Power Supplies

Linear Regulated Supplies

Switched Regulated Supplies

Batteries

Current

Direct

π/2 π 2 π

Im

-Im

Idc

π t

Current

Alternating

π t

π/2 π 2π

Im

-Im

The Power

Supply

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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

wavethe of value (dc) Average

wavethe of component galternatin of value (rms) Effective = or(r)ripplefact

100 xR

R = 100 V

V - V = (%) regulation LoadL

0FLML

⊗

100 xVV

V = ) V/ (% regulation InputONI

ONI •∆

∆

I x V

I x V = P

P = )( EfficiencyNINI

LO

NI

OUTη

2

0

21iaveragedti

TII

T

rmseff === ∫RIdtiT

RRdti

TP eff

TT

AV2

0

2

0

21 === ∫∫

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Current

Alternating

πt

π/2 π 2 π

Im

-Im

OutputStep-downtransformer

Rectifier Filter RegulatorDC

ACInput

AC linecomponents

Block diagram of a linear regulated power supply

CurrentDirect

π/2 π 2π

Im

-Im

Idc

πt

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127 Vrms

60 Hz

+

Fuse

Unregulateddc output

Transientsuppressor

ac line filter

Snubber

ac line transformer

Bridge rectifier

Filter capacitor

Input

socket

Unregulated supply with ac line components

(a transient suppressor and line filter)

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127 Vrms

60 Hz

+

Fuse

Unregulateddc output

Transientsuppressor

ac line filter

Snubber

ac line transformer

Bridge rectifier

Filter capacitor

Input

socket

•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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127 Vrms

60 Hz

+

Fuse

Unregulateddc output

Transientsuppressor

ac line filter

Snubber

ac line transformer

Bridge rectifier

Filter capacitor

Input

socket

Transient suppressor and line filter

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127 Vrms

60 Hz

+

Fuse

Unregulateddc output

Transientsuppressor

ac line filter

Snubber

ac line transformer

Bridge rectifier

Filter capacitor

Input

socket

RC Snubbers

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Transformer

η 2V + V + V

= Vdiodesrippledunregulate

rtransforme

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Exercise – 2

• 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

η 2

V + V + V = V

diodesrippledunregulatertransforme

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Rectifier Diodes

Diode Maximum Current Maximum

Reverse Voltage

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)

Output: half-wave varying DC (Pulsating DC)

(using only half the AC wave)

127 Vrms

60 Hz

VmSinωt

D1

RL

+

-

Vo = Vm - Vd with Vd ≅ 1 volt.

Vdc = (Vm - Vd)/π , Vrms = (Vm - Vd)/2

yielding a ripple factor (r) = 1.21

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ππππ/2 ππππ 2ππππ

Vm

-Vm

Full-wave rectified

with a center tapped

transformer

D1

R LD2

+

-

Vo = Vm - Vd with

Vd ≅ 1 volt.

Vdc = 2*(Vm - Vd)/π,

Vrms = (Vm - Vd)/√2

yielding a much

reduced ripple factor

that is r = 0.483

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Bridge Rectifiers

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D1

D4D2

D3π/2 π 2 π

Vm

-VmPulsating DC

Full-wave rectified

with a bridge

rectifier

D1

R L

D2

D3

D4

+

-

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Smoothing Filters

Types of smoothing filters

• Capacitive

• Inductive

• L – Section

• π - Section

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Rectifier diodes

Filter

capacitor

IL

Amplitude

Time

Heavy load

Light load

Filtered DC

Capacitive filter

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

+ +

- -RLC

ViVo

Amplitude

Time

Heavy load

Light load

T1

T2

T

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/T

1yielding

Vr=IL*T

1/C. T

1≈ T; Vr=I

L/2fC for full-wave rectified; I

L= V

dc/R

L;

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 T

2↓⇒ 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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++

- -

L

RL

Vi Vo

Vi: Rectified input voltage

Vo: Filtered output volt

RL: Effective load resistance

Inductive Filters

• 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 » R

Lthe DC value of the

output is approximately 2Vi/π and the ripple factor r ≈0.118R

L/fL where R

Lis the effective load resistance, f is

the frequency of the ripple and L is the inductance (in Henry).

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+ +

- -

L

RLC1 C2

Vi

+ +

- -

L

RL

CVi Vo Vo

L - section π - section

L and ππππ section filters

r = 0.83/LC

Vdc = 0.636Vm

r = 3300/C1RLC2L

Vdc = Vm-4200*Idc/C

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VoVo

++

- -

L

RLVi

Inductive

+ +

- -RLC

ViVo

Capacitive

+ +

- -

L

RLCVi Vo

L - section

+ +

- -

L

RLC1 C2

Vi

π - section

RL

2400 0.83 3300

r ----- ------ ----- ---------1600L R

LC LC C

1R

LC2

L

Vdc 0.636Vm Vm-4200Idc/C 0.636Vm Vm-4200Idc/C

Vi: Rectified input voltage; Vo: Filtered output volt.

RL: Effective load resistance; Frequency: 60 Hz

Effective (rms) value of AC part

Ripple factor (r) = -----------------------------------------Average (dc) value of output

Summary of filter responses

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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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