ECE 1750

Week 3 (part 1)Week 3 (part 1)

RectifiersRectifiers

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Rectifier• Rectifiers convert ac into dc• Rectifiers convert ac into dc

• Some commercial rectifiers:

(Used to charge batteries like those on the right)

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Example of Assumed State AnalysisExample of Assumed State Analysis

++

+Vac

–RL–

•Consider the Vac > 0 case

W k i t lli t th t I i•We make an intelligent guess that I is flowing out of the source + node.

• If current is flowing, then the diode must be “on”g,

•We see that KVL (Vac = I • RL ) is satisfied

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•Thus, our assumed state is correct

Example of Assumed State Analysis

+ ++ − 1V +10V

–RL 11V

–

11V–

1V

•We make an intelligent guess that I is flowing out of the 11V source

• If current is flowing, then the top diode must be “on”

C t t fl b k d th h th b tt di d it t

•The bottom node of the load resistor is connected to the source f th i t th b k t th 11V

•Current cannot flow backward through the bottom diode, so it must be “off”

reference, so there is a current path back to the 11V source

•KVL dictates that the load resistor has 11V across it

•The bottom diode is reverse biased, and thus confirmed to be “off”

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The bottom diode is reverse biased, and thus confirmed to be off

•Thus our assumed state is correct

Assumed State Analysis+

1 3

RLWhat are the states of the diodes on or off?

–

+Vac–

4 2RL the diodes – on or off?

• Consider the Vac > 0 case

• We make an intelligent guess that I is flowing out of the source + node.g g g

• I cannot flow into diode #4, so diode #4 must be “off.” If current is flowing, then diode #1 must be “on.”

• I cannot flow into diode #3, so diode #3 must be “off.” I flows through RL.

• I comes to the junction of diodes #2 and #4. We have already determined

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that diode #4 is “off.” If current is flowing, then diode #2 must be “on,” and I continues to the –Vac terminal.

Assumed State Analysis, cont.

1

RL+

++−

+Vac > 0

–

2RL+

− −

• A check of voltages confirms that diode #4 is indeed reverse biased as we have assumed

• We see that KVL (V = I • R ) is satisfied

• A check of voltages confirms that diode #3 is indeed reverse biased as we have assumed

• We see that KVL (Vac = I • RL ) is satisfied

• Thus, our assumed states are correct

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• The same process can be repeated for Vac < 0, where it can be seen that diodes #3 and #4 are “on,” and diodes #1 and #2 are “off”

AC and DC Waveforms for a Resistive LoadC a d C a e o s o a es st e oad

1 +Vd

3 +Vd

+Vac > 0

–

2Vdc

– –Vac < 0

+

4Vdc

–

Vac

0

20

40

ts

Vdc

0

20

40

olts

-40

-20

00.00 8.33 16.67 25.00 33.33

Milliseconds

Vol

-40

-20

00.00 8.33 16.67 25.00 33.33

MillisecondsVo

With a resistive load, the ac and dc current waveforms have the same waveshapes as Vac and Vdc shown above

7Note – DC does not mean constant!

Half-wave rectifier

0 0

1 sin( )2

T p pdc R

V VV v dt d

T

VpV

tt dcV

+Vac RL

+

8

––

Full wave rectifier

• Waveforms

vvR

0 0

21 sin( )T p p

dc RV V

V v dt dT

t

vR v

dcVpV

ttt

R tifi 2 d dRectifier 2nd order Filter

Diode Bridge Rectifier(DBR)

v vV ≈ 120√2V ≈ 170 V

Vdc≈ 120√2Vdc ≈ 170Vdc

Vp≈ 120√2Vdc ≈ 170 V

t

t

DC link+

1 3+

√Iac

DC link

–

+≈ 120Vac rms

–

4 2≈ 120√2Vdc ≈ 170Vdc−

First order

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First order low pass

filter

DC-Side Voltage and Current for Two Different L d L lLoad Levels

Vdc

800W Load200W Load

T1

Ripple voltage increases

f

IdcAverage current increases (current pulse gets taller and wider)

+ Idc = |Iac|+

+

1

4

3

2

Iac

Idc |Iac|

11–

≈ 120Vac rms–

Approximate Formula for DC Ripple Voltagepp o ate o u a o C pp e o tage

22 11Energy consumed by constant load power P during the same time interval

With a first order low pass filter

tPCVCVpeak 2min

221

21

Energy given up by capacitor as its

power P during the same time interval

VpeakV

tPVV

222

Energy given up by capacitor as its voltage drops from Vpeak to Vmin minV

CVVpeak min

tP2))((C

tPVVVV peakpeak

2))(( minmin

2 tP

12)(

2)(min

min VVCtPVV

peakpeak

Approximate Formula for DC Ripple Voltage, tcont.

)(2)(

minmin VVC

tPVVpeak

peak

)( minpeak

2VVV For low ripple Ttand∆t

T/2

T 1

,2min peakpeak VVV For low ripple, 2

t and

f

PVVV )(peak

ripplepeaktopeakpeak fCVVVV

2)( min

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A second order low-pass filter realized by adding an inductor in the “dc-link” allows to reduce the required capacitance for a given ripple goal.

AC Current Waveform1f

T 1=

• The ac current waveform has significant harmonic content.

• High harmonic components circulating in the electric grid may create quality and technical problems (higher losses in cables and transformers)quality and technical problems (higher losses in cables and transformers).

• Harmonic content is measurements: total harmonic distortion (THD) and power factor

1 ( )T

t dt0 ( )Average Power. .Total Used (Apparent) Power RMS RMS

p t dtTp fV I

V I I

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60 , 60 , 60 ,

60 ,

. . Hz RMS Hz RMS Hz RMS

Hz RMS RMS RMS

V I Ip f

V I I

“Vampire” Loads

?= ?• “Vampire” loads have high leakage currents and low power factor.

Your new lab safety tool:

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Estimated Diode Conduction Losses

Estimate the average value I avg of the ac current over the conduction interval T cond

Estimate the average value Vavgof diode forward voltage drop over one conduction i t l T

i(t)

Tcond

interval Tcond

Since the forward voltage on the diode is approximately constant during the conduction

v(t)

condavgavgcondavgavg

avg TIVTIV

P 2404

Watts.

interval, the energy absorbed by the diode during the conduction interval is approximately V avg • I avg • T cond . Each diode has one conduction interval per 60Hz period, so the average power absorbed by all four diodes is then

condavgavgHz

avg T60

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Forward Voltage on One Diode

Zero Conducting Forward voltage on one diode

Zoom-InForward voltage on one diodeZero one diode

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AC Current WaveformC Cu e t a e o

One pulse like this passes through each diode, once

l f 60Hper cycle of 60Hz

The shape is nearly triangular, so the average g , gvalue is approximately one-half the peak

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