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--- dummy [2026/03/27 01:46] – mexleadmin
+++ dummy [2026/04/20 09:22] (current) – mexleadmin
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-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Machine-Vision Strobe: Capacitor Charging and Safe Discharge                    #@TaskText_HTML@#
+{{drawio>diagram99.png}}
-A high-speed machine-vision inspection system on a production line uses a short high-voltage flash pulse.
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Machine-Vision Strobe Unit: Charging and Safe Discharge of a Flash Capacitor #@TaskText_HTML@#
-For this purpose, an energy-storage capacitor is charged by a regulated DC high-voltage source and must be safely discharged before maintenance.
-Data:
+A machine-vision inspection system on a production line uses a short high-voltage flash pulse. For this purpose, an energy-storage capacitor is charged from a DC source and must be safely discharged before maintenance.
-\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.
+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*}
-. What voltage must be present across the capacitor so that it stores the required energy?
+. What voltage must the capacitor have so that it stores the required energy? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~101~@# <WRAP leftalign> \begin{align*} W_e &= \frac{1}{2} C U^2 \\ U &= \sqrt{\frac{2W_e}{C}} = \sqrt{\frac{2 \cdot 0.1~{\rm J}}{1 \cdot 10^{-6}~{\rm F}}} \\ &= \sqrt{200000}~{\rm V} \approx 447.2~{\rm V} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~101~@# \begin{align*} U = 447.2~{\rm V} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-#@PathBegin_HTML~1~@#
+. The charging current must not exceed $100~\rm mA$ at the start of charging. What charging resistor is required? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~102~@# <WRAP leftalign> At the beginning of charging, the capacitor behaves like a short circuit, so \begin{align*} i_{C{\rm 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}} \\ &\approx 4472~{\rm \Omega} = 4.47~{\rm k\Omega} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~102~@# \begin{align*} R \ge 4.47~{\rm k\Omega} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-\begin{align*}
-W &= \frac{1}{2} C U^2 \\
-U &= \sqrt{\frac{2W}{C}}
-   = \sqrt{\frac{2 \cdot 0.1~{\rm J}}{1 \cdot 10^{-6}~{\rm F}}} \\
-  &= \sqrt{200000}~{\rm V}
-   \approx 447.2~{\rm V}
-\end{align*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~1~@#
+. How long does the charging process take until the capacitor is practically fully charged? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~103~@# <WRAP leftalign> The time constant is \begin{align*} T = RC = 4.47~{\rm k\Omega} \cdot 1~{\rm \mu F} = 4.47~{\rm ms} \end{align*} In engineering practice, a capacitor is considered practically fully charged after about $5T$: \begin{align*} t \approx 5T = 5 \cdot 4.47~{\rm ms} = 22.35~{\rm ms} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~103~@# \begin{align*} t \approx 22.35~{\rm ms} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-\begin{align*}
-U_C \approx 447~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-. The charging current must not exceed $100~\rm mA$ at the start of charging. What series charging resistor is required?
+. Give the time-dependent capacitor voltage and the voltage across the charging resistor. <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~104~@# <WRAP leftalign> For the charging process: \begin{align*} u_C(t) &= U\left(1-e^{-t/T}\right) \\ u_R(t) &= Ue^{-t/T} \end{align*} with \begin{align*} U &= 447.2~{\rm V} \\ T &= 4.47~{\rm ms} \end{align*} So the capacitor voltage rises exponentially from $0$ to $447.2~\rm V$, while the resistor voltage falls exponentially from $447.2~\rm V$ to $0$. </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~104~@# \begin{align*} u_C(t) &= 447.2\left(1-e^{-t/4.47{\rm ms}}\right)~{\rm V} \\ u_R(t) &= 447.2\,e^{-t/4.47{\rm ms}}~{\rm V} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-#@PathBegin_HTML~2~@#
+. 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? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~105~@# <WRAP leftalign> Half the energy means \begin{align*} W_e' = 0.5W_e \end{align*} Since \begin{align*} W_e = \frac{1}{2}CU^2 \end{align*} the voltage at half energy is \begin{align*} U' = \frac{U}{\sqrt{2}} = \frac{447.2~{\rm V}}{\sqrt{2}} = 316.2~{\rm V} \end{align*} For the discharge through the internal resistance: \begin{align*} u_C(t) = Ue^{-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) \\ &\approx 3.47~{\rm s} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~105~@# \begin{align*} U' = 316.2~{\rm V} \\ t = 3.47~{\rm s} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-\begin{align*}
-I_{\rm max} &= \frac{U}{R} \\
-R &= \frac{U}{I_{\rm max}}
-   = \frac{447.2~{\rm V}}{0.1~{\rm A}}
-   \approx 4472~{\rm \Omega}
-\end{align*}
-\end{align*}
-A practical standard value would be about $4.7~{\rm k\Omega}$.
