RECTIFIERS CIRCUITS

Textbook section 4.5

INEL 4201 Electronics I

Half-Wave Rectifier Circuit with Resistive Load

For positive half-cycle of input, source forces positive current through diode, diode is on, vo = vs.

During negative half cycle, negative current can�t exist in diode, diode is off, current in resistor is zero and vo =0 .

( assuming vD = 0. Otherwise vO = vS - vD when vS > vD )

Using CVD model, during on state of diode vo =(VP sinωt)- Von. Output voltage is zero when diode is off.

Often a step-up or step-down transformer is used to convert 120 V-60 Hz voltage available from power line to desired ac voltage level as shown.

Time-varying components in circuit output are removed using filter capacitor.

CVD = constant voltage drop

Example: Find ripple voltage if the secondary voltage VS,rms = 12.6V ,

f = 60Hz, R = 15⌦, C = 25, 000µF , and the diode’s voltage drop VON = 1V .

VS,peak =

p2⇥ VS,rms = 12.82V

vL,peak = VS,peak � VON = 11.82V

fRC = 60⇥ 15⌦⇥ 25⇥ 10

�3F = 22.5

Vr =

vL,peak

fRC

1

1 +

12fRC

=

11.82V

22.5

1

1 +

12⇥22.5

= 0.52V

Example: Find C necessary to obtain Vr = 0.2V if VS,rms = 12.6V , f =

60Hz, R = 15⌦, and VON = 1V .

C =

VL,peak � Vr/2

fRVr=

11.82V � 0.1V

60⇥ 15⌦⇥ 0.2V' 65000µF

Example: Find C necessary to obtain Vr = 0.2V if VS,rms = 12.6V , f =

60Hz, R = 1k⌦, and VON = 1V .

C =

VL,peak � Vr/2

fRVr=

11.82V � 0.1V

60⇥ 1k⌦⇥ 0.2V' 1000µF

9.0V

Example: Find ripple voltage if the secondary voltage VS,rms = 12.6V ,

f = 60Hz, R = 15⌦, C = 25, 000µF , and the diode’s voltage drop VON = 1V .

VS,peak =

p2⇥ VS,rms = 12.82V

vL,peak = VS,peak � VON = 11.82V

fRC = 60⇥ 15⌦⇥ 25⇥ 10

�3F = 22.5

Vr =

vL,peak

fRC

1

1 +

12fRC

=

11.82V

22.5

1

1 +

12⇥22.5

= 0.52V

Example: Find C necessary to obtain Vr = 0.2V if VS,rms = 12.6V , f =

60Hz, R = 15⌦, and VON = 1V .

C =

VL,peak � Vr/2

fRVr=

11.82V � 0.1V

60⇥ 15⌦⇥ 0.2V' 65000µF

Example: Find C necessary to obtain Vr = 0.2V if VS,rms = 12.6V , f =

60Hz, R = 1k⌦, and VON = 1V .

C =

VL,peak � Vr/2

fRVr=

11.82V � 0.1V

60⇥ 1k⌦⇥ 0.2V' 1000µF

9.0V

vL,peak =�

2 � 9V � 1V = 11.73V

iL,ave =vL,peak � vr/2

RL

� C�vC

�t= C

vr

1/f= fCvr

vL,peak

RL= 11.73V/15� = 0.78A

= fCvr +vr/2

RL= vr(60(0.025F ) + 1/30�)

= 1.53��1vr

vr =0.78

1.53V = 0.51V

Ripple voltage Vr: peak-to-peak value of variations of load voltage.

iC = C�vC�t

=

vL,dc

R

Let vL,dc ' VL,peak � Vr/2 and �t = 1/f where f is the frequency

Vr ' �vC =

vL,dc ⇥�t

RC' vL,peak � Vr/2

fRC

Vr ' vL,peak

fRC

1

1 +

12fRC

If Vr << VL,peak,

Vr ' vL,peak

fRC

To select C to obtain a specific Vr, use

C =

vL,peak � Vr/2

fRVr

In general,

Example: Find ripple voltage if the secondary voltage VS,rms = 12.6V ,

f = 60Hz, R = 15⌦, C = 25, 000µF , and the diode’s voltage drop VON = 1V .

