Single-phase fully controlled converter DC motor speed

In a single-phase fully controlled bridge (four thyristors) feeding a separately excited dc motor, the average armature voltage is set by the firing angle $\alpha$, and the speed follows from the dc motor equations.

Circuit

              P (+)
          +---+---+
          |       |
         T1      T3
          |       |
   A -----+       +----- B
   |      |       |      |
   |     T4      T2      |
   |      |       |      |
   |      +---+---+      |
   |          N (-)      |
   +------- ~ v_s -------+

   P ---[ Ra ]---[ La ]---( E )--- N    motor armature
   T1,T2 fired at alpha; T3,T4 at pi+alpha

Derivation

Supply: $v_s = V_m \sin\omega t$. With continuous conduction, T1–T2 conduct from $\alpha$ to $\pi+\alpha$, so the armature voltage follows $v_s$ in this interval:

$$\begin{aligned} V_a &= \frac{1}{\pi}\int_{\alpha}^{\pi+\alpha} V_m \sin\omega t\, d(\omega t) \\ &= \frac{V_m}{\pi}\left[-\cos\omega t\right]_{\alpha}^{\pi+\alpha} = \frac{2V_m}{\pi}\cos\alpha \end{aligned}$$

Motor equations (constant field, $K = K_b$):

$$\begin{aligned} V_a &= E_b + I_a R_a, \quad E_b = K\omega_m, \quad T = K I_a \end{aligned}$$

(the inductor has zero average voltage in steady state). Therefore

$$\begin{aligned} \omega_m &= \frac{V_a - I_a R_a}{K} = \frac{2V_m\cos\alpha}{\pi K} - \frac{R_a}{K}I_a \\ \omega_m &= \frac{2V_m\cos\alpha}{\pi K} - \frac{R_a}{K^2}T \end{aligned}$$

In rpm, $N = \dfrac{60}{2\pi}\omega_m$. With $V_m = \sqrt2 V_s$:

$$N = \frac{60}{2\pi K}\left(\frac{2\sqrt2 V_s}{\pi}\cos\alpha - I_a R_a\right)$$

• $0 \le \alpha < 90^\circ$: $V_a > 0$, motoring (quadrant I).
• $90^\circ < \alpha < 180^\circ$: $V_a < 0$; with reversed back emf (load driving) the converter inverts, giving regenerative braking (quadrant IV).
• Speed falls linearly with torque; the slope $R_a/K^2$ is the same for every $\alpha$, so the speed–torque lines are parallel.

Waveforms (continuous conduction, fixed speed)

 v_s
    |   .-.             .-.
    |  /   \           /   \
  --+-/-----\---------/-----\--------> wt
    |/       \       /       \
    |         '-._.-'         '-._
    0        pi             2pi

 v_a  (alpha about 45 deg)
    |     |'-.          |'-.
    |     |   \         |   \
  --+-----+----\--------+----\-------> wt
    |   alpha   \_|   pi+alpha \_|
 E_b - - - - - - - - - - - - - - - - -  (constant level)

 i_a
    |  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~  ripple about Ia, at 2f
  --+--------------------------------> wt  (never reaches zero)

• $v_a$ follows $+v_s$ from $\alpha$ to $\pi+\alpha$ and $-v_s$ from $\pi+\alpha$ to $2\pi+\alpha$; it goes negative for part of each half cycle (from $\pi$ to $\pi+\alpha$, and so on).
• The back emf $E_b$ is a constant line (fixed speed).
• $i_a$ rises when $v_a > E_b$ and falls when $v_a < E_b$, but never reaches zero (continuous conduction); the ripple frequency is $2f$.