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~2~@#
+. 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? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~106~@# <WRAP leftalign> The discharge time constant through the same resistor is again \begin{align*} T = RC = 4.47~{\rm ms} \end{align*} Thus the practical discharge time is \begin{align*} t \approx 5T = 22.35~{\rm ms} \end{align*} The complete stored capacitor energy is converted into heat in the resistor: \begin{align*} W_R = W_e = 0.1~{\rm Ws} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~106~@# \begin{align*} t \approx 22.35~{\rm ms} \\ W_R = 0.1~{\rm Ws} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-\begin{align*}
-R \approx 4.47~{\rm k\Omega}
-\end{align*}
-#@ResultEnd_HTML@#
-. How long does the charging process take until the capacitor is considered practically fully charged?
+#@TaskEnd_HTML@#
-#@PathBegin_HTML~3~@#
+{{tag>rc_circuit thevenin_equivalent transient_response sensor_interface industrial_electronics chapter1_1}}{{include_n>300}}
-\begin{align*}
-\tau &= RC
-     = 4.472~{\rm k\Omega} \cdot 1~{\rm \mu F}
-     \approx 4.47~{\rm ms}
-\end{align*}
-In engineering practice, a capacitor is usually considered fully charged after about $5\tau$.
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Sensor Input Buffer: Source, T-Network and Capacitor #@TaskText_HTML@#
-\begin{align*}
+A 12 V industrial sensor electronics unit feeds a buffered measurement node through a resistor T-network. A capacitor smooths the node voltage. At first, the load is disconnected. After the capacitor is fully charged, a measurement load is connected by a switch.
-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.
+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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~3~@#
+Initially, the capacitor is uncharged and the switch is open.
-\begin{align*}
-t_{\rm charge} \approx 22.4~{\rm ms} \qquad \text{(practically, }5\tau\text{)}
-\end{align*}
-#@ResultEnd_HTML@#
-. Sketch the capacitor voltage and the voltage across the charging resistor during charging.
+. What is the capacitor voltage after it is fully charged? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~201~@# <WRAP leftalign> Using the equivalent voltage source of the network on the left-hand side, the open-circuit voltage is \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*} After full charging, the capacitor voltage equals this voltage. </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~201~@# \begin{align*} U_C = U_{0e} = 10~{\rm V} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-#@PathBegin_HTML~4~@#
+. How long does the charging process take? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~202~@# <WRAP leftalign> The internal resistance seen by the capacitor is \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} \\ &= 5.00~{\rm k\Omega} \end{align*} So the time constant is \begin{align*} T &= R_{ie}C = 5.00~{\rm k\Omega}\cdot 2~{\rm \mu F} = 10~{\rm ms} \end{align*} Practical charging time: \begin{align*} t \approx 5T = 50~{\rm ms} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~202~@# \begin{align*} R_{ie} = 5.00~{\rm k\Omega} \\ t \approx 50~{\rm ms} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-\begin{align*}
-u_C(t) &= U_0 \left(1 - e^{-t/(RC)}\right) \\
-u_R(t) &= U_0 e^{-t/(RC)}
-\end{align*}
-with
+. Give the time-dependent capacitor voltage. <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~203~@# <WRAP leftalign> The charging law is \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*} So the capacitor voltage rises exponentially from $0~\rm V$ to $10~\rm V$. </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~203~@# \begin{align*} u_C(t) = 10\left(1-e^{-t/10{\rm ms}}\right)~{\rm V} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-\begin{align*}
+. After the capacitor is fully charged, the switch is closed and the load resistor is connected. What is the stationary load voltage? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~204~@# <WRAP leftalign> Now use a second equivalent voltage-source step. The Thevenin source seen by the load has \begin{align*} U_{0e} &= 10~{\rm V} \\ R_{ie} &= 5.00~{\rm k\Omega} \end{align*} Thus, the stationary load voltage is \begin{align*} U_C' = 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*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~204~@# \begin{align*} U_L = 5~{\rm V} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-U_0 &= 447.2~{\rm V} \\