VS,peak =

p2⇥ VS,rms = 12.82V

vL,peak = VS,peak � VON = 11.82V

fRC = 60⇥ 15⌦⇥ 25⇥ 10

�3F = 22.5

Vr =

vL,peak

fRC

1

1 +

12fRC

=

11.82V

22.5

1

1 +

12⇥22.5

= 0.52V

Example: Find C necessary to obtain Vr = 0.2V if VS,rms = 12.6V , f =

60Hz, R = 15⌦, and VON = 1V .

C =

VL,peak � Vr/2

fRVr=

11.82V � 0.1V

60⇥ 15⌦⇥ 0.2V' 65000µF

Example: Find C necessary to obtain Vr = 0.2V if VS,rms = 12.6V , f =

60Hz, R = 1k⌦, and VON = 1V .

C =

VL,peak � Vr/2

fRVr=

11.82V � 0.1V

60⇥ 1k⌦⇥ 0.2V' 1000µF

9.0V

Example: Find ripple voltage if the secondary voltage VS,rms = 12.6V ,

f = 60Hz, R = 15⌦, C = 25, 000µF , and the diode’s voltage drop VON = 1V .

VS,peak =

p2⇥ VS,rms = 12.82V

vL,peak = VS,peak � VON = 11.82V

fRC = 60⇥ 15⌦⇥ 25⇥ 10

�3F = 22.5

Vr =

vL,peak

fRC

1

1 +

12fRC

=

11.82V

22.5

1

1 +

12⇥22.5

= 0.52V

Example: Find C necessary to obtain Vr = 0.2V if VS,rms = 12.6V , f =

60Hz, R = 15⌦, and VON = 1V .

C =

VL,peak � Vr/2

fRVr=

11.82V � 0.1V

60⇥ 15⌦⇥ 0.2V' 65000µF

Example: Find C necessary to obtain Vr = 0.2V if VS,rms = 12.6V , f =

60Hz, R = 1k⌦, and VON = 1V .

C =

VL,peak � Vr/2

fRVr=

11.82V � 0.1V

60⇥ 1k⌦⇥ 0.2V' 1000µF

9.0V

Peak inverse voltage (PIV) rating of the rectifier diode gives the breakdown voltage.

When diode is off, reverse-bias across diode is Vdc - vs. When vs reaches negative peak,

PIV value corresponds to minimum value of Zener breakdown voltage for rectifier diode.

PIV ≃ 2 VS, PEAK

Full-wave rectifiers cut capacitor discharge time in half and require half the filter capacitance to achieve given ripple voltage. All specifications are same as for half-wave rectifiers. Reversing polarity of diodes gives a full-wave rectifier with negative output voltage.

To find C or Vr use same formula with f = 120HzPIV ≃ 2VS, PEAK

Full-wave rectifier with center-tap transformer

Requirement for a center-tapped transformer in the full-wave rectifier is eliminated through use of 2 extra diodes.All other specifications are same as for a half-wave rectifier except PIV=VP.

Full-wave rectifier with bridge circuit

ID,av = IL(1 + ��

Vp/2Vr)

ID,max = IL(1 + 2��

Vp/2Vr)

ISC = 2�fCVp

Repetitive diode current (after first cycle)

Diode surge current: can flow initially, when power is first applied.• cap is discharged• Sec. voltage can be the peak

DIODE CURRENT SPECS

Jaeger/Blalock 7/1/03

Microelectronic Circuit Design McGraw-Hill

Rectifier Topology Comparison

•  Filter capacitor is a major factor in determining cost, size and weight in design of rectifiers.

•  For given ripple voltage, full-wave rectifier requires half the filter capacitance as that in half-wave rectifier. Reduced peak current can reduce heat dissipation in diodes. Benefits of full-wave rectification outweigh increased expenses and circuit complexity (a extra diode and center-tapped transformer).

•  Bridge rectifier eliminates center-tapped transformer, PIV rating of diodes is reduced. Cost of extra diodes is negligible.

Chap 3 -59

User name: Manuel Toledo Book: Microelectronic Circuits, Sixth Edition, Instructor’s Edition Page: 225. No part of any book may be reproduced or transmitted byany means without the publisher's prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will beprosecuted to the full extent of the law.

User name: Manuel Toledo Book: Microelectronic Circuits, Sixth Edition, Instructor’s Edition Page: 225. No part of any book may be reproduced or transmitted byany means without the publisher's prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will beprosecuted to the full extent of the law.