Differences

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--- dummy [2026/03/27 01:46] – mexleadmin
+++ dummy [2026/05/26 13:48] (current) – removed mexleadmin
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-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Machine-Vision Strobe: Capacitor Charging and Safe Discharge                    #@TaskText_HTML@#
-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.
-. What voltage must be present across the capacitor so that it stores the required energy?
-#@PathBegin_HTML~1~@#
-\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~@#
-\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?
-#@PathBegin_HTML~2~@#
-\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~@#
-\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?
-#@PathBegin_HTML~3~@#
-\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 $5\tau$.
-\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.
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~3~@#
-\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.
-#@PathBegin_HTML~4~@#
-\begin{align*}
-u_C(t) &= U_0 \left(1 - e^{-t/(RC)}\right) \\
-u_R(t) &= U_0 e^{-t/(RC)}
-\end{align*}
-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.
-#@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@#
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Industrial Sensor Interface: Buffered Measurement Node                    #@TaskText_HTML@#
-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.
-. What is the voltage across the capacitor after it is fully charged?
-#@PathBegin_HTML~1~@#
-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*}
-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~@#
-\begin{align*}
-U_C(\infty) = 10~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-. How long does the charging process take?
-#@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@#