Series full converters: average and rms output

Series operation of two full converters is used to obtain a high dc output voltage and a better input power factor. Two single-phase full converters are fed from two secondary windings of one transformer (each secondary $v_s = V_m\sin\omega t$), and their dc outputs are connected in series.

 Sec 1 --> [Converter 1, a1] (+)--------+
                 (-)                    |
                  |    vo1              |
                 (+)                  Load
 Sec 2 --> [Converter 2, a2]          (Io)
                 (-)   vo2              |
                  +---------------------+
 Sec 1, Sec 2: two secondaries of one transformer
 vo = vo1 + vo2

Operation

• Output: $v_o = v_{o1} + v_{o2}$, each converter carrying the same load current $I_a$.
• Rectification: converter 1 is held at $\alpha_1 = 0$ (full output, acts like a diode bridge); converter 2's $\alpha_2$ is varied from $0$ to $\pi$. Output goes from $4V_m/\pi$ down to $0$.
• Inversion: converter 1 held at $\alpha_1 = \pi$, $\alpha_2$ varied from $0$ to $\pi$; output goes from $0$ to $-4V_m/\pi$.
• Since only one converter is phase controlled at a time, the reactive power and harmonics are lower than in a single converter giving the same voltage.

With $\alpha_1 = 0$, over one interval from $\alpha_2$ to $\pi + \alpha_2$:

 vo     ___            ___
       |   \  2Vm     |   \     (2Vm sin wt from a2
 ------+----\---------+----\---  to 180, zero from
       a2  180      180+a2       180 to 180+a2)

Average output voltage

General case (continuous current):

$$V_{dc} = V_{dc1} + V_{dc2} = \frac{2V_m}{\pi}\left(\cos\alpha_1 + \cos\alpha_2\right)$$

With $\alpha_1 = 0$:

$$V_{dc} = \frac{1}{\pi}\int_{\alpha_2}^{\pi}2V_m\sin\omega t\,d(\omega t) = \frac{2V_m}{\pi}\left(1 + \cos\alpha_2\right)$$

Maximum value (both $\alpha = 0$): $V_{dm} = \frac{4V_m}{\pi}$.

RMS output voltage

With $\alpha_1 = 0$:

$$\begin{aligned} V_{rms}^2 &= \frac{1}{\pi}\int_{\alpha_2}^{\pi}4V_m^2\sin^2\omega t\,d(\omega t) = \frac{2V_m^2}{\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{\alpha_2}^{\pi} \\ &= \frac{2V_m^2}{\pi}\left(\pi - \alpha_2 + \frac{\sin2\alpha_2}{2}\right) \end{aligned}$$ $$V_{rms} = \sqrt2\,V_m\left[\frac{1}{\pi}\left(\pi - \alpha_2 + \frac{\sin2\alpha_2}{2}\right)\right]^{1/2}$$

Special case $\alpha_1 = \alpha_2 = \alpha$: $v_o = 2v_{o1}$, so $V_{dc} = \frac{4V_m}{\pi}\cos\alpha$ and $V_{rms} = 2 \times \frac{V_m}{\sqrt2} = \sqrt2\,V_m$.

Input power factor

With $\alpha_1 = 0$, the fundamental input current is $I_{s1} = \frac{4\sqrt2 I_a}{\pi}\cos\frac{\alpha_2}{2}$ (referred to the primary, 1:1 ratio) lagging by $\alpha_2/2$, so the displacement factor is $\cos(\alpha_2/2)$, much better than $\cos\alpha$ of a single converter.