{{tag>capacitors rc_circuit transient_response energy_storage industrial_safety chapter1_1}}
{{include_n>300}}

#@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*}

1. 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>

2. 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>

3. 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>

4. 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>

5. 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>

6. 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>

#@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.

1. 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>

2. 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>

3. 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>

4. 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>

5. 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>

6. 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>

#@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.

1. 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>

2. 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>

3. 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>

4. 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>

5. 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>

6. 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>

7. 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>

#@TaskEnd_HTML@#