Half-wave diode battery charger, 10 A

In a half-wave battery charger, the diode conducts only when the supply voltage is greater than the battery emf $E$, i.e. from $\alpha$ to $\pi-\alpha$ in the positive half cycle, where $V_m\sin\alpha = E$. The resistor limits the current.

       R = 5 ohm      D
 +----/\/\/\----->|----+
 |                     | +
(~) 24 V, 50 Hz       === E = 12 V
 |                     | -
 +---------------------+

 vs  ___
    / | \           --- E = 12 V level
 --/--+--\-----------/--
 io   /\       (current pulse from a to 180-a,
 ----/  \------  zero for the rest of the cycle)
     a  180-a      360

Note on data: with $V_s = 24$ V and $R = 5\ \Omega$ the peak current is only $(33.94-12)/5 = 4.39$ A, so an average of 10 A is impossible. The quantities below are therefore calculated for the given $R = 5\ \Omega$; the resistance that would give 10 A is shown at the end.

i) Conduction angle of the diode

$$\begin{aligned} V_m &= \sqrt2 \times 24 = 33.94\ \text{V} \\ \alpha &= \sin^{-1}\frac{12}{33.94} = 20.70^\circ \\ \theta_c &= 180^\circ - 2\alpha = 138.59^\circ\ (2.4189\ \text{rad}) \end{aligned}$$

The diode conducts from $20.70^\circ$ to $159.30^\circ$.

iii) Average charging current

$$\begin{aligned} I_o &= \frac{1}{2\pi R}\left[2V_m\cos\alpha - E(\pi-2\alpha)\right] \\ &= \frac{1}{2\pi(5)}\left[63.498 - 29.026\right] = 1.097\ \text{A} \end{aligned}$$

ii) Power supplied to the battery

$$P_{batt} = E\,I_o = 12 \times 1.097 = 13.17\ \text{W}$$

iv) Rectifier efficiency

RMS current:

$$\begin{aligned} I_{rms}^2 &= \frac{1}{2\pi R^2}\left[\left(\frac{V_m^2}{2}+E^2\right)(\pi-2\alpha) + \frac{V_m^2}{2}\sin2\alpha - 4V_mE\cos\alpha\right] \\ &= \frac{1741.58 + 380.99 - 1523.95}{157.08} = 3.8109 \\ I_{rms} &= 1.952\ \text{A}, \qquad P_R = I_{rms}^2R = 19.05\ \text{W} \end{aligned}$$ $$\eta = \frac{P_{batt}}{P_{batt}+P_R} = \frac{13.17}{13.17 + 19.05} = 40.86\ \%$$

Answer: conduction angle $= 138.59^\circ$; $I_o = 1.097$ A; $P_{batt} = 13.17$ W; $\eta = 40.86$ %.

If an average current of 10 A is really required, the conduction angle stays $138.59^\circ$, but $R$ must be reduced to $\frac{63.498 - 29.026}{2\pi \times 10} = 0.549\ \Omega$; then $P_{batt} = 12 \times 10 = 120$ W.