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by Mohamed A El-Sharkawi.

### Transcript of Presentation 2 : Power Electronics

Power ElectronicsProfessor Mohamed A. El-Sharkawi

Power ControlPowerOn-time (ton)

ton Ps = P

P Ps

Off-time

(toff)Period (

Time

)2

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Off-time

(toff)Period (

Time (t)

)

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toff3

Energy Consumption (E)Power POn-time (ton)

PsOff-time

(toff)Period (

Time (t)

)

Duty Ratio (K)

E Pt

ton Es = Ps t = Pt [email protected] of Washington 4

vswivs

Ideal Switch+vs R

vt-

R

i

vsVsw

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Bi-polar Transistor(C) Collector Base (B) N (B) P N Emitter (E) (E) (B) VBE IB (C) VCB

(C) IC

VCE

IE (E)

IC I B I E = I B + IC

VCE = VCB + VBE

(C) IC VCB IB (B) VBE IE VCE

Characteristics of Bi-polar TransistorSaturation Region IC IB1 < IB2 IB1 Linear Region IB= 0 V 0.6 BE Cut Off Region VCE

IB

(E)

Base Characteristics

Collector Characteristics

IC RL IB V CE V CC

IC VCC RL (1)

IB max

(2) IB = 0

VCC = VCE + RL I CAt point (1) VCE is very small Closed switch Open switch

VCC

VCE

At point (2) IC is very small

VCC IC RL

VCE VCC

Example Estimate the losses of the transistor at point 1 and 2. Also calculate the losses at a mid point in the linear region where IB=0.1A. The current gain in the saturation region is 4.9 and in the linear region is 50.

IC 10IB max=2A

IC VCC RL (1)

IB max

V CE 100V (2) IB = 0 [email protected] of Washington

VCE13

SolutionIC 10IB max=2A

IC Vcc RL

IB max 1 3 2 IB = 0 VCC VCE

VCE 100V

At point 1 Total losses = base loses + collector lossesTotal losses = I B max * VBE + I C1 * VCE1

2* 0.7 + [4.9 * 2]* (100 4.9 * 2 * 10 ) = 21 W

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SolutionIC 10IB max=2A

IC Vcc RL

IB max 1 3 2 IB = 0 VCC VCE

VCE 100V

At point 2 Total losses = collector losses Assume VCE=0.99 VCCTotal losses = I C 2 * VCE 2

100 0.99 * 100 * (0.99* 100) = 10 W 10 [email protected] of Washington 15

SolutionIC 10IB max=0.1A

IC Vcc RL

IB max 1 3 2 IB = 0 VCC VCE

VCE 100V

At point 3 Total losses = base loses + collector losses

Total losses = I B 3 * VBE + I C 3 * VCE 3

0.1* 0.7 + [50 * 0.1]* (100 50 * 0.1* 10 ) = 250.07 W

Power transistors cannot operate in the linear [email protected] of Washington 16

Thyristors [Silicon Controlled Rectifier (SCR)]Anode (A)IA

Ig = max Ig > 0 Gate (G)VRB

Ig = 0

IhV AK V TO VBO

Cathode (K)

Closing Conditions of SCR1. Positive anode to cathode voltage (VAK) 2. Maximum triggering pulse is applied (Ig)Anode (A)

Gate (G)

Cathode (K)

Closing angle is 18

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Opening Conditions of SCR1. Anode current is below the holding value (Ih)VRB IA

Ig = 0 IhV AK

Opening angle is [email protected] of Washington 19

Power Converters

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Power Converters

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AC/DC Converters

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Single-Phase, Half-Wave+ vt -

vs

i

R

vs = Vmax sin( t )

vs i = ( only when SCR is closed ) Rvt = i R = vs ( only when SCR is closed )

vt i= Rvt

i

vs

i

+ vt -

R

tvs24

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Average Voltage Across the Load1 Vave = 22

1 vt d t = 2 vs d t 0

vt

i

tvs25

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1 1 Vave = vs d t = 2 vs d t 2

vt

i

1 Vave = Vmax sin ( t ) dt 2

Vmax Vave = ( 1 + cos ) 2

I ave Vave = R

tvs26

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VaveVmax

Vmax Vave = ( 1 + cos ) 2

Vmax 2

2

Root-Mean-Squares (RMS)

(.)1 22

2

0

. dt

Root Mean Squares of f1 Step 2: 222

0

( f ) dt2

Step 1: Step 3:

(f)

2

1 2

( f ) d t 2 0

Concept of RMSv2Average of v2

Square root of the average of [email protected] of Washington

tvAverage of v=030

Root-Mean-Squares (RMS) of a sinusoidal voltage1 Vmean = Vave = 21 Vrms = 22 0

2 0

v( t ) dt

[v( t )]

