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Exercise E1 Machine-Vision Strobe: Capacitor Charging and Safe Discharge

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.

1. What voltage must be present across the capacitor so that it stores the required energy?

Solution

\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*}

Result

\begin{align*} U_C \approx 447~{\rm V} \end{align*}

2. The charging current must not exceed $100~\rm mA$ at the start of charging. What series charging resistor is required?

Solution

\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 $4.7~{\rm k\Omega}$.

Result

\begin{align*} R \approx 4.47~{\rm k\Omega} \end{align*}

3. How long does the charging process take until the capacitor is considered practically fully charged?

Solution

\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.

Result

\begin{align*} t_{\rm charge} \approx 22.4~{\rm ms} \qquad \text{(practically, }5\tau\text{)} \end{align*}

4. Sketch the capacitor voltage and the voltage across the charging resistor during charging.

Solution

\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.

Result

\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*}

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?

Solution

\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*}

Result

\begin{align*} t_{\frac{W_0}{2}} \approx 3.47~{\rm s} \\ u_C \approx 316~{\rm V} \end{align*}

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?

Solution

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.

Result

\begin{align*} t_{\rm discharge} \approx 22.4~{\rm ms} \qquad \text{(practically, }5\tau\text{)} \\ W_R = 0.1~{\rm J} \end{align*}