Mumbai University > Electronics Engineering > Sem7 > Power Electronics 2

Marks: 10M

A separately excited dc motor, fed through single phase half wave converter, is shown in Figure (a) Motor field circuit is fed through a single phase semi converter in order to reduce the ripple content in the field circuit. Single phase half wave converter feeding a dc motor offers one quadrant drive, Figure (b)

1. The waveforms for source voltage $V_s$ , armature terminal voltage $V_t$ , armature current ia , source current is and freewheeling diode current ifd are sketched in Fig (c) . Note that thyristor current $ i_T = i_s$ . The armature current is assumed ripple free.

2. For single phase half wave converter , average output voltage of converter, Vo = armature terminal voltage, $V_t$

   $V_0=V_t=\frac{Vm}{2\pi}(1+cos\alpha_1) \; \; \; \; for \ \ \ 0\lt \alpha_1\lt\pi $

   Where $V_m$ = maximum value of source voltage

3. For single phase semi converter in the field circuit, the average output voltage is given as $V_f$

   $V_f=\frac{Vm}{\pi}(1+cos\alpha_2) \; \; \; \; for \ \ \ 0\lt \alpha_2\lt\pi $

4. It is seen from the waveforms of fig 12.7 (c) that

   rms value of armature current, $I_{ar} = I_a$

   rms value of source or thyristor current

$I_{\alpha t}=\sqrt{I^2 α \frac{π-α}{2π}}=I_α\big(\frac{π-α}{2π}\big)^{1/2}$

5) rms value of freewheeling diode current,

$$I_{fdr}=\sqrt{I^2 \alpha \frac{\pi +\alpha}{2\pi}}=I_\alpha \big(\frac{π+α}{2π}\big)^{1/2}$$

6) Apparent input power = ( rms source voltage)*(rms source current) $\hspace{2cm}= V_sI_{sr}$

7) Power delivered to motor = $E_aI_a + I_a^2 * r_a = (E_a + I_ar_a )I_a = V_t * I_a$

8) Input supply pf = $p_f=\frac{E_aI_a+I_a^2r_a}{V_a.I_{sr}}=\frac{V_t.I_a}{V_a.I_{sr}}$