2

1 dt = 2

2 0

[Vmax

sin( t )]2 dt

RMS of load voltage2 Vmax Vrms = [ sin( t )]2 d t =

vt

i

vs

t

2

2 Vmax [ 1 cos( 2 t ] dt

4

Vrms

Vmax = 2

sin ( 2 ) [1 + ] 2RMS of Supply Voltage

Vmax =

2 Vs rms

VrmsVmax 2

Vrms I rms = R

Example.2: An ac source of 110V (rms) is connected to a resistive element of 2 through a single SCR. o o For = 45 and 90 , calculate the followings: a) b) c) rms voltage across the load resistance rms current of the resistance Average voltage drop across the SCR

vs

i

Solution: For = 45o

+ vt -

R

a)

Vrms =

Vsrms 2

sin(2 ) 110 1 + 2 = 2

45 1 180 + sin( 90 ) = 74.13 V 2

(

)

b)

V 74.13 I rms = rms = = 37.07 A R 21 v s dt + v s dt VSCR = 2 0 VSCR = 2

c)

V = max ( 1 + cos ) 2

vt

2 110 [ 1 + cos( 45 )] = 42.27 V 2

This looks like the negative i of the average voltage across the load. Why?tvs

Electric Power2 Vrms 2 P = = I rms R

R

2 Vmax P = [2( ) + sin( 2 )]

8 R

Single-Phase, Full-Wave, AC-to-DC Conversion for Resistive Loadsi1 S1 vs A S4 B S2 D S3 i2 R vt C

Single-Phase, Full-Wave, AC-to-DC 2-SCRs and 2 Diodesi1 S1 vs A D2 B D1 D S2 i2 R vt C

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i1 S1 vs A S4 B S2 S3 i2

C

R

vt

D

vt

i1

vt

i2

vs

t

vt

i1

vt

i2

vs

t

Vave =

1

v

t

dt =2

1

2

V

max

sin( t ) dt =

Vmax

( 1 + cos )

Vrms =

1 2

v(t ) dt =0

1

[Vmax sin(t )]2 dt

Vrms =

2 Vmax sin( t )2 dt =

2 Vmax [ 1 cos( 2t )] dt

2

Vmax Vrms = 2

sin( 2 ) 1 + 2

2 2 Vrms Vmax P = = [2( ) + sin( 2 )]

R

4 R

Half Wave Versus Full WaveHalf WaveAverage Voltage RMS Voltage Power

Full WaveVave =Vrms

Vmax Vave = (1 + cos ) 2Vrms = Vmax 2 sin( 2 ) 1 + 2

Vmax

(1 + cos )

Vmax = 2

sin( 2 ) 1 + 2

2 Vmax P = [2( ) + sin ( 2 )] 8 R

2 Vmax P = [2( ) + sin ( 2 )] 4 R

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ExampleA full-wave, ac/dc converter is connected to a resistive load of 5 . The voltage of the ac source is 110 V(rms). It is required that the rms voltage across the load to be 55 V. Calculate the triggering angle, and the load power.

SolutionVrms = Vsrms55 = 110

[1

sin( 2 ) [1 + ] 2 sin( 2 ) + 2 ]

sin( 2 ) 2.25 = 180 2

112.5 o

2 Vrms ( 55 )2 P= = = 605 W R 5

DC/DC Converters

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DC-to-DC Conversion1. Step-down (Buck) converter: where the output voltage of the converter is lower than the input voltage. 2. Step-up (Boost) converter: where the voltage is higher than the input voltage. 3. Step-down/step-up (Buck-Boost) converter. output

Step Down (Buck converter)VS I Vl ton Time VS VCE

I

+ Vl -

ton

Time

Vave =

1

t on 0

Vs dt =

ton

Vs = K Vs

Examplef = 5 kHz ( switching frequency ) Vs = 12 V ; Vave = 5 V ; ton = ?Solution

1 1 = = = 0.2 ms f 5Vave = ton

Vs = K Vs

5 K = = o.417 12 ton = 0.417 0.2 = 0.0834 [email protected] of Washington 47

Step up (Boost converter)L

it

Is vsC

R

vt

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L

it

Is vsC

R

vt

Keep in mind Inductor current is unidirectional Voltage across inductor reverses Inductor cannot permanently store energy

L

L

ion vsvs

ioffC

R

vt

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L

L

vs

ion

i off

vs

C

R

vt

i

ion

ioff

ton

toff

Time

50

Inductor current

i

tonInductor voltagevon

toff

Time

voffTime

Energy is acquired by inductor Energy is released by inductor51

L

L

i on

i off

VS

VS

C

R

vt

Vs = L

ionton

V s = vt L

ioff t off

At steady state ion = ioff

vt