power electronics - philadelphia university...3 dr. firas obeidat faculty of engineering...

Power Electronics The Buck (Step-Down)

Converter

1

Dr. Firas Obeidat

2

Table of contents

1 • Introduction

2

• Step Down Chopper with Resistive Load

3

• Step Down Chopper with RL Load

4

• Step Down Chopper with Low Pass Filter

Dr. Firas Obeidat Faculty of Engineering Philadelphia University

3 Dr. Firas Obeidat Faculty of Engineering Philadelphia University

Introduction

DC-DC converters are power electronic circuits that convert a DC voltage to a different DC voltage level, often providing a regulated output.

LOAD

Vcontrol (derived from

feedback circuit)

DC supply

(from rectifier-

filter, battery,

fuel cell etc.)

DC output

General block diagram

Applications: – Switched-mode power supply (SMPS), DC motor control,

battery chargers, subway cars, trolley buses, vehicles, etc.


Introduction

Main Types of Choppers

1- Step-down DC-DC converter.

In step down chopper output voltage is less than input

voltage.

2- Step-up DC-DC converter.

In step up chopper output voltage is more than input

voltage.

3- Buck-Boost converter (Step-down/step-up converter).

4- Cuk converter.


The Buck (Step-Down) Converter

Step down chopper as Buck

converted is used to reduce the input

voltage level at the output side.

Circuit diagram of a step down

chopper is shown in the figure.

When CH is turned ON, Vs directly

appears across the load as shown in

figure. So VO=VS.

When CH is turned OFF, Vs is

disconnected from the load. So output

voltage VO = 0.

The voltage waveform of step down

chopper


The Buck (Step-Down) Converter TON → It is the interval in which chopper is in ON state.

TOFF → It is the interval in which chopper is in OFF state.

VS → Source or input voltage.

VO → Output or load voltage.

T → Chopping period = TON + TOFF

F=1/T is the frequency of chopper switching or chopping frequency

Operation of Step Down Chopper with Resistive Load

When CH is ON, VO = VS When CH is OFF, VO = 0

The Average output voltage is

𝑉𝑑𝑐 = 𝑉𝑜 =1

𝑇 𝑉𝑠𝑑𝑡

𝑇𝑂𝑁

0

=𝑉𝑠𝑇𝑂𝑁𝑇

= 𝐷𝑉𝑠

𝐼𝑑𝑐 =𝑉𝑑𝑐𝑅=𝐷𝑉𝑠𝑅



The rms output voltage is

Where,

D is duty cycle = TON/T. TON can be varied from 0 to T, so 0 ≤ D ≤ 1.

The output voltage VO can be varied from 0 to VS.

𝑉𝑟𝑚𝑠 =1

𝑇 𝑉𝑠

2𝑑𝑡

𝑇𝑂𝑁

0

= 𝑉𝑠𝑇𝑂𝑁𝑇= 𝐷𝑉𝑠

The output voltage is always less than the

input voltage and hence the name step

down chopper is justified.

𝐷 =𝑇𝑂𝑁𝑇

𝑇 = 𝑇𝑂𝑁 + 𝑇𝑂𝐹𝐹

𝑃𝑜 = 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠 =𝑉𝑟𝑚𝑠

2

𝑅= 𝐷

𝑉𝑠2

𝑅

𝐼𝑟𝑚𝑠 =𝑉𝑟𝑚𝑠𝑅=

𝐷𝑉𝑠𝑅

Step Down Chopper with Resistive Load



Ripple factor (RF) can be found from

𝑅𝐹 =𝑉𝑟𝑚𝑠𝑉𝑑𝑐

2

− 1 =𝐷𝑉𝑠

2

𝐷2𝑉𝑠2 − 1 =

1

𝐷− 1 =

1 − 𝐷

𝐷

Methods of Control

1- Pulse Width Modulation

• tON is varied keeping chopping frequency ‘f’ & chopping

period ‘T’ constant.

• Output voltage is varied by varying the ON time tON

2- Variable Frequency Control

• Chopping frequency ‘f’ is varied keeping either tON or tOFF

constant.

• To obtain full output voltage range, frequency has to be

varied over a wide range.

