Single-phase diode bridge with constant load current

Given: single-phase full-bridge diode rectifier, $V_s = 230$ V, 50 Hz; $R = 10\ \Omega$ with a large inductance, so the load current is constant ($I_o$).

              P (+)
        +-------+-------+
        |               |
       D1              D3
        |               |
 a o----+               +----o b
        |               |
       D4              D2
        |               |
        +-------+-------+
              N (-)
 ac supply vs between a and b (is into a)
 Load R = 10 ohm + large L (Io) between P and N

$D_1, D_2$ conduct in the positive half cycle and $D_3, D_4$ in the negative half cycle, each for $180^\circ$. The input current is a square wave of $\pm I_o$, in phase with $v_s$.

$$V_m = \sqrt{2} \times 230 = 325.27\ \text{V}$$

(i) Average output voltage and current

$$\begin{aligned} V_o &= \frac{2V_m}{\pi} = \frac{2 \times 325.27}{\pi} = 207.07\ \text{V} \\ I_o &= \frac{V_o}{R} = \frac{207.07}{10} = 20.71\ \text{A} \end{aligned}$$

(ii) Diode currents

Each diode carries $I_o$ for half the period:

$$\begin{aligned} I_{D(avg)} &= \frac{I_o}{2} = 10.35\ \text{A} \\ I_{D(rms)} &= \frac{I_o}{\sqrt{2}} = \frac{20.71}{\sqrt{2}} = 14.64\ \text{A} \end{aligned}$$

(iii) RMS output and input currents

• The output current is constant: $I_{o(rms)} = I_o = 20.71$ A.
• The input current is a square wave of amplitude $I_o$: $I_s = I_o = 20.71$ A.

(iv) Supply power factor

$$\begin{aligned} I_{s1} &= \frac{2\sqrt{2}}{\pi}I_o = 0.9003 \times 20.71 = 18.64\ \text{A} \\ \text{DF} &= \cos 0^\circ = 1 \\ \text{PF} &= \frac{I_{s1}}{I_s}\cos\phi_1 = 0.9003 \times 1 = 0.9003 \end{aligned}$$

Check: $P = V_o I_o = 207.07 \times 20.71 = 4288$ W and $V_s I_s \times \text{PF} = 230 \times 20.71 \times 0.9003 = 4288$ W.

Answer: $V_o = 207.07$ V, $I_o = 20.71$ A; diode $I_{avg} = 10.35$ A, $I_{rms} = 14.64$ A; output and input rms currents $= 20.71$ A; supply PF $= 0.90$.