# E7 Half-Wave Rectifi.. - Electrical Engineering

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Oct 7, 2013 (4 years and 9 months ago)

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Experiment E7 DC Power Supply

Worst
-
Case Design for Half
-
Wave Rectifier Circuit

James J. Whalen

Fall 2000

Experiment No. 7 DC Power Supply (Half
-
Wave Rectifier Circuit) provides an opportunity
to perform a worst
-
case design. Since design is the distingui
shing characteristic of the engineering
profession, ABET recommends the teaching of design at every opportunity. Unfortunately, there is
no easy way to teach design. Design, by its very nature, requires iteration. Usually, several designs
are necessary. Th
e design values given in the Lecture Slides for EE 312 & EE 352 in Fall 1999 are
but a first step. The initial power supply design might not perform well for a +10% &
-
10%
variation in ac voltage and a 100% variation in dc load current. The Zener Diode co
uld burn up at
110% ac voltage when the load resistor is removed. That is not good. The Zener Diode could cut
off at 90% ac voltage when the load resistor connected. That is not good. A second design was
done in Fall 1999 to deal with these problems. The

design procedure developed is called a worst
-
case design.

Important Design Rules

1.

Avoid destroying the Zener diode.

2.

The design must work for any Zener Diode with V
Z

= 12 V

10%.

3.

The design must work with the load resistor connected or disconnected.

4.

The design must work for an isolation transformer rms ac output voltage = 18 VAC

10%.

5.

The design must work for resistors with a

10% tolerance.

6.

The best design would have the best combination of the following factors:

A.

The highest allowable value of dc l
oad current at 100% ac voltage.

B.

Low peak
-
to
-
peak ripple voltage under all conditions of

load variation and ac voltage variation.

C.

Small change in dc load voltage under all conditions of

load variation and ac voltage variation.

Procedure

1.

Design the dc powe
possible. Include the circuit schematic, design equations, and design values. Note that the
circuit schematic for a half
-
wave rectifier circuit is given in the References.

2.

notebook create a table of design values that includes values for the following
components: R
L
, R
F
, R
S
, C
I
, & C
F.
You will be required to show your table to the Staff before
you are allowed to assemble the dc power supply. You may be required to explain so
me aspects

3.

Assemble the entire dc power supply and test it as an assembly. Note that this procedure is
different from the procedure used previously in Exp. No. 8 in which the dc power supply was
asse
mbled and tested section by section.

4.

Test the entire dc power supply at 100% ac voltage (18 VAC) with the load resistor connected.
Measure the dc voltages and ac peak
-
to
-
peak ripple voltages at the nodes 1, 2 & 3 in Figure 1.
ults. Staff must witness all six values of voltages and sign off on
each in a space below the entry. Staff must also check and initial calculations for I
Z

+ I
L
, I
Z
, & +
I
L
.

5.

Repeat Step 4 with load resistor removed. No demonstration required.

6.

Repeat Step 4

at 90% ac voltage (16.2 VAC) with the load resistor connected. Staff must
witness dc voltage and ac peak
-
to
-
peak ripple voltage at the node 3 and sign off on each in a
space below the entry. Staff must also check and initial calculations for I
Z

+ I
L
, I
Z
,
& + I
L
. Is the
Zener Diode cutoff?

7.

Repeat Step 4 at 90% ac voltage with the load resistor removed. No demonstration required.

8.

Repeat Step 4 at 110% ac voltage (19.8 VAC) with the load resistor connected. No
demonstration required.

9.

Repeat Step 4 at 110% ac
voltage with the load resistor removed. Staff must witness dc voltage
and ac peak
-
to
-
peak ripple voltage at the node 3 and sign off on each in a space below the entry.
Staff must also check and initial calculations for I
Z

+ I
L
, I
Z
, & + I
L
. Is the Zener Dio
de
maximum current rating exceeded?

10.

Enter your result from Steps 4
-
9 in a TABLE: DC & AC RIPPLE VOLTAGES

11.

Also enter in the table the percent dc load voltage regulation. The percent dc load voltage
regulation is calculated using the dc load voltage obtained

in Step 4 as the standard and using
the following equation:

L

(Step N))
-

V
L

(Step 4)}

V
L

(Step 4) X 100%

where N = 5 to 9.

12.

Summarize in 250 words or less the important results.

13.

Submit the copies of your lab notebook pages and any

PSPICE or ELECTRONIC
WORKBENCH simulations done in support of the design done prior to the lab.

Zener Diode Information

1N4742

Maximum Ratings

DC Power Dissipation

1 Watt

Derating Factor

6.67 mW per degree C

Junction Temperature

200 degree C

El
ectrical Characteristics at 25 C

Nominal Zener Voltage V
Z
@ I
ZT

12 V

Test Current I
ZT

21 mA

Max Zener Impedance Z
ZT

@ I
ZT

9 ohms

Test Current I
ZK

0.25 mA

Max Zener Impedance Z
ZT

@ I
ZK

700 ohms

Max Reverse Leakage Current I
R

5 uA @ V
R
= 9.1 V

Ma
x DC Zener Current
IZM

76 mA

TABLE 1 VALUES OF DC VOLTAGES AT NODES 1
-
3 AS A FUNCTION OF VAC

%

VAC

(rms)

Peak

Rectifier

Diode

Voltage

V1(dc)

0.9 X (Peak

-

0.7)

V3(dc)