• This method produces harmonics in the output and for large

tOFF load current may become discontinuous

V0

V

V

V0

t

ttON

tON tOFF

tOFF

T

v0

V

V

v0

t

t

tON

tON

T

T

tOFF

tOFF

Pulse Width Modulation Method

Variable Frequency Control Method




Examlpe: A transistor dc chopper circuit (Buck converter) is supplied with

power form an ideal battery of 100 V. The load voltage waveform consists

of rectangular pulses of duration 1 ms in an overall cycle time of 2.5 ms.

Calculate, for resistive load of 10 Ω.

(a) The duty cycle D.

(b) The average value of the output voltage Vdc.

(c) The rms value of the output voltage Vrms.

(d) The ripple factor RF.

(e) The output dc power.

𝐷 =𝑡𝑂𝑁𝑇=1𝑚𝑠𝑒𝑐

2.5𝑚𝑠𝑒𝑐= 0.4

(a)

𝑉𝑑𝑐 = 𝐷𝑉𝑠 = 0.4 × 100 = 40 V (b)

(c) 𝑉𝑟𝑚𝑠 = 𝐷𝑉𝑠 = 0.4 × 100 = 63.2 V

(d) 𝑅𝐹 =1−𝐷

𝐷 =

1−0.4

0.4= 1.225

𝑃𝑜 =𝑉𝑑𝑐

2

𝑅=402

10= 160 W (e)




When chopper is ON, supply is connected

across load. Current flows from supply to

load.

When chopper is OFF, load current

continues to flow in the same direction

through FWD due to energy stored in

inductor ‘L’.

Step Down Chopper with RL Load

Load current can be continuous or

discontinuous depending on the values of

‘L’ and duty cycle ‘D’

For a continuous current operation, load

current varies between two limits Imax and

Imin.

When current becomes equal to Imax the

chopper is turned-off and it is turned-on

when current reduces to Imin.

Outputvoltage

Outputcurrent

v0

V

i0

Imax

Imin

t

t

tON

T

tOFF

Continuouscurrent

Outputcurrent

t

Discontinuouscurrent

i0



When the switch is closed in the buck

converter, the circuit will be as shown in

the figure, the diode is reverse-biased.

Continuous Current Operation When Chopper Is ON (0 ≤t ≤

tON)

The voltage across the inductor is

𝑉𝑠 = 𝑉𝑅 + 𝑉𝐿

𝑉𝑠 = 𝑉𝑅 + 𝐿𝑑𝑖

𝑑𝑡 →

𝑑𝑖

𝑑𝑡=𝑉𝑠 − 𝑉𝑅𝐿

∆𝑖 = 𝑉𝑠 − 𝑉𝑅𝐿

𝑑𝑡

𝐷𝑇

0

=𝑉𝑠 − 𝑉𝑅𝐿

𝐷𝑇 =𝑉𝑠 − 𝑉𝑅𝐿

𝑡𝑂𝑁

Outputvoltage

Outputcurrent

v0

V

i0

Imax

Imin

t

t

tON

T

tOFF

Continuouscurrent

Outputcurrent

t


i0

𝑑𝑖

𝑑𝑡=∆𝑖

𝑡𝑂𝑁=𝐼𝑚𝑎𝑥 − 𝐼𝑚𝑖𝑛

𝑡𝑂𝑁=𝑉𝑠 − 𝑉𝑅𝐿

𝑖𝑜1 = 𝐼𝑚𝑖𝑛 +𝐼𝑚𝑎𝑥 − 𝐼𝑚𝑖𝑛

𝑡𝑂𝑁𝑡 = 𝐼𝑚𝑖𝑛 +

𝐼𝑚𝑎𝑥 − 𝐼𝑚𝑖𝑛𝐷𝑇

𝑡 = 𝐼𝑚𝑖𝑛 +𝑉𝑠 − 𝑉𝑅𝐿

𝑡 From straight line equation

(1)

(2)




Continuous Current Operation When Chopper Is OFF (tON ≤t ≤ T)

