Differences

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--- dummy [2026/03/27 02:16] – mexleadmin
+++ dummy [2026/05/26 13:48] (current) – removed mexleadmin
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-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Machine-Vision Strobe Unit: Charging and Safe Discharge of a Flash Capacitor                    #@TaskText_HTML@#
-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.
-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 the capacitor have so that it stores the required energy?
-<WRAP group>
-<WRAP half column rightalign>
-#@PathBegin_HTML~101~@#
-\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*}
-#@PathEnd_HTML@#
-</WRAP>
-<WRAP half column>
-#@ResultBegin_HTML~101~@#
-\begin{align*}
-U = 447.2~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-</WRAP>
-</WRAP>
-. 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~@#
-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*}
-#@PathEnd_HTML@#
-</WRAP>
-<WRAP half column>
-#@ResultBegin_HTML~102~@#
-\begin{align*}
-R \ge 4.47~{\rm k\Omega}
-\end{align*}
-#@ResultEnd_HTML@#
-</WRAP>
-</WRAP>
-. How long does the charging process take until the capacitor is practically fully charged?
-<WRAP group>
-<WRAP half column rightalign>
-#@PathBegin_HTML~103~@#
-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*}
-#@PathEnd_HTML@#
-</WRAP>
-<WRAP half column>
-#@ResultBegin_HTML~103~@#
-\begin{align*}
-t \approx 22.35~{\rm ms}
-\end{align*}
-#@ResultEnd_HTML@#
-</WRAP>
-</WRAP>
-. Give the time-dependent capacitor voltage and the voltage across the charging resistor.
-<WRAP group>
-<WRAP half column rightalign>
-#@PathBegin_HTML~104~@#
-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 $0$.
-#@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>
-. 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~@#
-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*}
-#@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>
-. 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~@#
-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*}
-#@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>
-#@TaskEnd_HTML@#
-{{tag>rc_circuit thevenin_equivalent transient_response sensor_interface industrial_electronics chapter1_1}}
-{{include_n>300}}
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Sensor Input Buffer: Source, T-Network and Capacitor                    #@TaskText_HTML@#
-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.
-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.
-. What is the capacitor voltage after it is fully charged?
-<WRAP group>
-<WRAP half column rightalign>
-#@PathBegin_HTML~201~@#
-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.
-#@PathEnd_HTML@#
-</WRAP>
-<WRAP half column>
-#@ResultBegin_HTML~201~@#
-\begin{align*}
-U_C = U_{0e} = 10~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-</WRAP>
-</WRAP>
-. How long does the charging process take?
-<WRAP group>
-<WRAP half column rightalign>
-#@PathBegin_HTML~202~@#
-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*}
-#@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>
-. Give the time-dependent capacitor voltage.
-<WRAP group>
-<WRAP half column rightalign>
-#@PathBegin_HTML~203~@#
-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 $10~\rm V$.
-#@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>
-. 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~@#
-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*}
-#@PathEnd_HTML@#
-</WRAP>
-<WRAP half column>
-#@ResultBegin_HTML~204~@#
-\begin{align*}
-U_L = 5~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-</WRAP>
-</WRAP>
-. How long does it take until this new stationary state is practically reached?
-<WRAP group>
-<WRAP half column rightalign>
-#@PathBegin_HTML~205~@#
-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*}
-#@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>
-. Give the time-dependent load voltage after the switch is closed.
-<WRAP group>
-<WRAP half column rightalign>
-#@PathBegin_HTML~206~@#
-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*}
-#@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>
-#@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.
-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 DC current of $1~\rm A$ shall flow through the coil.
-. Calculate the coil resistance $R$ at room temperature.
-<WRAP group>
-<WRAP half column rightalign>
-#@PathBegin_HTML~301~@#
-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*}
-#@PathEnd_HTML@#
-</WRAP>
-<WRAP half column>
-#@ResultBegin_HTML~301~@#
-\begin{align*}
-R = 55.6~{\rm m\Omega}
-\end{align*}
-#@ResultEnd_HTML@#
-</WRAP>
-</WRAP>
-. Calculate the coil inductance $L$.
-<WRAP group>
-<WRAP half column rightalign>
-#@PathBegin_HTML~302~@#
-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*}
-#@PathEnd_HTML@#
-</WRAP>
-<WRAP half column>
-#@ResultBegin_HTML~302~@#
-\begin{align*}
-L = 7.71~{\rm \mu H}
-\end{align*}
-#@ResultEnd_HTML@#
-</WRAP>
-</WRAP>
-. 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~@#
-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*}
-#@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>
-. How much magnetic energy is stored in the coil in the stationary state?
-<WRAP group>
-<WRAP half column rightalign>
-#@PathBegin_HTML~304~@#
-\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*}
-#@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>
-. Give the time-dependent coil current $i(t)$ when the coil is switched on.
-<WRAP group>
-<WRAP half column rightalign>
-#@PathBegin_HTML~305~@#
-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 $1~\rm A$.
-#@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>
-. How long does it take until the current has practically reached its stationary value?
-<WRAP group>
-<WRAP half column rightalign>
-#@PathBegin_HTML~306~@#
-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*}
-#@PathEnd_HTML@#
-</WRAP>
-<WRAP half column>
-#@ResultBegin_HTML~306~@#
-\begin{align*}
-t \approx 695~{\rm \mu s}
-\end{align*}
-#@ResultEnd_HTML@#
-</WRAP>
-</WRAP>
-. 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~@#
-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*}
-#@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>
-#@TaskEnd_HTML@#