-RC  &= 4.47~{\rm ms}
-\end{align*}
-Initial and final values:
+. How long does it take until this new stationary state is practically reached? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~205~@# <WRAP leftalign> The new internal resistance is \begin{align*} R_{ie}' &= R_{ie}\parallel R_L \\ &= 5.00~{\rm k\Omega}\parallel 5.00~{\rm k\Omega} \\ &= 2.50~{\rm k\Omega} \end{align*} Hence the new time constant is \begin{align*} T' &= R_{ie}'C = 2.50~{\rm k\Omega}\cdot 2~{\rm \mu F} = 5~{\rm ms} \end{align*} Practical settling time: \begin{align*} t \approx 5T' = 25~{\rm ms} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~205~@# \begin{align*} R_{ie}' = 2.50~{\rm k\Omega} \\ t \approx 25~{\rm ms} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-\begin{align*}
+. Give the time-dependent load voltage after the switch is closed. <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~206~@# <WRAP leftalign> At the switching instant, the capacitor voltage cannot jump. Therefore: \begin{align*} u_L(0^+) &= 10~{\rm V} \\ u_L(\infty) &= 5~{\rm V} \end{align*} The voltage therefore decays exponentially toward the new final value: \begin{align*} u_L(t) &= u_L(\infty) + \left(u_L(0^+)-u_L(\infty)\right)e^{-t/T'} \\ &= 5 + 5e^{-t/5{\rm ms}}~{\rm V} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~206~@# \begin{align*} u_L(t) = 5 + 5e^{-t/5{\rm ms}}~{\rm V} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-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.
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~4~@#
-\begin{align*}
-u_C(t) &= 447.2 \left(1 - e^{-t/4.47{\rm ms}}\right)~{\rm V} \\
-u_R(t) &= 447.2\, e^{-t/4.47{\rm ms}}~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-. 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?
-#@PathBegin_HTML~5~@#
-\begin{align*}
-u_C(t) &= U_0 e^{-t/(R_i C)}
-\end{align*}
-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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~5~@#
-\begin{align*}
-t_{\frac{W_0}{2}} \approx 3.47~{\rm s} \\
-u_C \approx 316~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-. 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?
-#@PathBegin_HTML~6~@#
-For discharge through the same 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.
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~6~@#
-\begin{align*}
-t_{\rm discharge} \approx 22.4~{\rm ms} \qquad \text{(practically, }5\tau\text{)} \\
-W_R = 0.1~{\rm J}
-\end{align*}
-#@ResultEnd_HTML@#
 #@TaskEnd_HTML@#
+{{tag>inductors air_core_coil magnetic_field hall_sensor transient_response current_density chapter1_1}}{{include_n>300}}
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Hall-Sensor Calibration Coil: Short Air-Core Coil #@TaskText_HTML@#
+A Hall-sensor calibration bench uses a short air-core coil to create a defined magnetic field. An air-core coil is chosen because it avoids hysteresis and remanence effects. The coil is wound as a short cylindrical coil.
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Industrial Sensor Interface: Buffered Measurement Node                    #@TaskText_HTML@#
+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*}
-A 12 V industrial sensor module feeds a buffered measurement node through a resistor T-network.
+A DC current of $1~\rm A$ shall flow through the coil.
-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:
+. Calculate the coil resistance $R$ at room temperature. <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~301~@# <WRAP leftalign> The wire cross section is \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 approximated by the number of turns times the circumference: \begin{align*} l_{\rm Cu} &= N\pi d \\ &= 25\pi \cdot 20~{\rm mm} \\ &= 1570.8~{\rm mm} = 1.571~{\rm m} \end{align*} Thus, \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*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~301~@# \begin{align*} R = 55.6~{\rm m\Omega} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-\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.
