Revision problems
2. The buck regulator in Figure Q2 has an input voltage of Vi = 12 V. The
required average output voltage is Vo = 5 V and the peak-to-peak ripple
voltage is 20 mV. The switching frequency is 25 kHz. If the peak-to-peak
ripple current of inductor is limited to 0.8 A, determine (a) the duty cycle D,
(b) the filter inductance L, and (c) the filter capacitor CL.
1
2
vC =
1
i dt +v C (t 0 )
CL ∫ C
t
1 2
⇒ ΔVC = v C − v C (0) = ∫ I C dt
C L t1
where I C is the average capacitor current
in the time duration t 1 to t 2.
The average capacitor current, which flows for
T , is given by I = ΔI L .
C
2
4
T
ΔI L
ΔI LT
dt
=
∫0 4
8C L f S
0.8
⇒ CL =
= 200 µF
8× 25×103 × 20 ×10 −3
1
⇒ ΔVC =
CL
2
3
4. In a buck converter, consider all components to be ideal. Assume that the output
voltage Vo is 5 V, the switching frequency fs is 20 kHz, the inductance L is 1 mH,
and the filter capacitance CL is 470 μF.
a.  Calculate the peak-to-peak output voltage ripple ΔVo if the input voltage
Vi is 12.6 V and the output current Io is 200 mA.
b.  Calculate the rms value of the ripple current through L and, hence,
through, CL.
c.  Calculate ΔVo if Io is equal to half of the output current measured at the
boundary of continuous-discontinuous condition.
4
The rms value of the capacitor current is given by:
1 ΔI L
C L,rms =
×
2 2
with the assumptions :
1. All AC ripple currents are through the capacitor.
2. The ripple is approximately sinusoidal.
5
Revision problem
8. A boost regulator in Figure Q8 has an input voltage of Vi = 5V. The average output
voltage is Vo = 15 V and the average load current is Io = 0.5 A. The switching
frequency is 25 KHz. If L = 150 µH and CL = 220 µF, determine (a) the duty cycle
D, (b) the ripple current of inductor ∆IL, (c) the peak current of inductor IL(pk), and
(d) the ripple voltage of filter capacitor ∆VCL.
6
By inductor volt -second balance, we have :
Vi DT = (Vi −Vo )(1− D)T
⇒
Vo
1
V
=
⇒ D = 1− i = 0.6667
Vi 1− D
Vo
ΔI L =
VLt on Vi D
5× 0.6667
=
=
= 0.89 A
L
Lf s 150 ×10 −6 × 25×103
Assume a lossless circuit, we have :
VI
I
Vi I i = Vi I L = Vo I o = i o ⇒ I L = o
1− D
1− D
ΔI
I
ΔI
I L,peak = I L + L = o + L = 1.945 A
2 1− D 2
When the switch is ON, the capacitor supplies the load current.
The average capacitor current during that period is I C = I o .
Vi
Vi −Vo
tON
t
1 ON
IoD
I
dt
=
I
dt
=
C
o
∫0
C L ∫0
CL f S
0.5× 0.6667
⇒ CL =
= 60.61 µF
25×103 × 220 ×10 −6
1
ΔVC =
CL
7
9. In a boost converter, consider all components to be ideal. If the input voltage
Vi is 8 to 16 V, the output voltage Vo be held constant at 24 V, the switching
frequency fs = 20 kHz, and the filter capacitance CL is 470 μF. Calculate the
minimum inductance L that will keep the converter operating in a continuousconduction mode if the output power Po is greater than 5 W.
8
When the switch is OFF, the inductor current supplies
the load current, we have :
ΔABC 1 Vi
Io =
= ×
× DT (1− D)
T
2 Lcrit
1 (1− D)Vo
= ×
× DT (1− D)
2
Lcrit
1 (1− D)Vo
⇒ Lcrit = ×
× DT (1− D)
2
Io
Vo (1− D)2 D
⇒ Lcrit =
2 f sI o
Vo (1− Dmax )2 Dmax
Lcrit =
2 f sI o
where Vi,min = (1− Dmax )Vo
9