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Exercise E1 Machine-Vision Strobe: Capacitor Charging and Safe Discharge
A machine-vision inspection unit on a production line uses a high-voltage flash pulse. For this purpose, an energy-storage capacitor is charged by a DC source and later discharged safely before maintenance.
Data: \begin{align*} C &= 1~{\rm \mu F} \\ W_e &= 0.1~{\rm J} \\ I_{\rm max} &= 100~{\rm mA} \\ R_i &= 10~{\rm M\Omega} \end{align*}
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 beginning of charging. What minimum charging resistor is required?
\begin{align*} i_{C,\max} = i_C(t=0) = \frac{U}{R} \end{align*}
Thus,
\begin{align*} R &\ge \frac{U}{I_{\rm max}} = \frac{447.2~{\rm V}}{0.1~{\rm A}} \\ &= 4472~{\rm \Omega} = 4.47~{\rm k\Omega} \end{align*}
3. How long does the charging process take in practice?
\begin{align*} T = RC = 4.47~{\rm k\Omega} \cdot 1~{\rm \mu F} = 4.47~{\rm ms} \end{align*}
A practical engineering approximation is that the capacitor is essentially charged after about $5T$:
\begin{align*} t_{\rm charge} \approx 5T = 5RC = 22.35~{\rm ms} \end{align*}
4. Sketch the capacitor voltage and the resistor voltage during charging.
\begin{align*} u_C(t) = U \left(1-e^{-t/T}\right) \end{align*}
The resistor voltage falls exponentially:
\begin{align*} u_R(t) = Ue^{-t/T} \end{align*}
with
\begin{align*} T = RC = 4.47~{\rm ms} \end{align*}
Thus, $u_C$ starts at $0~\rm V$ and approaches $447.2~\rm V$, while $u_R$ starts at $447.2~\rm V$ and falls to $0~\rm V$.
5. After charging, the capacitor is disconnected from the source. Due to leakage, it can be modeled with an internal resistance of $10~\rm M\Omega$. After what time has the stored energy fallen to one half, and what is the capacitor voltage then?
\begin{align*} W_e' = 0.5W_e \end{align*}
Since capacitor energy is proportional to $U^2$:
\begin{align*} U' = \frac{U}{\sqrt{2}} = \frac{447.2~{\rm V}}{\sqrt{2}} = 316.2~{\rm V} \end{align*}
For discharge:
\begin{align*} u_C(t) = U e^{-t/T_2} \end{align*}
with
\begin{align*} T_2 = R_iC = 10~{\rm M\Omega}\cdot 1~{\rm \mu F} = 10~{\rm s} \end{align*}
Set $u_C(t)=U'$:
\begin{align*} Ue^{-t/T_2} &= U' \\ t &= T_2 \ln\left(\frac{U}{U'}\right) \\ &= 10~{\rm s}\cdot \ln\left(\frac{447.2}{316.2}\right) \\ &= 3.47~{\rm s} \end{align*}
6. The fully charged capacitor is discharged through the charging resistor before maintenance. How long does the discharge take in practice, and how much energy is converted into heat in the resistor?
\begin{align*} t_{\rm discharge} \approx 5T = 22.35~{\rm ms} \end{align*}
The complete initially stored capacitor energy is dissipated as heat in the resistor:
\begin{align*} W_R = W_e = 0.1~{\rm J} = 0.1~{\rm Ws} \end{align*}
Exercise E2 Industrial Sensor Interface: Source, T-Network and Capacitor
An industrial sensor module feeds a buffered measurement node through a T-network. A capacitor smooths the node voltage. At first, the load is disconnected. After the capacitor has been charged, a load resistor is connected by a switch.
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 is open.
1. What is the capacitor voltage after it is fully charged?
\begin{align*} U_{0e} = \frac{R_2}{R_1+R_2}\,U = \frac{10~{\rm k\Omega}}{2~{\rm k\Omega}+10~{\rm k\Omega}}\cdot 12~{\rm V} = 10~{\rm V} \end{align*}
2. How long does the charging process take in practice?
\begin{align*} R_{ie} &= R_3 + (R_1 \parallel R_2) \\ &= 3.33~{\rm k\Omega} + \frac{2~{\rm k\Omega}\cdot 10~{\rm k\Omega}}{2~{\rm k\Omega}+10~{\rm k\Omega}} \\ &= 3.33~{\rm k\Omega} + 1.67~{\rm k\Omega} \\ &= 5~{\rm k\Omega} \end{align*}
Thus,
\begin{align*} T &= R_{ie}C = 5~{\rm k\Omega}\cdot 2~{\rm \mu F} = 10~{\rm ms} \end{align*}
Practical charging time:
\begin{align*} t_{\rm charge} \approx 5T = 50~{\rm ms} \end{align*}
3. Give the time-dependent capacitor voltage during charging.
\begin{align*} u_C(t) = U_{0e}\left(1-e^{-t/T}\right) = 10\left(1-e^{-t/10{\rm ms}}\right)~{\rm V} \end{align*}
4. After the capacitor is fully charged, the switch is closed and the load resistor is connected. What is the new stationary capacitor voltage?
