Chapter 2, Problem 2.1EP

Repeat Example 2.1 if the input voltage is ${\upsilon }_s\left(t\right)=12\sin \text{ω}t\left(\text{V}\right)$ , $V_B=4.5\text{V}$ , and $R=250\Omega$ . (Ans. $i_D\left(\text{peak}\right)=27.6\text{mA}$ , ${\upsilon }_R\left(\max \right)=16.5\text{V}$ , 36.0%)

To determine

The peak diode current, the maximum reverse-biased voltage of the diode and the fraction of the cycle over which diode conducts for a half-wave battery charging circuit.

Answer to Problem 2.1EP

The peak diode current is $i_{d(peak)}=27.6\text{ mA}$

Maximum reversed biased voltage of the diode is $V_{R(\max )}=16.5\text{ V}$

Fraction of the cycle diode is conducting is $36.02\%$

Explanation of Solution

Given Information:

The given values are

  $V_B=4.5\text{ V}$

  $v_s\left(t\right)=12\sin (\omega t)\text{ V}$

  $R=250\Omega$

  $V_{\gamma }=0.6\text{ V}$

Calculation:

Apply Kirchhoff’s voltage law,

  $v_{s(\text{peak})}-i_{d(\text{peak})}\cdot R-V_B-V_{\gamma }=0$

Substituting the given values,

  $\begin{array}{c}{i}_{d(\text{peak})}=\frac{12-4.5-0.6}{250}\text{ A}\\ =\text{0}\text{.0276}\\ =27.6\times {10}^{-3}\\ =27.6\text{mA}\end{array}$

Maximum reversed biased voltage of the diode,

  $\begin{array}{c}{V}_{R(\max )}=v_{s(\text{peak})}+V_B\\ =\left(12+4.5\right)\text{ V}\\ \text{=16}\text{.5}\text{V}\end{array}$

Consider the time at which diode start conducting as $\left(t_1\right)$ . When the diode is forward biasedthen

  $\begin{array}{c}\omega t_1={\sin }^{-1}\left(\frac{V_B+V_{\gamma }}{v_{s(\text{peak})}}\right)\\ ={\sin }^{-1}\left(\frac{5.1}{12}\right)\\ =25.15°\end{array}$

By symmetry,

  $\begin{array}{c}\omega t_2=180°-25.15°\\ =154.85°\end{array}$

Fraction of the cycle diode is conducting can be calculated as,

  $\begin{array}{c}\frac{\omega t_2-\omega t_1}{360°}\times 100\%=\frac{154.85°-25.15°}{360°}\times 100\%\\ =\frac{129.7°}{360°}\times 100\%\\ =0.3602\times 100\%\\ =36.02\%\end{array}$

Conclusion:

Therefore, the peak diode current is $i_{d(peak)}=27.6\text{ mA}$ , maximum reversed biased voltage of the diode is $V_{R(\max )}=16.5\text{ V}$ and the fraction of the cycle diode is conducting is $36.02\%$ .