Power Supplies
Linear Regulated Supplies
Switched Regulated Supplies
Batteries
Alternating
Current
Im
The Power
Supply
πt
π /2
π
2π
Direct
-Im
Im
Current
Idc
π/2
-Im
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π
2π
πt
Exercise 1
The current in a 10 Ω resistor is 5*sin(314t) A
• Draw the waveform of the current
• Define and calculate the following values for the
current:
–
–
–
–
Peak
Peak to peak
Average
Root Mean Square (RMS)
• Calculate the value of the power dissipated by the
resistor
• How much would be the current if it would be DC to
generate the same power on the resistor?
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Power supply terminology
PAV
1 T 2
R T 2
= ∫ i Rdt = ∫ i dt = I eff2 R
T 0
T 0
I eff = I rms
1 T 2
2
=
i
dt
=
average
i
T ∫0
Effective (rms) value of alternating component of the wave
ripplefactor(r) =
Average (dc) value of the wave
V ML - V FL
Load regulation (%) =
100 = R0 x 100
V⊗
RL
Input regulation (% / V I N
∆V O
)=
x 100
∆V I N •V O
VO x IL
POUT
Efficiency ( η ) =
=
PI N V I N x I I N
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Alternating
Current
Im
Im
Idc
π/2
π
πt
2π
Direct
Current
π/2
π
2π
πt
-Im
-Im
AC line
components
AC
Input
Step-down
transformer
Rectifier
Filter
Regulator
DC
Output
Block diagram of a linear regulated power supply
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Unregulated supply with ac line components
(a transient suppressor and line filter)
Fuse
ac line filter
ac line transformer
127 Vrms
Filter capacitor
+
60 Hz
Input
socket
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Transient
suppressor
Snubber
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Bridge rectifier
Unregulated
dc output
Fuse
ac line filter
ac line transformer Filter capacitor
127 Vrms
+
Unregulated
dc output
60 Hz
Input Transient
socket suppressor
Snubber
Bridge rectifier
•fast blow fuses cut the power
as quick as they can
•slow blow fuses tolerate
more short term overload
•wire link fuses are just an
open piece of wire, and have
poorer overload
characteristics than glass and
ceramic fuses
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Transient suppressor and line filter
Fuse
ac line filter
ac line transformer
127 Vrms
Filter capacitor
+
60 Hz
Input
socket
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Transient
suppressor
Snubber
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Bridge rectifier
Unregulated
dc output
Fuse
ac line filter
ac line transformer
127 Vrms
Filter capacitor
+
Unregulated
dc output
60 Hz
Input
socket
Transient
suppressor
Snubber
Bridge rectifier
RC Snubbers
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Transformer
+Vdiodes
Vunregulate
d+Vripple
Vtransforme
r=
2η
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V transformer =
V unregulated +V ripple+V diodes
Exercise – 2
2η
• For transformer in a power supply
–
–
–
–
–
Required average output voltage = 10 V
Ripple voltage = 1 V
Diode drops = 2 V
Output current (average) = 1 A
Efficiency (η) of the transformer = 0.8
• Find the required output voltage of the transformer
• Find the input current of the transformer if the input
voltage is 220 V
• Find the output power delivered by the power supply
• Find the power loss by the transformer
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Rectifier Diodes
Diode
Maximum Current Maximum
Reverse Voltage
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1N4001
1A
50V
1N4002
1A
100V
1N4007
1A
1000V
1N5401
3A
100V
1N5408
3A
1000V
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Half-Wave Rectified (Single Diode
Rectifier)
Vo = Vm - Vd with Vd ≅ 1 volt.
Vdc = (Vm - Vd)/π , Vrms = (Vm - Vd)/2
yielding a ripple factor (r) = 1.21
VmSinωt
127 Vrms
60 Hz
+
D1
RL
-
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Output: half-wave varying DC (Pulsating DC)
(using only half the AChttp://bkaragoz.kau.edu.sa
wave)
D1
Vo = Vm - Vd with
+
Vd ≅ 1 volt.
