align*} u_C(0) &= 0, & u_C(\infty) &= U_0 \\ u_R(0) &= U_0, & u_R(\infty) &= 0 \end{align*}
Thus, the capacitor voltage rises exponentially, while the resistor voltage falls exponentially.
5. After charging, the capacitor is disconnected from the source. Its leakage can be modeled by an internal resistance of $10~\rm M\Omega$. After what time has the stored energy dropped to one half, and what is the capacitor voltage then?
Since capacitor energy is
\begin{align*} W(t) &= \frac{1}{2} C u_C^2(t) \end{align*}
it follows that
\begin{align*} \frac{W(t)}{W_0} &= e^{-2t/(R_i C)} \end{align*}
For half the initial energy:
\begin{align*} \frac{1}{2} &= e^{-2t/(R_i C)} \\ t &= \frac{R_i C}{2}\ln(2) \end{align*}
With
\begin{align*} R_i C = 10~{\rm M\Omega} \cdot 1~{\rm \mu F} = 10~{\rm s} \end{align*}
we get
\begin{align*} t &= \frac{10~{\rm s}}{2}\ln(2) \approx 3.47~{\rm s} \end{align*}
The voltage at this instant is
\begin{align*} u_C &= U_0 \cdot \frac{1}{\sqrt{2}} = \frac{447.2~{\rm V}}{\sqrt{2}} \approx 316.2~{\rm V} \end{align*}
6. The fully charged capacitor is discharged through the charging resistor before maintenance. How long does the discharge take, and how much energy is converted into heat in the resistor?
\begin{align*} u_C(t) &= U_0 e^{-t/(RC)} \end{align*}
with
\begin{align*} \tau &= RC \approx 4.47~{\rm ms} \end{align*}
Again, practical discharge time is about
\begin{align*} t_{\rm discharge} &\approx 5\tau \approx 22.4~{\rm ms} \end{align*}
The energy stored initially in the capacitor is
\begin{align*} W_0 &= 0.1~{\rm J} \end{align*}
After full discharge, the capacitor energy is zero, so the entire initial energy is converted into heat in the resistor:
\begin{align*} W_R &= W_0 = 0.1~{\rm J} \end{align*}
In a real design, the resistor must therefore be checked for pulse-load capability.
Exercise E2 Industrial Sensor Interface: Buffered Measurement Node
A 12 V industrial sensor module feeds a buffered measurement node through a resistor T-network. The capacitor at the output node is used to smooth the signal and to provide a stable voltage for a short measurement cycle. At first, the measurement electronics are disconnected. Once the capacitor is fully charged, a switch closes and the measurement load is connected.
Data: \begin{align*} U &= 12~{\rm V} \\ R_1 &= 2~{\rm k\Omega} \\ R_2 &= 10~{\rm k\Omega} \\ R_3 &= 3.33~{\rm k\Omega} \\ C &= 2~{\rm \mu F} \\ R_L &= 5~{\rm k\Omega} \end{align*}
Initially, the capacitor is uncharged and the switch $S$ is open.
1. What is the voltage across the capacitor after it is fully charged?
\begin{align*} U_C(\infty) &= U \cdot \frac{R_2}{R_1 + R_2} \\ &= 12~{\rm V} \cdot \frac{10~{\rm k\Omega}}{2~{\rm k\Omega} + 10~{\rm k\Omega}} \\ &= 12~{\rm V} \cdot \frac{10}{12} \\ &= 10~{\rm V} \end{align*}
2. How long does the charging process take?
First, calculate the parallel combination of $R_1$ and $R_2$:
\begin{align*} R_1 \parallel R_2 &= \frac{R_1 R_2}{R_1 + R_2} = \frac{2~{\rm k\Omega} \cdot 10~{\rm k\Omega}}{2~{\rm k\Omega} + 10~{\rm k\Omega}} \\ &= \frac{20}{12}~{\rm k\Omega} \approx 1.67~{\rm k\Omega} \end{align*}
Thus, the Thevenin resistance is
\begin{align*} R_{\rm th} &= R_3 + (R_1 \parallel R_2) \\ &= 3.33~{\rm k\Omega} + 1.67~{\rm k\Omega} \\ &\approx 5.00~{\rm k\Omega} \end{align*}
The charging time constant is
\begin{align*} \tau_1 &= R_{\rm th} C \\ &= 5.00~{\rm k\Omega} \cdot 2~{\rm \mu F} \\ &= 10~{\rm ms} \end{align*}
In practice, the capacitor is considered fully charged after about $5\tau_1$:
\begin{align*} t_{\rm charge} &\approx 5\tau_1 = 5 \cdot 10~{\rm ms} = 50~{\rm ms} \end{align*}
