Battery charger bridge 15 V: limit current to 0.5 A

In a battery-charging rectifier, a diode conducts only when the instantaneous supply voltage exceeds the battery emf $E$. In each half cycle current flows from $\alpha$ to $\pi-\alpha$, where $V_m\sin\alpha = E$, and is limited by $R$:

$$i_o = \frac{V_m\sin\omega t - E}{R}, \qquad \alpha < \omega t < \pi - \alpha$$

     io     R
 P o--->---/\/\/---+
                   |
                 + |
                 ===  Eb (battery)
                 - |
 N o---------------+
 (P, N = output of the diode bridge)

Data: $V_s = 15$ V (rms), $E_b = 12$ V, $R = 2\ \Omega$, ideal diodes.

Conduction angle

$$\begin{aligned} V_m &= \sqrt2 \times 15 = 21.21\ \text{V} \\ \alpha &= \sin^{-1}\frac{E}{V_m} = \sin^{-1}\frac{12}{21.21} = 34.45^\circ = 0.6013\ \text{rad} \end{aligned}$$

Current flows for $\pi - 2\alpha = 111.10^\circ$ in each half cycle.

i) Average charging current

$$\begin{aligned} I_o &= \frac{1}{\pi}\int_{\alpha}^{\pi-\alpha}\frac{V_m\sin\omega t - E}{R}\,d(\omega t) = \frac{1}{\pi R}\left[2V_m\cos\alpha - E(\pi-2\alpha)\right] \\ &= \frac{1}{\pi \times 2}\left[2(21.21)(0.8246) - 12(1.9391)\right] = \frac{34.986 - 23.269}{6.2832} = 1.865\ \text{A} \end{aligned}$$

ii) Power supplied to the battery

$$P_{batt} = E_b I_o = 12 \times 1.865 = 22.38\ \text{W}$$

iii) Gross output power of the rectifier

Gross output = power to battery + loss in $R$. First the rms current:

$$\begin{aligned} I_{rms}^2 &= \frac{1}{\pi R^2}\left[\left(\frac{V_m^2}{2}+E^2\right)(\pi-2\alpha) + \frac{V_m^2}{2}\sin2\alpha - 4V_mE\cos\alpha\right] \\ &= \frac{715.51 + 209.91 - 839.66}{12.566} = 6.826 \\ I_{rms} &= 2.613\ \text{A} \end{aligned}$$ $$\begin{aligned} P_{batt} &= E\,I_o = 12 \times 1.865 = 22.38\ \text{W} \\ P_R &= I_{rms}^2R = 6.826 \times 2 = 13.65\ \text{W} \\ P_{gross} &= P_{batt} + P_R = 22.38 + 13.65 = 36.03\ \text{W} \end{aligned}$$

iv) Extra resistance for $I_o = 0.5$ A

The conduction angle does not depend on $R$, so the bracket $\left[2V_m\cos\alpha - E(\pi-2\alpha)\right] = 11.717$ V stays the same:

$$\begin{aligned} R_{total} &= \frac{2V_m\cos\alpha - E(\pi-2\alpha)}{\pi I_o} = \frac{11.717}{\pi \times 0.5} = 7.459\ \Omega \\ R_{extra} &= 7.459 - 2 = 5.459\ \Omega \end{aligned}$$

Answer: (i) $I_o = 1.865$ A, (ii) $P_{batt} = 22.38$ W, (iii) gross output $= 36.03$ W, (iv) extra series resistance $= 5.46\ \Omega$.