\begin{align*} U_{0e}' &= \frac{R_L}{R_{ie}+R_L}\,U_{0e} \\ &= \frac{5~{\rm k\Omega}}{5~{\rm k\Omega}+5~{\rm k\Omega}}\cdot 10~{\rm V} \\ &= 5~{\rm V} \end{align*}
5. How long does it take in practice until the new stationary state is reached?
\begin{align*} R_{ie}' = R_{ie}\parallel R_L = 5~{\rm k\Omega}\parallel 5~{\rm k\Omega} = 2.5~{\rm k\Omega} \end{align*}
Hence,
\begin{align*} T' &= R_{ie}'C = 2.5~{\rm k\Omega}\cdot 2~{\rm \mu F} = 5~{\rm ms} \end{align*}
Practical settling time:
\begin{align*} t_{\rm settle} \approx 5T' = 25~{\rm ms} \end{align*}
6. Give the time-dependent capacitor voltage after the load is connected.
\begin{align*} u_C(0^+) = 10~{\rm V} \end{align*}
The final value is
\begin{align*} u_C(\infty) = 5~{\rm V} \end{align*}
Therefore, the capacitor voltage decays exponentially:
\begin{align*} u_C(t) &= U_{0e}' + \left(U_{0e}-U_{0e}'\right)e^{-t/T'} \\ &= 5 + (10-5)e^{-t/5{\rm ms}} \\ &= 5 + 5e^{-t/5{\rm ms}}~{\rm V} \end{align*}
Exercise E3 Hall-Sensor Test Bench: Air-Core Calibration Coil
A small air-core cylindrical coil is used in a Hall-sensor test bench to generate a reproducible magnetic field. Because there is no iron core, the magnetic behaviour is easy to predict and no remanence occurs.
Data: \begin{align*} l &= 22~{\rm mm} \\ d &= 20~{\rm mm} \\ d_{\rm Cu} &= 0.8~{\rm mm} \\ N &= 25 \\ \rho_{\rm Cu} &= 0.0178~{\rm \Omega\,mm^2/m} \end{align*}
Use the following approximation for the inductance of the short air-core coil:
\begin{align*} L = N^2 \cdot \frac{\mu_0 A}{l} \cdot \frac{1}{1+\frac{d}{2l}} \end{align*}
The coil is then connected to a DC source.
1. Calculate the coil resistance $R$ at room temperature.
\begin{align*} A_{\rm Cu} &= \pi\left(\frac{d_{\rm Cu}}{2}\right)^2 = \pi(0.4~{\rm mm})^2 \\ &= 0.503~{\rm mm^2} \end{align*}
The total wire length is approximately
\begin{align*} l_{\rm Cu} &= N\pi d = 25\pi \cdot 20~{\rm mm} \\ &= 1570.8~{\rm mm} = 1.571~{\rm m} \end{align*}
Now the resistance is
\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}} \\ &= 0.0556~{\rm \Omega} \end{align*}
2. Calculate the coil inductance $L$.
\begin{align*} A &= \pi\left(\frac{d}{2}\right)^2 = \pi(10~{\rm mm})^2 \\ &= 314.16~{\rm mm^2} = 3.1416\cdot 10^{-4}~{\rm m^2} \end{align*}
With
\begin{align*} \mu_0 = 4\pi\cdot 10^{-7}~{\rm Vs/(Am)} \end{align*}
the inductance becomes
\begin{align*} L &= N^2 \cdot \frac{\mu_0 A}{l} \cdot \frac{1}{1+\frac{d}{2l}} \\ &= 25^2 \cdot \frac{4\pi\cdot 10^{-7}~{\rm Vs/(Am)} \cdot 3.1416\cdot 10^{-4}~{\rm m^2}}{22\cdot 10^{-3}~{\rm m}} \cdot \frac{1}{1+\frac{20}{2\cdot 22}} \\ &= 7.71\cdot 10^{-6}~{\rm H} \end{align*}
3. Which DC voltage must be applied so that the stationary current becomes $I=1~\rm A$? How large is the current density in the copper wire?
\begin{align*} U &= RI = 0.0556~{\rm \Omega}\cdot 1~{\rm A} = 55.6~{\rm mV} \end{align*}
The current density is
\begin{align*} j &= \frac{I}{A_{\rm Cu}} = \frac{1~{\rm A}}{0.503~{\rm mm^2}} = 1.99~{\rm A/mm^2} \end{align*}
4. How much magnetic energy is stored in the coil in the stationary state?
5. Sketch the coil current $i(t)$ when the coil is switched on.
For a DC step, the current is
\begin{align*} i(t) = I\left(1-e^{-t/T}\right) \end{align*}
with the time constant
\begin{align*} T = \frac{L}{R} \end{align*}
So the current starts at $0~\rm A$ and approaches $1~\rm A$ exponentially.
6. How long does it take in practice until the coil current has reached its final value?
\begin{align*} T &= \frac{L}{R} = \frac{7.71~{\rm \mu H}}{55.6~{\rm m\Omega}} \\ &\approx 138.7~{\rm \mu s} \end{align*}
In practice, the current is essentially at its final value after about $5T$:
\begin{align*} t_{\rm practical} \approx 5T \approx 5\cdot 138.7~{\rm \mu s} = 695~{\rm \mu s} \end{align*}
7. How much energy is dissipated as heat in the winding resistance during the current build-up?
\begin{align*} W_R &= \int_0^{5T} R\,i^2(t)\,dt \end{align*}
With
\begin{align*} i(t)=I\left(1-e^{-t/T}\right) \end{align*}
this becomes
\begin{align*} W_R &= R I^2 \int_0^{5T}\left(1-e^{-t/T}\right)^2 dt \end{align*}
Using the engineering approximation from the solution sheet,
\begin{align*} \int_0^{5T}\left(1-e^{-t/T}\right)^2 dt \approx \frac{7}{2}T \end{align*}
therefore
\begin{align*} W_R &\approx RI^2 \cdot \frac{7}{2}T \\ &= 0.0556~{\rm \Omega}\cdot (1~{\rm A})^2 \cdot \frac{7}{2}\cdot 138.7~{\rm \mu s} \\ &= 27.05\cdot 10^{-6}~{\rm J} \end{align*}