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06/27/22 1 POWER ELECTRONICS POWER COMPUTATIONS DET 309

11/24/20151 POWER ELECTRONICS POWER COMPUTATIONS DET 309

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Page 1: 11/24/20151 POWER ELECTRONICS POWER COMPUTATIONS DET 309

04/21/23 1

POWER ELECTRONICS

POWER COMPUTATIONS

DET 309

Page 2: 11/24/20151 POWER ELECTRONICS POWER COMPUTATIONS DET 309

04/21/23 2

2.1 INTRODUCTIONo Power computations are essential in analyzing and designing

power electronics circuits. o Basic power concepts are reviewed in this chapter, with

particular emphasis on power calculations for circuits with non-sinusoidal voltages and currents.

o Extra treatment is given to some special cases that are encountered frequently in power electronics.

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2.2 POWER AND ENERGY

o The instantaneous power for any device is computed from the voltage across it and the current in it

o Instantaneous power is

Instantaneous Power

(2.1))()()( titvtp

This relationship is valid for any device or circuit. Instantaneous power is generally a time-varying quantity.

o If v(t) is in volts and i(t) is in amperes, power has units of

watts, W

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Observe the passive sign convention illustrated in Fig. 2.1a

o The device is absorbing power if p(t) is positive at a specified time, t.

o The device is supplying power if p(t) is negative.

Sources frequently have an assumed current direction consistent with supplying power. With the convention of Fig. 2.1b, a positive indicates that the source is supplying power.

i(t)

v(t)

+

-

v(t)

+

-

i(t)

(a) (b)

Figure 2.1 (a) Passive sign convention: p(t) > 0 indicates power is being absorbed. (b) p(t) > 0 indicates power is being supplied by the source.

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(2.2)

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Energy

o Energy or work, is the integral of instantaneous power.

o Observing the passive sign convention, the energy absorbed by

a component in the time interval from t1 to t2 is

2

1

)(t

tdttpW

o If v(t) is in volts and i(t) is in amperes, energy has units of joules, J.

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(2.3)

6

Average Power

o Periodic voltage and current functions produce a periodic instantaneous power

function.

o Average power is the time average of p(t) over one or more periods.

o Average power, P is computed from

Tt

t

Tt

t

o

o

o

o

dttitvT

dttpT

P )()(1

)(1

Where T is the period of the power waveform. Combining Eqs. 2.3 and 2.2, power is also computed from energy per period:

(2.4)T

WP

Average power is sometimes called real power or active power, especially in ac circuits. The term power usually means average power. The total average power absorbed in a circuit equals the total average power supplied.

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Example 2-1 Power and EnergyVoltage and current (consistent with the passive sign convention) for a device are shown in Figs. 2.2a and 2.2b.

(a) Determine the instantaneous power absorbed by the device,

(b) Determine the energy absorbed by the device in one period,

(c) Determine the average power absorbed by the device.

10 ms 20 ms

(a)

v(t)

(t)

20 V

(b)

6 ms 20 ms

i(t)

(t)

20 A

- 15 A

10 ms 20 ms

(c)

p(t)

(t)

6 ms

400 W

-300 W

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Solution (a) The instantaneous power is computed from Eq. 2.1. The voltage and current are expressed as

Instantaneous power, shown in Fig. 2.2c, is the product of voltage and current and is expressed as

20ms10ms 0

10ms0 20)(

tV

tVtv

20msms 6 15

ms 60 20)(

tA

tAti

ms 20ms 10 0

ms 10ms 6 300

ms 60 400

)(

t

tW

tW

tp

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(b) Energy absorbed by the device in one period is determined from Eq. 2.2:

J

dtdtdtdttpWms

ms

ms

ms

t

t

ms

1.2

0300400)( 20

10

10

6

6

0

2

1

(c) Average power is determined from Eq. 2.3:

W

dtdtdtms

dttpT

PTt

t

ms

ms

ms

ms

mso

o

60

030040020

1)(

110

6

20

10

6

0

Average power could also be computed from Eq. 2.4 using the energy per period from part (b):

WmsJ

TW

P 6020

2.1

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2.5

10

o A special case that is frequently encountered in power electronics is the power absorbed or supplied from a dc source. Applications include battery-charging circuits and dc power supplies.

The average power absorbed by a dc voltage source v(t) which has a periodic current i(t) is derived from the basic definition of average power in Eq. 2.3:

Tt

t

Tt

t dcdco

o

o

o

dttiVT

dttitvT

P )(1

)()(1

Bringing the constant Vdc outside of the integral,

Tt

tdcdco

o

dttiT

VP )(1

The term in brackets is the average of the current waveform

Therefore, average power absorbed by a dc voltage source is the product of the voltage and the average current:

avgdcdc IVP

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(2.6)

11

Similarly, average power absorbed by a dc current source i(t) = idc is

dcavgdc IVP

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(2.7)

12

2.3 INDUCTORS AND CAPACITORS

o Inductors and capacitors have some particular characteristics

that are important in power electronics applications.

o For periodic currents and voltages,

o For an inductor, the stored energy is

(2.8)

2.3.1 Inductors

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If the inductor current is periodic, the stored energy at the end of one period is the same as at the beginning.

No net energy transfer indicates that the average power absorbed by an inductor is zero for steady-state periodic operation:

(2.9)

2.3.1 Inductors…cont.

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Instantaneous power is not necessarily zero because power may be absorbed during one part of the period and returned to the circuit during another part of the period. Furthermore, from the voltage-current relationship for the inductor,

2.3.1 Inductors…cont.

(2.10)

Rearranging and recognizing that the starting and ending values are the same for periodic currents,

(2.11)

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Multiplying by L/T yields an expression equivalent to the average voltage across the inductor over one period:

(2.12)

Therefore, for periodic currents, the average voltage across an inductor is zero.

2.3.1 Inductors…cont.

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For a capacitor, stored energy is

2.3 INDUCTORS AND CAPACITORS…cont.

2.3.2 Capacitors

If the capacitor voltage is periodic, the stored energy is the same at the end of a period as at the beginning.

Therefore, the average power absorbed by the capacitor is zero for steady-state periodic operation:

(2.14)

(2.13)

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From the voltage-current relationship for the capacitor,

2.3.2 Capacitors…cont.

Rearranging the preceding equation and recognizing that the starting and ending values are the same for periodic voltages,

(2.16)

(2.15)

Multiplying by C/T yields an expression for average current in the capacitor over one period:

(2.17)

Therefore, for periodic voltages, the average current in a capacitor is zero.

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Example 2.2 Power and Voltage for an Inductor

The current in the 5-mH inductor of Fig. 2.3a is the periodic triangular wave shown in Fig. 2.3b. Determine the voltage, instantaneous power, and average power for the inductor.

Figure 2.3 (a) Circuit for Example 2.2. (b) Inductor current, (c) Inductor voltage, (d) Inductor instantaneous power.

i(t)

+

_

v(t)5 mH

(a)1 ms 2 ms 3 ms 4 ms

t

4 A

i(t)

(b)

t

20 V

v(t)

- 20 V

p(t)

t

80 W

- 80 W

(c) (d)

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The voltage across the inductor is computed from and is shown in Fig. 2.3c.

The average inductor voltage is zero, as can be determined from Fig. 2.3c by inspection.

The instantaneous power in the inductor is determined from and is shown in Fig. 2.3d.

When is positive, the inductor is absorbing power, and when is negative, the inductor is supplying power. The average inductor power is zero.

Solution

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2.4 ENERGY RECOVERY

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Thank You