1) The rotary conversion is replaced by an uncontrolled diode bridge and the DC motor is replaced by a three phase SCR bridge.

2) The slip power is drawn from the rotor of motor. The three phase rotor induced voltage is rectified by a three phase diode bridge. The output of a diode bridge is rectified by choke ‘L’ and fed to line commutated inverter bridge.

3) The SCR bridge operates in the inverting mode of the firing angle ‘α’ is adjusted to be greater than 90°

$$90° ≤ α ≤ 180°$$

4) The inverter bridge returns the “slip power “ extracted from the rotor to A.C. supply through suitable transformer.

5) Due to the use of diode bridge, the power can only be extracted from the rotor (power cannot be supplied to rotor). Therefore the motor can operate in the subsynchronous region only.

6) Torque and speed expressions for this drive can be derived as follows:

• Rotor voltage per phase = s$E_2$

  Where $E_2$ = per phase rotor e.m.f. at standstill

  And s = slip

• Voltage s$E_2$ is rectified to $V_d$ by diode bridge. Uncontrolled output voltage of diode rectifier,

$$V_d=\frac{3\sqrt 6}{\pi}.sa_1=2.339saV_1$$ $$V_1 = \text{supply voltage per phage}$$

• For three phase line commutated inverter, average dc output voltage ( with no transformer) is

$$V_{dc}=-\frac{3 \sqrt 6}{\pi}V_1.cos \alpha \ \ \ \ \ 90^0 ≤ \alpha≤180^0 \\ V_{dc}=-2.339V_1cos\alpha $$

Here negative sign is used to confirm to the firing angle range

• Case 1 : At NO load

  a) Electric torque Te is negligible and dc link current Id is almost zero.

$$2.339SaV_1=-2.339V_1cos \alpha \\ Slip, \hspace{2 cm} s=-\frac{1}{\alpha} \cos \alpha$$

i. If α = 1, s = -cosα

ii. For α = 90° , s = 0 , speed synchronous

iii. For α = 180° , s = 1 , speed zero

b) This shows that no load speed of the motor can be controlled from near –stand-still to full speed as firing angle α is varied from almost 180° to 90°

c) In practice, rotor circuit is less than stator voltage and therefore, α<1 . Thus, a three phase transformer is often required between the ac supply network and the inverter in order tostep down the supply voltage to a level that is appropriate to $V_d$ .

d) Let the transformer turn ratio be αT where

$$\alpha_T=\dfrac {\text{per phase voltage to inverter,} V_2}{\text{per phase supply voltage,}V_1}$$

AC voltage across inverter terminals, $V_2 = α_T.V_1$

The inverter dc voltage $V_{dc}$,

$$V_{dc}=-\dfrac{3\sqrt{6}}{\pi}.\alpha _T.V_1 \cos \alpha \\ \ \ \ \ =-2.339. \alpha _T.V_1 \cos \alpha \\ 2.339 sa V_1=-2.339 \alpha _T. V_1 \cos \alpha \\ slip , s=-\dfrac{\alpha_T}{\alpha} \cos \alpha$$

• Case2 : with Load

  a) In order to develop motor torque, a rotor current $I_2$ is required and the rectifier rotor voltage $V_d$ must force a current Id against the inverter dc voltage $V_{dc}$ .

  b) When the motor is loaded, speed falls and the increased rotor voltage $sE_2$ can overcome the voltage drops in the rotor windings, in the dc link circuit and the converters.

  c) If resistance of the rotor circuit and inductor $L_d$ are neglected, then total slip power,

$\begin{bmatrix} \ i_L\\ \ \bar{v_0}\\ \end{bmatrix} =\begin{bmatrix} \ 0-1(1- \delta(t))/L_i \\ \ (1-\delta(T))/C_0-1/(R_0C_0) \\ \end{bmatrix} \begin{bmatrix} \ i_L\\ \ v_0\\ \end{bmatrix}+\begin{bmatrix} \ \delta(t)/L_i \\ \ 0\\ \end{bmatrix} V_{DC}$

$\begin{bmatrix} \ v_0\\ \ i_L\\ \end{bmatrix}=\begin{bmatrix} \ 0 & 1\\ \ 1 & 0\\ \end{bmatrix}\begin{bmatrix} \ i_L\\ \ \bar{v_0}\\ \end{bmatrix}+\begin{bmatrix} \ 0\\ \ 0\\ \end{bmatrix}[V_{DC}]$

