0195136047.pdf

16.1 POWER SEMICONDUCTOR-CONTROLLED DRIVES 757

of the previous second-order homogeneous differential equation, we can solve for the natural

component. The characteristic equation in the frequency domain is

s^2 +

R

L

s+

1

LC

= 0

whose roots are given by

s 1 , 2 =−

R

2 L

±

√(

R

2 L

) 2

−

1

LC

For the second-order circuit, the damping factorαand the resonant frequencyω 0 are given by

α=

R

2 L

; ω 0 =

1

√

LC

(Note: the ratio ofα/ω 0 is known asdamping ratioδ.) Substituting, we get

s 1 , 2 =−α±

√

α^2 −ω^20

Three possible cases arise for the solution of the current, which will depend on the values of

αandω 0 :

Case 1—α=ω 0 : The roots are then equal,s 1 =s 2. The current is said to becritically damped.

The solution of the current is of the form

i(t)=(A 1 +A 2 t)es^1 t

Case 2—α>ω 0 : The roots are unequal and real. The circuit is said to beoverdamped. The

solution is then

i(t)=A 1 es^1 t+A 2 es^2 t

Case 3—α<ω 0 : the roots are complex. The circuit isunderdamped. Lets 1 , 2 =−α±jωr,

whereωris known as the damped resonant frequency orringing frequency, given by

√

ω^20 −α^2.

The solution takes the form

i(t)=e−αt(A 1 cosωrt+A 2 sinωrt)

Observe that the current consists of a damped or decaying sinusoid. The constantsA 1 andA 2 are

determined from the initial conditions of the circuit.

The current waveform can be sketched, taking the conduction time of the diode into account.

Let us now consider a single-phase half-wave rectifier circuit as shown in Figure 16.1.10(a),
with a purely resistive loadR. We shall introduce and calculate the following quantities:

1. Efficiency


2. Form factor


3. Ripple factor


4. Transformer utilization factor


5. Peak inverse voltage (PIV) of diode


6. Displacement factor


7. Harmonic factor


8. Input power factor.
   During the positive half-cycle of the input voltage, the diodeDconducts and the input voltage
   appears across the load. During the negative half-cycle of the input voltage, the output voltage is