Chapter Chapter 33Diode ApplicationsDiode ApplicationsDiode ApplicationsDiode Applications

Presented by Sopapun Suwansawang

t tttt

sv+

−

Block diagram of a dc power supply

Rectifier CircuitRectifier Circuit

HalfHalf--wave rectifierwave rectifier

FullFull--wave rectifierwave rectifier

R

D

Sv ov

+

−

RSv ov

+

−

DrDV

,0, =< oDS vVv

Half-wave rectifier

1,0, =< oDS vVv

DDs

DoDS rR

RVv

rR

RvVv

+−

+=≥ 0,

Dso Vvv −≈

1

In many applications, RrD <<

θ

sV

sv

ov

t

v

osD vvV −=

for simplicity, ,sin tVv ss ω= θω =tand

s

D

V

V1sin−=θ

svov

tω

v

π π2

We can also find the average value of the waveform of ov

Period

AreaV aveo =)(

+= ∫∫

π

π

π

ωωωπ

2

0)( 0)sin(

2

1tdtdtVV maveo

We can also find the average value of the waveform of ov

θ

sV

sv

ov

t

v

osD vvV −=

.

tdVtVPeriod

AreaV Dsaveo ωω

π

θπ

θ∫−

−==)(

)( )sin(2

1

θ

θ πθπ ≈− )2(Generally, the angle Generally, the angle is small and thus is small and thus 1cos ≈θ and and

Therefore, the last relation above reduces toTherefore, the last relation above reduces to

2)(Ds

aveo

VVV −≈

π

The peak current through each diode isThe peak current through each diode isThe peak current through each diode isThe peak current through each diode is

R

VVI Ds

peakD

−=)(

When using diodes in rectifier circuits we must calculate When using diodes in rectifier circuits we must calculate

2 2 important parameters:important parameters:

a. the maximum current that the diode will allow without being

damaged, and

b. the peak inverse voltage (PIV) that the diode can withstand

without reaching the reverse biased breakdown region.

,sin tVv ω=If the applied voltage is ,sin tVv ss ω=

sVPIV >

If the applied voltage is

sVPIV =

In practice we must use diodes whose reverse-biased breakdown voltages

70% or greater than the value of sV

Test your understanding

For the half-wave rectifier in Figure ,

R

D

Sv ov

+

−

define the input voltage is 12 V(rms) sinusoidal, and

(Neglecting the effect of ) Find the following:

(a) The half-cycles during which the diode conducts.

(b) The average value (dc component) of .

(c) The peak diode current.(d) The peak inverse voltage.

VVD 7.0≈ Ω=100R

Dr

Full-wave rectifier

+

−+

−

sv

sv

1D

2D

+

−

220 Vac50 Hz

Center-tappedR

+

−ov

v v−vv

t

sv sv−ovsV

DV

θ θ2

OND ,1 OND ,2 OND ,1 OND ,2

We can also find the average value of the waveform of ov

tdVtVPeriod

AreaV Dsaveo ωω

π

θπ

θ∫−

−==)(

)( )sin(1

θ πθπ ≈− )2(Generally, the angle Generally, the angle is small and thus is small and thus 1cos ≈θ and and

Therefore, the last relation above reduces toTherefore, the last relation above reduces to

V2D

saveo V

VV −≈

π

2)(

The peak current through each diode isThe peak current through each diode is

R

VVI Ds

peakD

−=)(

VVPIV −= 2 Ds VVPIV −= 2

Test your understanding

For the full-wave rectifier in Figure ,

+

−+

−

sv

sv

1D

2D

+

−

220 Vac50 Hz

Center-tappedR

+

−ov

define the input voltage is 12 V(rms) sinusoidal, and

(Neglecting the effect of ) Find the following:

(a) The average value (dc component) of .

(b) The peak diode current.(c) The peak inverse voltage.

VVD 7.0≈ Ω=100R

Dr

The bridge rectifier

sv

+

−

1D

2D3D

4D

R

+

−

ov

t

sv sv−ov

sVDV2

θ θ2

v

Because two diodes are in series in the conduction path, In each half-cycle,

the current flows through two diodes, so the output amplitude is two voltage

drops (about 1.4 V) less than the input amplitude. The average (or dc component)

of the output voltage is

Ds

aveo VV

V 22

)( −≈π

and the peak current of diode isand the peak current of diode is

R

VVI Ds

peakD

2)(

−=

)()( 23 fowardvvrevesev DoD +=

DsDDs VVVVVPIV −=+−= 2

Test your understanding

For the full-wave rectifier in Figure ,

sv

+

−

1D

2D3D

4D

R

+

−

ov

define the input voltage is 12 V(rms) sinusoidal, and

(Neglecting the effect of ) Find the following:

(a) The average value (dc component) of .

(b) The peak diode current.(c) The peak inverse voltage.

VVD 7.0≈ Ω=100R

Dr

The Rectifier with a Filter CapacitorThe Rectifier with a Filter Capacitor--The Peak Rectifier The Peak Rectifier

Sv oV

+

−

pVsv

t

oV

v

Sv oV

+

−

RC

DDi

Ci Ri

pV

t

ov

v

T ′

t∆

1t 2t

rV

Di

i

T

vdvCiii

R

vi

oCRCD

oR

+=+=

=

DiRi

t

RdtCiii oC

RCD +=+=

rpaveo VVV21

)( −=

pV

t

ov

v

T ′

t∆

1t 2t

rV

T Half-wave

pV

ov

t

v

VrT ′T

full-wave

Tf

21

=

TT

For full-wave rectifier with an RC filter is one-half the signal period.

to the signal frequency,Therefore, we can relate

The peak-to-peak ripple voltage can be derived using a procedure

identical to that above, thus identical to that above, thus

fRC

VV p

r 2=

and

+=

r

pRavgD V

VIi

21)( π

+=

r

pRD V

VIi

221(max) π