+. Calculate the coil inductance $L$. <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~302~@# <WRAP leftalign> For this short air-core coil, use \begin{align*} L = N^2 \cdot \frac{\mu_0 A}{l}\cdot\frac{1}{1+\frac{d}{2l}} \end{align*} with \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} \\ \mu_0 &= 4\pi\cdot 10^{-7}~{\rm Vs/(Am)} \end{align*} Therefore, \begin{align*} L &= 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}} \\ &\approx 7.71\cdot 10^{-6}~{\rm H} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~302~@# \begin{align*} L = 7.71~{\rm \mu H} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-. What is the voltage across the capacitor after it is fully charged?
+. Which DC voltage must be applied so that the stationary current becomes $I=1~\rm A$? How large is the current density $j$ in the copper wire? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~303~@# <WRAP leftalign> In the stationary DC state, the coil behaves like its ohmic resistance: \begin{align*} U &= RI \\ &= 55.6~{\rm m\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}} \\ &\approx 1.99~{\rm A/mm^2} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~303~@# \begin{align*} U = 55.6~{\rm mV} \\ j = 1.99~{\rm A/mm^2} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-#@PathBegin_HTML~1~@#
+. How much magnetic energy is stored in the coil in the stationary state? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~304~@# <WRAP leftalign> \begin{align*} W_m &= \frac{1}{2}LI^2 \\ &= \frac{1}{2}\cdot 7.71\cdot 10^{-6}~{\rm H}\cdot (1~{\rm A})^2 \\ &= 3.86\cdot 10^{-6}~{\rm Ws} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~304~@# \begin{align*} W_m = 3.86\cdot 10^{-6}~{\rm Ws} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-With the switch open, no DC current flows through $R_3$ after the capacitor is fully charged.
-Therefore, there is no voltage drop across $R_3$, and the capacitor voltage is equal to the divider voltage of $R_1$ and $R_2$:
-\begin{align*}
+. Give the time-dependent coil current $i(t)$ when the coil is switched on. <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~305~@# <WRAP leftalign> A coil current cannot jump instantly. It starts at $0$ and approaches the final value $I=1~\rm A$ exponentially: \begin{align*} i(t) = I\left(1-e^{-t/T}\right) \end{align*} So the sketch starts at $0~\rm A$, rises quickly, and then slowly approaches $1~\rm A$. </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~305~@# \begin{align*} i(t) = 1\left(1-e^{-t/T}\right)~{\rm A} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~1~@#
+. How long does it take until the current has practically reached its stationary value? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~306~@# <WRAP leftalign> The time constant is \begin{align*} T &= \frac{L}{R} = \frac{7.71~{\rm \mu H}}{55.6~{\rm m\Omega}} \\ &\approx 138.9~{\rm \mu s} \end{align*} A practical final value is reached after about $5T$: \begin{align*} t \approx 5T = 5\cdot 138.9~{\rm \mu s} \approx 695~{\rm \mu s} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~306~@# \begin{align*} t \approx 695~{\rm \mu s} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-\begin{align*}
-U_C(\infty) = 10~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-. How long does the charging process take?
+. How much energy is dissipated as heat in the coil resistance during the current build-up? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~307~@# <WRAP leftalign> Using the current from task 5, \begin{align*} i(t) = I\left(1-e^{-t/T}\right) \end{align*} the heat dissipated in the winding resistance up to the practical final time $5T$ is \begin{align*} W_R &= \int_0^{5T} R\,i^2(t)\,dt \\ &= R I^2 \int_0^{5T} \left(1-e^{-t/T}\right)^2 dt \end{align*} For this interval, the integral is approximately \begin{align*} \int_0^{5T} \left(1-e^{-t/T}\right)^2 dt \approx \frac{7}{2}T \end{align*} Thus, \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 138.9~{\rm \mu s} \\ &\approx 27.05\cdot 10^{-6}~{\rm Ws} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~307~@# \begin{align*} W_R \approx 27.05\cdot 10^{-6}~{\rm Ws} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP>
-#@PathBegin_HTML~2~@#
-For the charging process, the capacitor sees the Thevenin resistance of the left-hand network.
