Chapter 2, Problem 2.1TYU

Consider the circuit in Figure 2.4. The input voltage is ${\upsilon }_s\left(t\right)=15\sin \text{ω}t\left(V\right)$ and the diode cut−in voltage is $V_{\gamma }=0.7\text{V}$ . The voltage $V_B$ varies between $4\le V_B\le 8\text{V}$ . The peak current is to be limited to $i_D$ $i_D\left(\text{peak}\right)=1\text{8 mA}$ . (a) Determine the minimum value of R. (b) Using the results of part (a), determine the range in peak current and the range in duty cycle. (Ans. (a) $R=572\Omega$ ; (b) $11\le i_D\left(\text{peak}\right)\le 1\text{8 mA}$ , $\text{3}0.\text{3}\le \text{duty cycle}\le \text{39}.\text{9}\%$ ).

To determine

The minimum value of the resistance of a battery charger circuit for a given peak battery charging current.

Answer to Problem 2.1TYU

Minimum resistance is $R=572\Omega$ .

Explanation of Solution

Given Information:

The given values are:

  $\begin{array}{l}{v}_s=15\sin (\omega t)\text{ V}\\ V_{\gamma }=0.7\text{ V}\\ i_{D(peak)}=18\text{ mA}\end{array}$

Range of voltage $V_B$ is $4\le V_B\le \text{8 V}$

Calculation:

The battery charger circuit contains a half-wave rectifier as shown below.

  

Using the Kirchhoff’s Voltage Law, voltage across the resistor is given by,

  $v_R=V_S\sin (\omega t)-V_{\gamma }-V_B\to \text{(1)}$

The current through the resistor,

  $i_d=\frac{v_R}{R}$

The peak current through the resistor (or the diode),

  $i_{d(peak)}=\frac{v_{R(max)}}{R_{(\min )}}\to (2)$

The diode current will be maximum when the input voltage is maximum at $\omega t=\frac{\pi }{2}$ and the battery voltage is minimum,

From equation (1) $v_{R(max)}$ can be obtained by substituting the $\omega t=\frac{\pi }{2}$

  $v_{R(max)}=V_s-V_{\gamma }-V_{B(\min )}$

Then, equation (2) can be written as,

  $i_{d(peak)}=\frac{V_s-V_{\gamma }-V_{B(\min )}}{R_{(\min )}}$

Then the minimum value of the resistor, $R_{(\min )}=\frac{V_s-V_{\gamma }-V_{B(\min )}}{i_{d(peak)}}$

Substituting the values,

  $R_{(\min )}=572\Omega$

Conclusion:

The minimum resistance value is $R_{(\min )}=572\Omega$

To determine

The range in peak current and the range in a fraction of cycle diode conducts.

Answer to Problem 2.1TYU

The range in peak current is, $11\text{ mA}{\le }_{\gamma }{i}_{d(peak)}\le 18\text{ mA}$

The range in duty cycle is $30.3\%\le \text{Duty Cycle}\le 39.9\%$

Explanation of Solution

Given Information:

  $\begin{array}{l}4\le V_B\le \text{8 V}\\ v_s=15\sin (\omega t)\text{ V}\\ V_{\gamma }=0.7\text{ V}\\ R=572\Omega \end{array}$

Calculation:

Here the battery charger circuit contains a halfwave rectifier and the circuit can be drawn as below. The range for the battery voltage is $V_B$ . So, an equation for $V_B$ can be obtained using Kirchhoff’s voltage law. Then find out the range of peak current through the circuit use maximum supply voltage in the equations. To find out the fraction of diode conduction time (duty cycle) same expression for $V_B$ can be used with output voltage zero where the diode is getting on and off.

  

Using the Kirchhoff’s Voltage Law,

  $V_B=V_S\sin (\omega t)-V_{\gamma }-v_R$ ; where $v_R$ is the voltage across the resistor.

It is known that, $4\le V_B\le \text{8 V}$ , then

  $4\le V_S\sin (\omega t)-V_{\gamma }-v_R\le \text{8 }\to \text{ (1)}$

The voltage through the resistor,

  $v_R=i_{d}R$

Then equation (1) can be written as,

  $\frac{8-(V_S\sin (\omega t)-V_{\gamma })}{R}{\le }_{\gamma }{i}_d\le \frac{4\text{-(}{V}_S\sin (\omega t)-V_{\gamma })}{R}$

The range for the peak current through the diode,

  $\begin{array}{c}\frac{(15-0.7)-8}{572}\text{ A}{\le }_{\gamma }{i}_{d(peak)}\le \frac{\text{(}15-0.7)-4}{572}\text{ A}\\ 11\text{ mA}{\le }_{\gamma }{i}_{d(peak)}\le 18\text{ mA}\end{array}$

When the diode is forward biased and at the time diode start conducting, let’s say $\left(t_1\right)$ , $v_R=0$ and from the equation (1),

  $4\le V_S\sin (\omega t_1)-V_{\gamma }\le \text{8 }$

Substitute the values $V_s=15\text{ V and }{V}_{\gamma }=0.7$ in above equation,

  $4.7\le 15\sin (\omega t_1)\le \text{8}\text{.7 }$

  ${\sin }^{-1}\left(\frac{4.7}{15}\right)\le \omega t_1\le {\sin }^{-1}\left(\frac{8.7}{15}\right)$

  $\begin{array}{l}{\sin }^{-1}\left(\frac{4.7}{15}\right)\le \omega t_1\le {\sin }^{-1}\left(\frac{8.7}{15}\right)\\ \text{18}\text{.26}°\le \omega t_1\le 35.45°\end{array}$

By symmetry, the point where $v_R$ goes zero again $\left(t_2\right)$ is obtained as below,

  $\omega t_2=180°-\omega t_1$

Then fraction of the time diode is conducting can be calculated as,

  $\frac{\omega t_2-\omega t_1}{2\pi }\times 100\%$

Hence,

  $\begin{array}{c}\frac{(180°-35.45°)-35.45°}{360°}\times 100\%\le \left(\frac{\omega t_2-\omega t_1}{{360}^0}\right)\times 100\%\le \frac{(180°-\text{18}\text{.26}°)-\text{18}\text{.26}°}{360°}\times 100\%\\ (180°-35.45°)-35.45°\le \left(\frac{\omega t_2-\omega t_1}{{360}^0}\right)\times 100\%\le (180°-\text{18}\text{.26}°)-\text{18}\text{.26}°\\ 30.3\%\le \text{Duty Cycle}\le 39.9\%\end{array}$

Conclusion:

The range in peak current is, $11\text{ mA}{\le }_{\gamma }{i}_{d(peak)}\le 18\text{ mA}$

The range in the duty cycle is $30.3\%\le \text{Duty Cycle}\le 39.9\%$