3. Draw the time-dependent capacitor voltage.
\begin{align*} u_C(t) &= U_C(\infty)\left(1 - e^{-t/\tau_1}\right) \end{align*}
With
\begin{align*} U_C(\infty) &= 10~{\rm V} \\ \tau_1 &= 10~{\rm ms} \end{align*}
the time-dependent capacitor voltage is
\begin{align*} u_C(t) = 10\left(1 - e^{-t/(10~{\rm ms})}\right)~{\rm V} \end{align*}
It starts at $0~{\rm V}$ and rises exponentially toward $10~{\rm V}$.
4. Once the capacitor is fully charged, the switch is closed and the load resistor is connected. What is the load voltage in the stationary state?
\begin{align*} U_{\rm th} &= 10~{\rm V} \\ R_{\rm th} &= 5.00~{\rm k\Omega} \end{align*}
After closing the switch, the load voltage is the divider voltage of $R_{\rm th}$ and $R_L$:
\begin{align*} U_L(\infty) &= U_{\rm th} \cdot \frac{R_L}{R_{\rm th} + R_L} \\ &= 10~{\rm V} \cdot \frac{5~{\rm k\Omega}}{5~{\rm k\Omega} + 5~{\rm k\Omega}} \\ &= 10~{\rm V} \cdot \frac{1}{2} \\ &= 5~{\rm V} \end{align*}
5. How long does it take until this stationary state is reached?
\begin{align*} R_{\rm eq} &= R_{\rm th} \parallel R_L \\ &= 5.00~{\rm k\Omega} \parallel 5.00~{\rm k\Omega} \\ &= 2.50~{\rm k\Omega} \end{align*}
The new time constant is
\begin{align*} \tau_2 &= R_{\rm eq} C \\ &= 2.50~{\rm k\Omega} \cdot 2~{\rm \mu F} \\ &= 5~{\rm ms} \end{align*}
In practice, the new stationary state is reached after about $5\tau_2$:
\begin{align*} t_{\rm settle} &\approx 5\tau_2 = 5 \cdot 5~{\rm ms} = 25~{\rm ms} \end{align*}
6. Draw the time-dependent load voltage.
\begin{align*} u_L(0^+) = 10~{\rm V} \end{align*}
The final stationary value is
\begin{align*} u_L(\infty) = 5~{\rm V} \end{align*}
Thus, the transient load voltage is
\begin{align*} u_L(t) &= u_L(\infty) + \left(u_L(0^+) - u_L(\infty)\right)e^{-t/\tau_2} \\ &= 5~{\rm V} + \left(10~{\rm V} - 5~{\rm V}\right)e^{-t/(5~{\rm ms})} \\ &= 5 + 5e^{-t/(5~{\rm ms})}~{\rm V} \end{align*}
So the load voltage starts at $10~{\rm V}$ and decays exponentially to $5~{\rm V}$.
Exercise E3 Hall-Sensor Test Bench: Air-Core Calibration Coil
In a production test bench for Hall sensors, a small air-core coil is used to generate a reproducible magnetic field. An air-core design is chosen because it avoids hysteresis and remanence effects of iron cores. The coil is wound as a short, single-layer cylindrical coil.
Data: \begin{align*} l &= 22~{\rm mm} \\ d &= 20~{\rm mm} \\ d_{\rm Cu} &= 0.8~{\rm mm} \\ N &= 25 \\ \rho_{\rm Cu,20^\circ C} &= 0.0178~{\rm \Omega\,mm^2/m} \end{align*}
For the inductance of the short air-core coil, use Wheeler's approximation: \begin{align*} L[{\rm \mu H}] \approx \frac{r^2 N^2}{9r + 10l} \end{align*} with $r$ and $l$ entered in inches.
The coil is then connected to a DC supply.