-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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~2~@#
-\begin{align*}
-\tau_1 = 10~{\rm ms} \\
-t_{\rm charge} \approx 50~{\rm ms}
-\end{align*}
-#@ResultEnd_HTML@#
-. Draw the time-dependent capacitor voltage.
-#@PathBegin_HTML~3~@#
-The charging equation of the capacitor is
-\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 $10~{\rm V}$.
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~3~@#
-\begin{align*}
-u_C(t) = 10\left(1 - e^{-t/(10~{\rm ms})}\right)~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-. 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?
-#@PathBegin_HTML~4~@#
-Use the Thevenin equivalent of the left-hand network as seen from the capacitor/load node:
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~4~@#
-\begin{align*}
-U_L(\infty) = 5~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-. How long does it take until this stationary state is reached?
-#@PathBegin_HTML~5~@#
-After the switch is closed, the capacitor sees the equivalent resistance
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~5~@#
-\begin{align*}
-\tau_2 = 5~{\rm ms} \\
-t_{\rm settle} \approx 25~{\rm ms}
-\end{align*}
-#@ResultEnd_HTML@#
-. Draw the time-dependent load voltage.
-#@PathBegin_HTML~6~@#
-At the instant the switch closes, the capacitor voltage cannot jump.
-Therefore, the initial load voltage is
-\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 $5~{\rm V}$.
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~6~@#
-\begin{align*}
-u_L(t) = 5 + 5e^{-t/(5~{\rm ms})}~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
 #@TaskEnd_HTML@#
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Hall-Sensor Test Bench: Air-Core Calibration Coil                    #@TaskText_HTML@#
-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.
-. Calculate the coil resistance $R$ at room temperature.
-#@PathBegin_HTML~1~@#
-First, determine the copper cross-sectional area:
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~1~@#
-\begin{align*}
-R \approx 0.0556~{\rm \Omega} = 55.6~{\rm m\Omega}
-\end{align*}
-#@ResultEnd_HTML@#
-. Calculate the coil inductance $L$.
-#@PathBegin_HTML~2~@#
-First convert radius and coil length to inches:
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~2~@#
-\begin{align*}
-L \approx 7.94~{\rm \mu H}
-\end{align*}
-#@ResultEnd_HTML@#
-. 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?
-#@PathBegin_HTML~3~@#
-In the stationary DC state, the inductance no longer affects the current.
-Then only the ohmic resistance remains:
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~3~@#
-\begin{align*}
-U \approx 55.6~{\rm mV} \\
-j \approx 1.99~{\rm A/mm^2}
-\end{align*}
-#@ResultEnd_HTML@#
-. How much energy is stored in the coil in the stationary state?
-#@PathBegin_HTML~4~@#
-The magnetic energy stored in an inductor is
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~4~@#
-\begin{align*}
-W_{\rm mag} \approx 3.97~{\rm \mu J}
-\end{align*}
-#@ResultEnd_HTML@#
-. Sketch the coil current $i(t)$ when the coil is switched on.
-#@PathBegin_HTML~5~@#
-A coil does not allow its current to jump instantly.
-Therefore:
-\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.
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~5~@#
-\begin{align*}
-i(t) = 1\left(1-e^{-t/\tau}\right)~{\rm A}
-\end{align*}
-with an exponential rise from $0~\rm A$ to $1~\rm A$.
-#@ResultEnd_HTML@#
-. How long does it take until the current has practically reached its stationary final value?
-#@PathBegin_HTML~6~@#
-First compute the time constant:
-\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.
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~6~@#
-\begin{align*}
-\tau \approx 0.143~{\rm ms} \\
-t_{\rm practical} \approx 0.714~{\rm ms}
-\end{align*}
-#@ResultEnd_HTML@#
-. How large is the energy dissipated as heat in the coil resistance during the current build-up?
-#@PathBegin_HTML~7~@#
-During the switching-on process, part of the supplied energy is stored in the magnetic field and part is dissipated as heat in the winding resistance.
-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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~7~@#
-\begin{align*}
-W_{\rm loss} \approx 3.97~{\rm \mu J}
-\end{align*}
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#

rc circuit thevenin equivalent transient response sensor interface industrial electronics chapter1 1 inductors air core coil magnetic field hall sensor current density