Differences

This shows you the differences between two versions of the page.

--- dummy [2026/03/27 01:53] – mexleadmin
+++ dummy [2026/05/26 13:48] (current) – removed mexleadmin
@@ Line 1: / Line 1: @@
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Machine-Vision Strobe: Capacitor Charging and Safe Discharge                    #@TaskText_HTML@#
-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*}
-. What voltage must be present across the capacitor so that it stores the required energy?
-#@PathBegin_HTML~1~@#
-\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}}} \\
-  &= 447.2~{\rm V}
-\end{align*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~1~@#
-\begin{align*}
-U = 447.2~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-. The charging current must not exceed $100~\rm mA$ at the beginning of charging. What minimum charging resistor is required?
-#@PathBegin_HTML~2~@#
-At $t=0$, the capacitor is uncharged, therefore the maximum charging current is
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~2~@#
-\begin{align*}
-R \ge 4.47~{\rm k\Omega}
-\end{align*}
-#@ResultEnd_HTML@#
-. How long does the charging process take in practice?
-#@PathBegin_HTML~3~@#
-The time constant is
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~3~@#
-\begin{align*}
-t_{\rm charge} \approx 22.35~{\rm ms}
-\end{align*}
-#@ResultEnd_HTML@#
-. Sketch the capacitor voltage and the resistor voltage during charging.
-#@PathBegin_HTML~4~@#
-The capacitor voltage rises exponentially:
-\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 $0~\rm V$.
-#@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) &= 447.2e^{-t/4.47{\rm ms}}~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-. 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?
-#@PathBegin_HTML~5~@#
-Half energy means
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~5~@#
-\begin{align*}
-U' &= 316.2~{\rm V} \\
-t  &= 3.47~{\rm s}
-\end{align*}
-#@ResultEnd_HTML@#
-. 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?
-#@PathBegin_HTML~6~@#
-Using the same resistor, the discharge time is again approximately
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~6~@#
-\begin{align*}
-t_{\rm discharge} \approx 22.35~{\rm ms} \\
-W_R = 0.1~{\rm Ws}
-\end{align*}
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#
-{{tag>rc_circuit thevenin_equivalent transient_response sensor_interface industrial_electronics chapter1_1}}
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Industrial Sensor Interface: Source, T-Network and Capacitor                    #@TaskText_HTML@#
-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.
-. What is the capacitor voltage after it is fully charged?
-#@PathBegin_HTML~1~@#
-With the switch open, use the equivalent source voltage:
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~1~@#
-\begin{align*}
-U_C = U_{0e} = 10~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-. How long does the charging process take in practice?
-#@PathBegin_HTML~2~@#
-The equivalent internal resistance 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~{\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~2~@#
-\begin{align*}
-R_{ie} = 5~{\rm k\Omega} \\
-t_{\rm charge} \approx 50~{\rm ms}
-\end{align*}
-#@ResultEnd_HTML@#
-. Give the time-dependent capacitor voltage during charging.
-#@PathBegin_HTML~3~@#
-The capacitor charges exponentially toward $U_{0e}=10~\rm V$:
-\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*}
-#@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@#
-. After the capacitor is fully charged, the switch is closed and the load resistor is connected.
-What is the new stationary capacitor voltage?
-#@PathBegin_HTML~4~@#
-Now use a second equivalent source.
-With the load resistor connected, the new stationary voltage is
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~4~@#
-\begin{align*}
-U_C = U_{0e}' = 5~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-. How long does it take in practice until the new stationary state is reached?
-#@PathBegin_HTML~5~@#
-The new equivalent internal resistance is
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~5~@#
-\begin{align*}
-R_{ie}' = 2.5~{\rm k\Omega} \\
-t_{\rm settle} \approx 25~{\rm ms}
-\end{align*}
-#@ResultEnd_HTML@#
-. Give the time-dependent capacitor voltage after the load is connected.
-#@PathBegin_HTML~6~@#
-At the switching instant, the capacitor voltage cannot jump.
-So immediately after closing the switch:
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~6~@#
-\begin{align*}
-u_C(t) = 5 + 5e^{-t/5{\rm ms}}~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#
-{{tag>inductors air_core_coil magnetic_field sensor_calibration transient_response current_density chapter1_1}}
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Hall-Sensor Test Bench: Air-Core Calibration Coil                    #@TaskText_HTML@#
-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.
-. Calculate the coil resistance $R$ at room temperature.
-#@PathBegin_HTML~1~@#
-The cross-sectional area of the copper wire 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 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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~1~@#
-\begin{align*}
-R = 55.6~{\rm m\Omega}
-\end{align*}
-#@ResultEnd_HTML@#
-. Calculate the coil inductance $L$.
-#@PathBegin_HTML~2~@#
-The coil cross-sectional area is
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~2~@#
-\begin{align*}
-L = 7.71~{\rm \mu H}
-\end{align*}
-#@ResultEnd_HTML@#
-. 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?
-#@PathBegin_HTML~3~@#
-In the stationary DC state, the coil behaves like a resistor:
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~3~@#
-\begin{align*}
-U &= 55.6~{\rm mV} \\
-j &= 1.99~{\rm A/mm^2}
-\end{align*}
-#@ResultEnd_HTML@#
-. How much magnetic energy is stored in the coil in the stationary state?
-#@PathBegin_HTML~4~@#
-\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 J}
-\end{align*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~4~@#
-\begin{align*}
-W_m = 3.86\cdot 10^{-6}~{\rm Ws}
-\end{align*}
-#@ResultEnd_HTML@#
-. Sketch the coil current $i(t)$ when the coil is switched on.
-#@PathBegin_HTML~5~@#
-A coil current cannot jump instantaneously.
-It starts at zero and rises gradually toward the final value $I$.
-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.
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~5~@#
-\begin{align*}
-i(t) = I\left(1-e^{-t/T}\right)
-\end{align*}
-#@ResultEnd_HTML@#
-. How long does it take in practice until the coil current has reached its final value?
-#@PathBegin_HTML~6~@#
-First compute the time constant:
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~6~@#
-\begin{align*}
-t_{\rm practical} \approx 695~{\rm \mu s}
-\end{align*}
-#@ResultEnd_HTML@#
-. How much energy is dissipated as heat in the winding resistance during the current build-up?
-#@PathBegin_HTML~7~@#
-The resistor loss during switch-on is
-\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*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~7~@#
-\begin{align*}
-W_R \approx 27.05\cdot 10^{-6}~{\rm Ws}
-\end{align*}
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#