Battery charger bridge 30 V: conduction, power

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: $V_s = 30$ V (rms), $E_b = 24$ V, $R = 5\ \Omega$, ideal diodes. D1, D2 conduct in the positive half cycle and D3, D4 in the negative half cycle, but only while $|v_s| > 24$ V.

 |vs|   ___         ___
       / | \       / | \       --- 24 V level
 -----/--+--\-----/--+--\----
 io      /\          /\
 -------/  \--------/  \-----
      a  180-a  180+a  360-a

i) Conduction period

$$\begin{aligned} V_m &= \sqrt2 \times 30 = 42.43\ \text{V} \\ \alpha &= \sin^{-1}\frac{E}{V_m} = \sin^{-1}\frac{24}{42.43} = 34.45^\circ = 0.6013\ \text{rad} \end{aligned}$$ $$\begin{aligned} \theta_c &= \pi - 2\alpha = 180^\circ - 2(34.45^\circ) = 111.10^\circ = 1.9391\ \text{rad} \\ t_c &= \frac{111.10^\circ}{360^\circ} \times 20\ \text{ms} = 6.17\ \text{ms (in each half cycle)} \end{aligned}$$

ii) Average charging current

$$\begin{aligned} I_o &= \frac{1}{\pi}\int_{\alpha}^{\pi-\alpha}\frac{V_m\sin\omega t - E}{R}\,d(\omega t) = \frac{1}{\pi R}\left[2V_m\cos\alpha - E(\pi-2\alpha)\right] \\ &= \frac{1}{\pi \times 5}\left[2(42.43)(0.8246) - 24(1.9391)\right] = \frac{69.971 - 46.538}{15.7080} = 1.492\ \text{A} \end{aligned}$$

iii) Power to battery and gross power output

$$\begin{aligned} I_{rms}^2 &= \frac{1}{\pi R^2}\left[\left(\frac{V_m^2}{2}+E^2\right)(\pi-2\alpha) + \frac{V_m^2}{2}\sin2\alpha - 4V_mE\cos\alpha\right] \\ &= \frac{2862.06 + 839.66 - 3358.63}{78.540} = 4.368 \\ I_{rms} &= 2.090\ \text{A} \end{aligned}$$ $$\begin{aligned} P_{batt} &= E\,I_o = 24 \times 1.492 = 35.80\ \text{W} \\ P_R &= I_{rms}^2R = 4.368 \times 5 = 21.84\ \text{W} \\ P_{gross} &= P_{batt} + P_R = 35.80 + 21.84 = 57.65\ \text{W} \end{aligned}$$

Answer: (i) conduction 111.10° (6.17 ms) per half cycle, from 34.45° to 145.55°; (ii) $I_o = 1.492$ A; (iii) $P_{batt} = 35.80$ W, gross output $= 57.65$ W.