It can be obtained as

$\begin{bmatrix} \ \bar{i_L} \\ \ \bar{v_0} \\ \end{bmatrix}=\bigg[\begin{bmatrix} \ 0 \\ \ 0 \dfrac{-1}{R_0C_0} \\ \end{bmatrix} \delta_1+\begin{bmatrix} \ 0-\dfrac{-1}{L_i} \\ \ \dfrac{1}{C_0} \dfrac{-1}{R_0C_0} \\ \end{bmatrix} \delta_2 \bigg] \begin{bmatrix} \ \bar{i_L} \\ \ \bar{v_0}\\ \end{bmatrix}+\bigg[\begin{bmatrix} \ \dfrac{1}{i_L} \\ \ 0\\ \end{bmatrix} \delta_1+\begin{bmatrix} \ 0 \\ \ 0 \\ \end{bmatrix} \delta_2 \bigg][\bar{V}_{DC}]$

$\begin{bmatrix} \ \bar{v_0} \\ \ \bar{i_L} \\ \end{bmatrix}=\bigg[\begin{bmatrix} \ 0 & 1 \\ \ 1 & 0 \\ \end{bmatrix} \delta_1+\begin{bmatrix} \ 0 & 1 \\ \ 1 & 0 \\ \end{bmatrix} \delta_2 \bigg]\begin{bmatrix} \ \bar{i_L} \\ \ \bar{v_0} \\ \end{bmatrix}+\bigg[\begin{bmatrix} \ 0 \\ \ 0 \\ \end{bmatrix} \delta_1+\begin{bmatrix} \ 0 \\ \ 0 \\ \end{bmatrix} \delta_2\bigg][\bar{V}_{DC}]$

$Amperes=\dfrac{VA \times pf}{\eta \times V_{dc}}+A_1$

$P=(IN)^2 \frac{ρ.\pi d}{1. \delta}.....W \\ P_0=\frac{P}{\pi dl}=(IN)^2 \frac{\rho . \pi d}{1. \delta} \times \frac{1}{\pi dl} \\ \hspace{2 cm} \big(\frac{IN}{l}\big)^2.\frac{\rho}{\delta}.....W/m^2$

$V_0=V_t=\dfrac{V_m}{2{\pi}}(1+ \cos \alpha_1) \hspace{1cm} for \ \ 0 \lt \alpha_1 \lt {\pi} \\ V_f=\dfrac{V_m}{\pi}(1+ \cos \alpha_2) \hspace{1cm} 0 \lt \alpha_2 \lt {\pi} \\ I_{\alpha t}=\sqrt{I^2 \alpha \dfrac{{\pi}-{\alpha}}{2{\pi}}}=I_{\alpha}\bigg(\frac{{\pi}-\alpha}{2{\pi}}\bigg)^{\frac12} \\ I_{fdr}=\sqrt{I^2 \alpha \dfrac{{\pi}+{\alpha}}{2{\pi}}}=I_{\alpha}\bigg(\frac{{\pi}+\alpha}{2{\pi}}\bigg)^{\frac12}$

$I_d=\dfrac{V_d-V_{dc}}{Rd} \\ V_d=V_{dc}+I_d.R_d=2.339saV_1$

$\therefore Slip, s=\dfrac{V_d=V_{dc}+I_d.R_d}{2.339saV_1} \hspace{2cm} R_d- \text{resistance of dc link inductor}$

$s=\dfrac{-2.339. \alpha_T.V_1 \cos \alpha}{2.339 \alpha V}+ \dfrac{I_d.R_d}{2.339 \alpha V}$

Motor speed is given by

$ω_m=ω_a(1-s) \\ ω_m=ω_a \big[1+\frac{a_T}{a} cos\alpha-\frac{ω_a.R_d}{(2.339 aV_1)}\big]$

Under steady state, total torque 3Te is given by

$3T_e=T_L=2.339 \frac{aV_1.I_d}{ω_s} \\ I_d=\frac{ω_s.T_L}{2.339aV_1}$

Substituting the value of $I_d$ in motor speed equation

$ω_m=ω_a \big[1+\frac{a_T}{a}cos\alpha - \frac{ω_a.R.d}{(2.339aV_1)^2} .T_L\big] \\ ω_m=ω_a\big[1+\frac{a_T}{a}cos\alpha - K.T_L \big] \\ K=\frac{ω_aR_d}{(2.339aV_1)^2}$

