Figure 1 illustrates a 2 -pulse mid-point converter circuit with resistive-load. This type of full-wave rectifier circuit uses two SCRs connected to the centre-tapped secondary of a transformer, as shown in Fig.1. The input signal is coupled through the transformer to the centretapped secondary.

During the positive half-cycle of the a.c. supply, i.e. when terminal A of the transformer is positive with respect to terminal $B,$ or the secondary-winding terminal A is positive with respect to $N, \operatorname{SCR}_{1}\left(T_{1}\right)$ is forward-biased and $\mathrm{SCR}_{2}$ $\left(T_{2}\right)$ is reverse-biased. Since no triggering pulses are given to the gates of the SCRs, initially they are in off-state. When $\mathrm{SCR}_{1}$ is triggered at a firing-angle $\alpha$ current would flow from terminal A through $\mathrm{SCR}_{1},$ the resistive load R and back to the centre-tap of the transformer.This current path is also shown in Fig.1. This current continuous to flow up to angle $\pi$ when the line voltage reverses its polarity and $\mathrm{SCR}_{1}$ is turned-off. Depending upon the value of $\alpha$ and the load circuit parameters, the conduction angle of $\mathrm{SCR}_{1}$ may be any value between 0 and $\pi .$

During the negative half-cycle of the a.c. supply, the terminal B of the transformer is positive with respect to $N . \mathrm{SCR}_{2}$ is forward-biased. When $\mathrm{SCR}_{2}$ is triggered at an angle $(\pi+\alpha),$ current would flow from terminal $B,$ through $\mathrm{SCR}_{2},$ the resistive load and back to centre-tap of the transformer. This current continues till angle $2 \pi,$ then $\mathrm{SCR}_{2}$ is turned off. Here it is assumed that both thyristors are triggered with the same firing angle, hence they share the load current equally.

Each half of the input-wave is applied across the load. Thus, across the load, there are two pulses of current in the same direction. Hence the ripple frequency across the load is twice that of the input supply frequency. The voltage and current waveforms of this configuration is shown in Fig.2. It is is clear from Fig.2 that with purely resistive load, the load current is always discontinuous.

The voltage and current relations are derived as follows

(a) Average d.c. Output Voltage: The output d.c. voltage, $E_{\mathrm{dc}},$ across the resistive load is given by $$E_{\mathrm{dc}}=\frac{1}{\pi} \int_{\alpha}^{\pi} E_{m} \cdot \sin \omega t \cdot \mathrm{d}(\omega t)=\frac{E_{m}}{\pi}[-\cos \omega t]_{\alpha}^{\pi}=\frac{E_{m}}{\pi}[1+\cos \alpha]$$

(b) Average-load Current: The average-load current is given by $$I_{\mathrm{dc}}=\frac{E_{m}}{\pi \cdot R}[1+\cos \alpha]$$

(c) RMS Load-voltage: The RMS load-voltage for a given firing angle $\alpha$ is given by

$$E_{\mathrm{rms}}=\left[\frac{1}{\pi} \int_{\alpha}^{\pi} E_{m}^{2} \sin ^{2} \omega t \cdot \mathrm{d} \omega t\right]^{\frac{1}{2}}=E_{m} \cdot\left[\frac{1}{\pi} \int_{\alpha}^{\pi} \sin ^{2} \omega t \cdot \mathrm{d} \omega t\right]^{\frac{1}{2}}$$

$$=E_{m} \cdot\left[\frac{1}{\pi} \int_{\alpha}^{\pi}\left(\frac{1-\cos 2 \omega t}{2}\right) \mathrm{d} \omega t\right]^{\frac{1}{2}}=E_{m} \cdot\left[\frac{1}{2 \pi}\left(\omega t-\frac{\sin 2 \omega t}{2}\right)_{\alpha}^{\pi}\right]^{\frac{1}{2}}$$

$$=E_{m} \cdot\left[\frac{1}{2 \pi}\left(\pi-\alpha+\frac{\sin 2 \alpha-\sin 2 \pi}{2}\right)\right]^{\frac{1}{2}}=E_{m} \cdot\left[\frac{1}{2 \pi}\left(\pi-\alpha+\frac{\sin 2 \alpha}{2}-0\right)\right]^{\frac{1}{2}}$$

$$E_{\mathrm{rms}}=E_{m} \cdot\left[\frac{\pi-\alpha}{2 \pi}+\frac{\sin 2 \alpha}{4 \pi}\right]^{\frac{1}{2}}$$