1. Calculate the coil resistance $R$ at room temperature.
\begin{align*} A_{\rm Cu} &= \frac{\pi}{4} d_{\rm Cu}^2 = \frac{\pi}{4}(0.8~{\rm mm})^2 \\ &= 0.503~{\rm mm^2} \end{align*}
The mean length of one turn is approximately the circumference:
\begin{align*} l_{\rm turn} &\approx \pi d = \pi \cdot 20~{\rm mm} = 62.83~{\rm mm} \end{align*}
Thus, the total wire length is
\begin{align*} l_{\rm Cu} &= N \cdot l_{\rm turn} = 25 \cdot 62.83~{\rm mm} \\ &= 1570.8~{\rm mm} = 1.571~{\rm m} \end{align*}
Now calculate the resistance:
\begin{align*} R &= \rho_{\rm Cu}\frac{l_{\rm Cu}}{A_{\rm Cu}} \\ &= 0.0178~{\rm \Omega\,mm^2/m}\cdot\frac{1.571~{\rm m}}{0.503~{\rm mm^2}} \\ &\approx 0.0556~{\rm \Omega} \end{align*}
2. Calculate the coil inductance $L$.
\begin{align*} r &= 10~{\rm mm} = \frac{10}{25.4}~{\rm in} \approx 0.394~{\rm in} \\ l &= 22~{\rm mm} = \frac{22}{25.4}~{\rm in} \approx 0.866~{\rm in} \end{align*}
Using Wheeler's approximation:
\begin{align*} L[{\rm \mu H}] &\approx \frac{r^2 N^2}{9r + 10l} \\ &\approx \frac{(0.394)^2 \cdot 25^2}{9\cdot 0.394 + 10\cdot 0.866} \\ &\approx 7.94~{\rm \mu H} \end{align*}
3. Which DC voltage $U$ must be applied so that the stationary coil current becomes $I = 1~\rm A$? How large is the current density $j$ in the copper wire?
\begin{align*} U &= RI \\ &= 0.0556~{\rm \Omega} \cdot 1~{\rm A} \\ &= 0.0556~{\rm V} \end{align*}
The current density is
\begin{align*} j &= \frac{I}{A_{\rm Cu}} \\ &= \frac{1~{\rm A}}{0.503~{\rm mm^2}} \\ &\approx 1.99~{\rm A/mm^2} \end{align*}
4. How much energy is stored in the coil in the stationary state?
\begin{align*} W_{\rm mag} &= \frac{1}{2}LI^2 \\ &= \frac{1}{2}\cdot 7.94\cdot 10^{-6}~{\rm H}\cdot (1~{\rm A})^2 \\ &\approx 3.97\cdot 10^{-6}~{\rm J} \end{align*}
5. Sketch the coil current $i(t)$ when the coil is switched on.
\begin{align*} i(0) &= 0 \\ i(\infty) &= I = 1~{\rm A} \end{align*}
The current rises exponentially:
\begin{align*} i(t) &= I\left(1-e^{-t/\tau}\right) \end{align*}
with the time constant
\begin{align*} \tau = \frac{L}{R} \end{align*}
So the sketch starts at $0~\rm A$, rises steeply at first, and then approaches $1~\rm A$ asymptotically.
6. How long does it take until the current has practically reached its stationary final value?
\begin{align*} \tau &= \frac{L}{R} = \frac{7.94~{\rm \mu H}}{0.0556~{\rm \Omega}} \\ &\approx 1.43\cdot 10^{-4}~{\rm s} = 0.143~{\rm ms} \end{align*}
A useful engineering rule is: after about $5\tau$, the current has reached more than $99\%$ of its final value.
\begin{align*} t_{\rm practical} &\approx 5\tau \\ &\approx 5 \cdot 0.143~{\rm ms} \\ &\approx 0.714~{\rm ms} \end{align*}
Mathematically, the exact final value is reached only after infinite time.
7. How large is the energy dissipated as heat in the coil resistance during the current build-up?
For an RL switch-on process, the heat loss during current build-up is equal to the finally stored magnetic energy:
\begin{align*} W_{\rm loss} &= \frac{1}{2}LI^2 \\ &= \frac{1}{2}\cdot 7.94\cdot 10^{-6}~{\rm H}\cdot (1~{\rm A})^2 \\ &\approx 3.97\cdot 10^{-6}~{\rm J} \end{align*}