No load speed of the drive is given by

$ω_m=ω_s\big[1+\frac{a_T}{a}cos\alpha \big]$

Solution (a) Motor constant=0.5 V.sec/rad = 0.5Nm/A=$K_m$

But motor torque, $T_e=K_mI_a$

$\therefore$ Armature current= $\frac{15}{0.5}=30A$

Motor emf,

For 1-phase half –wave converter feeding a dc motor,

$V_t=\dfrac{V_m}{2 {\pi}}(1+ \cos \alpha_1)=E_{\alpha}+I_{\alpha}r_{\alpha} \\ Or \\ V_t=\dfrac{\sqrt{2}\times 230}{2{\pi}}(1+ \cos \alpha_1)=52.36+30 \times 0.7=73.36 V \\ \alpha_1=\cos^{-1}\bigg[\dfrac{73.36 \times 2{\pi}}{\sqrt{2 \times230}}-1\bigg]=63.336^0$

Thus, firing –angle dristor current elay of converter 1 is $65.336^0$

(b) Rms value of free wheeling –diode current, from equ. Is,

$I_{fdr}=I_a \bigg(\frac{\pi +\alpha}{2\pi}\bigg)^{1/2}=30\bigg(\frac{180+65.336}{260}\bigg)^{1/2}=24.766 A=I_{ar}$

(c) From equ. Input power factor of armature converter

$=\dfrac{V_t I_\alpha}{V_t.}=\dfrac{73.36 \times30}{230 \times 16.931}= 0.5651 \log \\ V_0=V_t=\dfrac{V_m}{\pi}(1+\cos \alpha_1) \hspace{1cm} V_f=\dfrac{V_m}{\pi}(1+\cos \alpha_2) \\ I_{ar}=I_{\alpha}\bigg[\dfrac{\pi - \alpha}{\pi}\bigg]^{\frac{1}{2}} \\ I_{fdr}=I_\alpha\bigg[\dfrac{\alpha}{\pi}\bigg]^{\frac12} \\ I_{Tr}=I_\alpha\bigg[\dfrac{\pi-\alpha}{2\pi}\bigg]^{\frac12} \\ =\dfrac{V_tI_\alpha}{V_\alpha I_{\alpha r}}$

For a single phase semi converter feeding a separately. Excited motor,

$$V_0=V_t=\frac{V_m}{\pi}(1+cos \alpha)=E_a+I_ar_a \\ \frac{330}{\pi}(1+cos 30^0)=80+I_a.4 \\ 196.01=80+I_a.4 $$

Therefore, Average armature current,

Motor emf,

$$I_a=\frac{196.01-80}{4}=23.003A \\ E_a=K_mω_m=K_m\frac{2\pi \times 1400}{60} \\ K_m=\frac{80 × 60}{2π × 1400} = 0.546V - s/rad $$

$\hspace{9cm} or 0.546Nm/A$

therefore Motor torque $T_e = K_m I_a = 0.546 x 29.003 = 15.836$

Solution: Under rated operating conditions of the seperately excited dc motor.

$V_t = E_a + I_a r_a = K_m w_m + I_ar_a \\ 220=K_m \frac{2\pi \times 1500}{60}+10 \times 1=50.\pi . K_m+10$

Therefore ,Motor constant

Or 1.337Nm/A

(a) for a torque of 5 Nm ,motor armature current,

$I_a =\frac{5}{1.337}=3.74A$

The equation giving the operation of converter –motor is

$V_0=V_t=E_\alpha+Ir_\alpha \\ \dfrac{2V_m}{\pi}\cos \alpha=K_m.\omega_m+I_\alpha.r_\alpha \\ \dfrac{2\sqrt{2}\times230}{\pi}\cos3\theta^0=1.337\omega_m+3.74\times1 \\ \omega_m=\dfrac{179.3-3.74}{1.337}=131.31rad/ \sec\\ \dfrac{2\pi.N}{60}=131.31rad/\sec \\ \text{Motor speed}=\dfrac{131.31\times60}{2\pi}=1253.92 rpm \\ (b)for \alpha=45^0 \\ \dfrac{2\sqrt{2}\times230}{\pi}\cos45^0=1.337\times\dfrac{2\pi\times1000}{60}+I_\alpha\times1 \\ 146.4=140.01+I_\alpha\times1 \\ I_\alpha=\dfrac{6.39}{1}=6.39A \\ T_c=K_mI_\alpha=1.337\times6.39=8.543Nm \hspace{2cm}----------\text{(motor developed Torque)}$