Outputvoltage

Outputcurrent

v0

V

i0

Imax

Imin

t

t

tON

T

tOFF

Continuouscurrent

Outputcurrent

t


i0

0 = 𝑉𝑅 + 𝑉𝐿

0 = 𝑉𝑅 + 𝐿𝑑𝑖

𝑑𝑡 →

𝑑𝑖

𝑑𝑡= −

𝑉𝑅𝐿

∆𝑖 = −𝑉𝑅𝐿𝑑𝑡

𝑡𝑂𝐹𝐹

0

= −𝑉𝑅𝐿𝑡𝑂𝐹𝐹

𝑑𝑖

𝑑𝑡=∆𝑖

𝑡𝑂𝐹𝐹=𝐼𝑚𝑖𝑛 − 𝐼𝑚𝑎𝑥𝑡𝑂𝐹𝐹

= −𝐼𝑚𝑎𝑥 − 𝐼𝑚𝑖𝑛𝑡𝑂𝐹𝐹

= −𝑉𝑅𝐿

(3)

𝑖𝑜2 = 𝐼𝑚𝑎𝑥 +𝐼𝑚𝑖𝑛 − 𝐼𝑚𝑎𝑥𝑡𝑂𝐹𝐹

𝑡 − 𝑡𝑂𝑁 = 𝐼𝑚𝑎𝑥 −𝑉𝑅𝐿(𝑡 − 𝑡𝑂𝑁)

From straight line equation

(4)




𝑉𝑠 − 𝑉𝑅𝐿

𝑡𝑂𝑁 −𝑉𝑅𝐿𝑡𝑂𝐹𝐹 = 0

𝑉𝑠 − 𝑉𝑅𝑉𝑅

=𝑡𝑂𝐹𝐹𝑡𝑂𝑁

𝑉𝑠𝑉𝑅− 1 =

𝑡𝑂𝐹𝐹𝑡𝑂𝑁

𝑉𝑠𝑉𝑅=𝑡𝑂𝐹𝐹𝑡𝑂𝑁

+ 1

𝑉𝑠𝑉𝑅=𝑡𝑂𝐹𝐹 + 𝑡𝑂𝑁𝑡𝑂𝑁

=𝑇

𝑡𝑂𝑁 𝑉𝑅 = 𝐷𝑉𝑠

From equation (1)

∆𝑖 =𝑉𝑠 − 𝐷𝑉𝑠𝐿

𝐷𝑇 =𝑉𝑠 1 − 𝐷 𝐷

𝐿𝑓

since

𝑓 =1

𝑇

𝐷 =𝑡𝑂𝑁𝑇


Steady-state operation requires that the inductor current at the end of the

switching cycle be the same as that at the beginning, meaning that the net change

in inductor current over one period is zero. This requires



At steady state operation, the average inductor current must be the same as

the average current in the load resistor.

𝐼𝐿 = 𝐼𝑅 =𝑉𝑅𝑅

The maximum and minimum values of the inductor current are computed as

𝐼𝑚𝑎𝑥 = 𝐼𝐿 +∆𝑖

2

𝐼𝑚𝑎𝑥 = 𝐼𝐿 +𝑉𝑠 1 − 𝐷 𝐷

2𝐿𝑓= 𝐼𝐿 +

𝑉𝑅 1 − 𝐷

2𝐿𝑓

𝐼𝑚𝑖𝑛 = 𝐼𝐿 −∆𝑖

2

𝐼𝑚𝑖𝑛 = 𝐼𝐿 −𝑉𝑠 1 − 𝐷 𝐷

2𝐿𝑓= 𝐼𝐿 −

𝑉𝑅 1 − 𝐷

2𝐿𝑓

The average dc output voltage and current can found as

𝑉𝑑𝑐 = 𝐷𝑉𝑠 𝐼𝑑𝑐 ≅𝐼𝑚𝑎𝑥 − 𝐼𝑚𝑖𝑛

2




Examlpe: A dc chopper has a resistive load of 20Ω and input voltage

VS=220V. When chopper is ON, its voltage drop is 1.5 volts and chopping

frequency is 10 kHz. If the duty cycle is 80%, determine the average output

voltage and the chopper on time.

𝑉𝑑𝑐 = 𝐷𝑉𝑠 =𝑡𝑂𝑁𝑇

𝑉𝑠 − 𝑉𝐶𝐻 = 0.8 220 − 1.5 = 174.8 V

𝑉𝑠 = 220V

𝑇 =1

𝑓=

1

10 × 10−3= 0.1m 𝑠𝑒𝑐

𝑡𝑂𝑁 = 𝐷𝑇 = 0.8 × 0.1 × 10−3 = 80μ 𝑠𝑒𝑐

𝐷 =𝑡𝑂𝑁𝑇= 0.8




Examlpe: A Chopper circuit is operating at a frequency of 2 kHz on a 460

V supply. If the load voltage is 350 volts, calculate the conduction period of

the thyristor in each cycle.

