Battery charger bridge 50 V: power and efficiency

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: $V_s = 50$ V, 50 Hz, $R = 5\ \Omega$, $E = 24$ V (as stated in the question; the figure shows 20 V, results for 20 V are given at the end). Diodes ideal.

Waveforms

 vs   ___
     /   \             /
 ---/-----\-----------/---
           \___/
 vo  ___       ___         vo = |vs| while a diode pair
    / | \     / | \        conducts, = E otherwise
 ==/==+==\===/==+==\== E
 io   /\        /\
 ----/  \------/  \-------
    a  180-a 180+a  360-a

• D1, D2 conduct from $\alpha$ to $\pi-\alpha$; D3, D4 from $\pi+\alpha$ to $2\pi-\alpha$.
• Outside these intervals all diodes are off and $v_o = E$.

$$\begin{aligned} V_m &= \sqrt2 \times 50 = 70.71\ \text{V} \\ \alpha &= \sin^{-1}\frac{E}{V_m} = \sin^{-1}\frac{24}{70.71} = 19.84^\circ = 0.3463\ \text{rad} \end{aligned}$$

Conduction angle $= \pi - 2\alpha = 140.32^\circ$ per half cycle.

i) Average charging current

$$\begin{aligned} I_o &= \frac{1}{\pi}\int_{\alpha}^{\pi-\alpha}\frac{V_m\sin\omega t - E}{R}\,d(\omega t) = \frac{1}{\pi R}\left[2V_m\cos\alpha - E(\pi-2\alpha)\right] \\ &= \frac{1}{\pi \times 5}\left[2(70.71)(0.9406) - 24(2.4490)\right] = \frac{133.026 - 58.776}{15.7080} = 4.727\ \text{A} \end{aligned}$$

ii) Power supplied to the battery and iii) gross output

$$\begin{aligned} I_{rms}^2 &= \frac{1}{\pi R^2}\left[\left(\frac{V_m^2}{2}+E^2\right)(\pi-2\alpha) + \frac{V_m^2}{2}\sin2\alpha - 4V_mE\cos\alpha\right] \\ &= \frac{7533.16 + 1596.32 - 6385.26}{78.540} = 34.940 \\ I_{rms} &= 5.911\ \text{A} \end{aligned}$$ $$\begin{aligned} P_{batt} &= E\,I_o = 24 \times 4.727 = 113.45\ \text{W} \\ P_R &= I_{rms}^2R = 34.940 \times 5 = 174.70\ \text{W} \\ P_{gross} &= P_{batt} + P_R = 113.45 + 174.70 = 288.15\ \text{W} \end{aligned}$$

iv) Efficiency of the rectifier

With ideal diodes all ac input power appears as gross output; the useful output is the power stored in the battery:

$$\eta = \frac{P_{batt}}{P_{gross}} = \frac{113.45}{288.15} = 39.37\ \%$$

Answer: $I_o = 4.727$ A, $P_{batt} = 113.45$ W, gross output $= 288.15$ W, $\eta = 39.37$ %.

If $E = 20$ V (figure value): $\alpha = 16.43^\circ$, $I_o = 5.366$ A, $P_{batt} = 107.32$ W, gross $= 322.37$ W, $\eta = 33.29$ %.