$I_2^1 = \frac{E_1}{\Bigg[(R_1+ R_2^1/s)^2 + (X_1+X_2^1)^2\Bigg]^{\frac{1}{2}}}$

The Rotor copper per phase = $(I_2^1)R_2^1$

Total Rotor copper loss = $3(I_2^1)R_2^1$

$P_c = (w_s - w)T $

$P_c = _sW_{ST}$

$\therefore P_c = (I_2)^1 R_2^1 x 3 \\$ $ \ \; \ $

$_s W_{sT} = \frac{E_1^2 × R_2^1 × 3}{\Bigg[(R_1+ R_2^1/s)^2 + (X_1+X_2^1)^2\Bigg]}$ $ \\ \; \\ $

Torque speed Char. With variable STATOR VOLTAGE

Rotor Input Power = Mechanical Power + Rotor copper loss

$\therefore P_R = P_M + P_c \\ P_R = \text{Rotor Input Power} \\ P_M = \text{Mechanical Power} \\ P_c = \text{Rotor copper loss} \\ P_R=W_sT \\ P_M=WT \\ \therefore P_c=P_R - P_M=(W_s-W)T$

But slip,

$s=\frac{W_s - W}{W_s} \\ = W_s - W = SW_s$

$\therefore P_c=W_sT=P_R-P_c \\ P_M=P_R-SP_R \\ P_M=(1-s)P_R \\ \eta=\dfrac{P_M}{P_R} \\ \eta=1-s$

$\text{If Rotor input Power remains constant then from (1)}$

1. It is observed from the above equation as follows

   i. Proportional to dc link current, $I_d$

   ii. Proportional to stator voltage, $V_1$

   iii. Proportional to effective rotor to stator turns ratio

   iv. Inversely proportional to synchronous speed, $ω_s$

   The dc link current $I_d$ is given by

   $I_d=\dfrac{V_d-V_{dc}}{Rd} \\ V-d=V_{dc}+I_d.R_d=2.339saV_1 \\ s=\dfrac{V_{dc}+I_d.R_d}{2.339 \alpha V_1} \hspace{1cm} R_d=\text{resistance of dc link inductor} \\ s=\dfrac{-2.339.\alpha_TV_1\cos\alpha}{2.339\alpha V}+\dfrac{I_d.R_d}{2.339\alpha V} \\ \text{Motor speed is given by} \\ \omega_m=\omega_\alpha(1-s) \\ \omega_m=\omega_\alpha\bigg[1+\dfrac{\alpha_T}{\alpha}\cos\alpha-\dfrac{\omega_\alpha.R_d}{(2.339\alpha V_1)}\bigg] \\ \text{Under steady state, total torque} \ \ 3T_e \ \ \text{is given by } \\ 3T_e=T_L=2.339\dfrac{\alpha V_1.I.d}{\omega_s} \\ I_d=\dfrac{\omega_s.T_L}{2.339 \alpha V_1} \\ \text{Substituting the value of} \ \ I_d \ \ \text{in motor speed equation} \\ \omega_m =\omega_a \bigg[1+\frac{a_T}{a}cos\alpha - \frac{\omega_a.R_d}{(2.339aV_1)^2.T_L}\bigg] \\ \omega_m =\omega_a \bigg[1+\frac{a_T}{a}cos\alpha - K.T_L\bigg] \\ K=\frac{\omega_a.R_d}{(2.339aV_1)^2.T_L} \\ \text{No load speed of the drive is given by} \\ \omega_m =\omega_s \bigg[1+\frac{a_T}{a}cos\alpha \bigg] \\ $

1) Speed torque characteristics of static Kramer drive for open loop system

2) Advantages

• very efficient
• the converted power rating is low because t has to handle only the slip power
• the drive system has D.C. machine like characteristics
• the control is very simple

3) Disadvantages

The major is the poor input power factor. Reasons of poor input factor as follows:

• The reactive power drawn by the motor to maintain the air gap flux.
• The reactive power drawn by the inverter (SCR bridge)

• The reactive power drawn by inverter is maximum at α=90° and reduces as α is increased from 90° to 180°

  The sum of all the reactive power drawn by the machine and inverter causes power factor becomes extremely poor at low speeds.

4) Improvement

• Force commutated inverter in place of line commutated inverter.
• For α = 180°, lowest value of controllable speed range, reactive power drawn is less.
• Suitable transformer required between the inverter and the supply line.