𝑉𝑑𝑐 = 𝐷𝑉𝑠 =𝑡𝑂𝑁𝑇𝑉𝑠

𝑉𝑠 = 460V

Chopping period

𝑇 =1

𝑓=

1

2 × 10−3= 0.5m 𝑠𝑒𝑐

𝑡𝑂𝑁 =𝑇𝑉𝑑𝑐𝑉𝑠

=0.5 × 10−3 × 350

460= 0.38m 𝑠𝑒𝑐




Step Down Chopper with Low Pass Filter

This converter is used if the

objective is to produce an output

that is purely DC.

If the low-pass filter is ideal, the

output voltage is the average of the

input voltage to the filter.

Analysis for the Switch Closed

When the switch is closed in the buck

converter circuit of fig. a, the diode is

reverse-biased and fig. b is an

equivalent circuit. The voltage across

the inductor is




Analysis for the Switch Closed

Since the derivative of the current

is a positive constant, the current

increases linearly. The change in

current while the switch is closed is

computed by modifying the

preceding equation.

(1)

(∆𝑖𝐿)𝑐𝑙𝑜𝑠𝑒𝑑= 𝑉𝑠 − 𝑉𝑜𝐿

𝐷𝑇

0

𝑑𝑡 =𝑉𝑠 − 𝑉𝑜𝐿

𝐷𝑇

or




Analysis for the Switch Opened

When the switch is open, the diode

becomes forward-biased to carry the

inductor current and the equivalent

circuit of fig. c applies. The voltage

across the inductor when the switch

is open is

The derivative of current in the inductor is a negative constant, and the

current decreases linearly. The change in inductor current when the switch is

open is

(2) (∆𝑖𝐿)𝑜𝑝𝑒𝑛𝑒𝑑=

−𝑉𝑜𝐿

(1−𝐷)𝑇

0

𝑑𝑡 =−𝑉𝑜𝐿(1 − 𝐷)𝑇 or




Steady-state operation requires that the

inductor current at the end of the switching

cycle be the same as that at the beginning,

meaning that the net change in inductor

current over one period is zero. This

requires

Using equations 1&2

The average inductor current must be the

same as the average current in the load

resistor, since the average capacitor current

must be zero for steady-state operation:

t

t

t

t

t

t

t

0

Vo

Ic

lV

Di

Si

Li

bev

onT offT

sT

oI

in oV V

oV

Q

Vo




The maximum and minimum values of the inductor current are computed as

Since Imin=0 is the boundary between continuous and discontinuous current,

The minimum combination of inductance and switching frequency for

continuous current in the buck converter is




where Lmin is the minimum inductance required for continuous current. In

practice, a value of inductance greater than Lmin is desirable to ensure

continuous current.

Since the converter components are assumed to be ideal, the power supplied by

the source must be the same as the power absorbed by the load resistor.

This relationship is similar to the voltage-current relationship

for a transformer in AC applications. Therefore, the buck

converter circuit is equivalent to a DC transformer.

In the preceding analysis, the capacitor was assumed to be very large to keep

the output voltage constant. In practice, the output voltage cannot be kept

perfectly constant with a finite capacitance. The variation in output voltage, or

ripple, is computed from the voltage-current relationship of the capacitor. The

current in the capacitor is




While the capacitor current is positive, the

capacitor is charging. From the definition of

capacitance,

The change in charge ∆Q is the area of the

triangle above the time axis

Substitute (∆iL)open in the above equation yields

∆Vo is the peak-to-peak ripple voltage at the output

The required capacitance in terms of specified voltage ripple:




Examlpe: buck dc-dc converter with Low Pass Filter has the following

parameters:

Assuming ideal components, calculate (a) the output voltage Vo, (b) the

maximum and minimum inductor current, and (c) the output voltage

ripple.

(a)

(b)

(c)

The average inductor current is 1 A, and ∆iL=1.5 A.

power electronics - philadelphia university...3 dr. firas obeidat faculty of engineering...

Documents