RL
D2
Vdc = 2*(Vm - Vd)/π,
Vrms = (Vm - Vd)/√2
yielding a much
-
reduced ripple factor
that is r = 0.483
Full-wave rectified
with a center tapped
transformer
Vm
π/2
-Vm
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π
2π
Bridge Rectifiers
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D1
+
D2
D3
RL
D4
Vm
Full-wave rectified
with a bridge
rectifier
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-Vm
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D1
D4
π/2
π
D2
D3
2π
Pulsating DC
Smoothing Filters
Types of smoothing filters
• Capacitive
• Inductive
• L – Section
• π - Section
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Capacitive filter
Amplitude
IL
Light load
Heavy load
Filter
capacitor
Rectifier diodes
Filtered DC
Time
C = smoothing capacitance in farads (F)
Io = output current from the supply in amps (A)
Vs = supply voltage in volts (V), this is the peak
value of the unsmoothed DC
f = frequency of the AC supply in hertz (Hz)
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Effect of Capacitance on Ripple
Amplitude
Light load
+
Vi
-
C RL
+
Vo
-
Heavy load
T1
T2
T
Time
Assuming that the load current stays constant, the ripple voltage
(peak to variation at the top of the waveform) can be approximately
from the charge lost by the capacitor as IL=C*Vr/T1 yielding
Vr=IL*T1/C. T1 ≈ T; Vr=IL/2fC for full-wave rectified; IL = Vdc/RL ;
r=2400/RLC and Vdc=(Vi - 4200Idc/C) where C is in μF and f = 60 Hz
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Some Facts on Smoothing Capacitors
• C ↑⇒ r ↓ ⇒ charging time T2 ↓ ⇒ larger currents to flow through the
rectifier diodes. Eventually, rectifier diodes and the transformer will be
afflicted by increased I2R heating.
• Care for:
– The ripple voltage we can tolerate → value of C and its tolerance
– Polarity
– The maximum DC voltage that the capacitor can withstand (the working DC
(WVDC)) >50% more than the maximum voltage is a good choice
– Appreciable series inductive components → may not behave as an effective
capacitive element for high frequency spikes ⇒ Add a small parallel capacitor
• Add a small series resistance (to conduction resistance of the diode and wire
resistance of the transformer)
– Improves the ripple factor considerably.
– Limit the forward current → extend the life of diodes and transformer
• The charged capacitor retains some charge → A (bleeder) resistor (around 1
kΩ, 0.25 or 0.5 W) connected across discharges the capacitor in a few
seconds. If a led indicator is connected, then no need for such a resistor.
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Inductive Filters
+
Vi
-
L
RL
+
Vi: Rectified input voltage
Vo
-
Vo: Filtered output volt
RL: Effective load resistance
• Inductive filters have better control of the ripple for
large load currents.
• The inductor behaves as a short circuit for the DC
component. Hence, when 2fL » RL the DC value of the
output is approximately 2Vi/π and the ripple factor r ≈
0.118RL/fL where RL is the effective load resistance, f is
the frequency of the ripple and L is the inductance (in
Henry).
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L and π section filters
+
Vi
+
L
C
RL
-
Vo
-
+
Vi
-
C1
+
Vo
C2
RL
π - section
L - section
r = 3300/C1RLC2L
Vdc = Vm-4200*Idc/C
r = 0.83/LC
Vdc = 0.636Vm
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L
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-
Summary of filter responses
+
+ + L
+
+ +
+ L
+
L
Vi
Vo
Vo
Vi RL Vo
Vi
R
Vi
C L
RL Vo
C1 C2 R L
C
- Capacitive
Inductive
L - section
π - section
Vi: Rectified input voltage;
Vo: Filtered output volt.
R L : Effective load resistance;
Frequency: 60 Hz
Effective (rms) value of AC part
Ripple factor (r) = ----------------------------------------Average (dc) value of output
r
Vdc
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R
L
----1600L
2400
-----R C
L
0.83
----LC
0.636Vm Vm-4200Idc/C
0.636Vm
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3300
--------C R C L
1 L 2
Vm-4200Idc/C
Exercise 3
A series R-L circuit has R = 0.1 kΩ and L = 10 mH.
The circuit is excited by Vi = 5 + 10 sin(1000t)
V
• Draw the circuit diagram
• Calculate the voltages across R and L
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Duties for next lecture
• Study linear (dissipative) regulated power
supplies from the lecture notes
• Solve the exercises in this lecture in detail at
home and bring it to the next lecture for a
discussion in class
• Be prepared for a quiz and active learning
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