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ZM

= 76 mA. The worst case is at 110% ac voltage. The dc voltage drop
across the resistors R
F
+ R
S

is then given by V1(dc)
-

V3(dc) = (R
F
+ R
S
) X I
Z
. The worst case
occurs at 110% ac voltage where V1 (dc)
-

V3 (dc) has a maximum value giv
en by max {V1 (dc)
-

V3 (dc)} = 12.6 + 1.2 = 13.8 V. To limit the dc current that flows through the Zener Diode to a
value less than I
ZM

= 76 mA, the value for (R
F
+ R
S
) must exceed a minimum value for R
F
+ R
S
given by

Min {R
F
+ R
S
} = max {V1 (dc)
-

V3 (dc
)}

I
ZM

= 13.8 V/76mA = 0.1816 kohm

Min {R
F
+ R
S
} = 13.8 V/76mA = 0.1816 kohm

If we allow for a 10% tolerance on the resistors, then

Min {R
F
+ R
S
} = 0.1816 kohm + 10% X 0.1816 kohm = 0.1997 kohm

Min {R
F
+ R
S
} = 0.1997 kohm

The nominal values avail
able include 100 ohms. Thus an available choice is

R
F
= R
S

= 0.100 kohm = 100 ohm

2. Another important consideration is that the dc current through the Zener Diode must not fall
below some minimum value I
Zmin
. The worst case occurs at 90% ac voltage wi
connected. The voltage V1(dc)
-

V3(dc) has a minimum value min{V1(dc)
-

V3(dc)} = 8.0
-

1.2 =
6.8 V. The current through R
F
+ R
S

is a minimum and is given by

Min {IZ + I
} = min {V1 (dc)
-

V3 (dc)}

(R
F
+ R
S
)

Min {IZ + I
} = 6
.8V

0.200k = 34 mA

3. A value for I
Z

= 0.25 mA where the Max Zener Impedance Z
ZT

= 700 ohms is obviously too low.
The dc voltage regulation will be poor, and the ac peak
-
to
-
peak ripple voltage will be high. As an
initial design decision a value for I
Zm
in

= 0.5 X I
ZT

= 0.5 X 21 mA = 10.5 mA is selected.

I
Zmin

= 10.5 mA

It should be noted that no design decision is final. The value selected for I
Zmin

can be changed in a
subsequent design.

4. The value for the load current I

can be determined fro
m

I

= min {IZ + I
}
-

I
Zmin

I

= 34 mA
-

10.5 mA = 23.5 mA

I

= 23.5 mA

5. The value for the load resistor R
L

can be determined from

R
L

= V
Z

I

R
L

= 12V

23.5mA = 0.5106 kohm

R
L

= 511 ohm

The variable resistor available for
R
L

should be set at 511 ohms. Use the DMM.

6. The resistors available have nominal values as given the Appendix.

7. The capacitors available have values 200

F, 100

F, and 50

F.

8. Initially the 200

F capacitor & 50

F capacitor are connected in p
arallel for the capacitor C
I
in
Figure 1, i. e. the capacitor C
I

= 250

F.

C
I

= 250

F = 0.250 mF

9. The maximum droop

V across the capacitor across C
I
in Figure 1 can be estimated. See figure ?
in Lecture Slides. Using the expression Q = CV where Q i
s the charge on a capacitor C and V the
across the capacitor, the droop

V can be calculated from

V =

Q/C = I

t/C

where I = I
Z

+ I
is the discharge current for the capacitor. The worst case droop would be
obtained where I = I
Z

+ I

has a maximu
m value max{I
Z

+ I
} . Since the value for the sum
of the resistors (R
F
+ R
S
) was chosen to limit the value of the current through them to a value I
Z

+
I

I
ZM

= 76 mA, the maximum value for I is I
ZM

= 76 mA. The maximum droop max{

V} is
given by

Max {

V} = I
ZM
/60C

Max {

V} = 76mA /(60 X 0.250mF) = 5.1 V

where

t has been set equal to 1/f where f = 60 Hz. Actually, as shown in the Lecture slides, the
value for

t will be less than 1/f. For example, for

V/V = 0.8 (20% droop), the value for

t is
given
by

t = 0.9/f. Thus the maximum droop calculated with the equations above would be over
-
estimated by 10%.

The filter capacitor also contributes charge to the current that flows through the Zener diode and the
R
L
. If the value of the filter capacito
r C
F

= 100

F = 0.100 mF is added to the value of C
F

= 250

F =
0.250 mF, the value obtained is

C
F

+ C
I

= 250

F + 100

F = 350

F = 0.35 mF

If the value C
F

+ C
I

= 0.35 mF is used to calculate the droop, the droop value is given by

Max {

V} = 76mA /(6
0 X 0.350mF) = 3.6 V

Which value is the better choice: 5.1 V or 3.6 V? That is a question best answered using PSPICE.

Since the capacitors have 20% tolerance, the droop calculated or predicted using PSPICE
could differ from that measured by 20%.

10. Th
e final design decision that must be made is how to divide the sum of the resistors (R
F
+ R
S
).
The filter capacitor C
F

and the filter resistor R
F

form a low pass RC filter and cause a reduction in
the ac voltage between the input at node 1 and the output a
t node 2. The reduction in the ratio
V2/V1 for a sinusoidal signal at frequency f is given by

V2/V1 = 1/{1 + (R
F

2

fC
F
)
2
}
0.5

One of the value used for f can be f = 60 Hz. Harmonics at 120 Hz and higher are also present.
They harmonics will be attenuated

more. Previously, an initial assignment R
F

= 100 ohms was
F

= 100 ohms, C
F

= 100

F, & f = 60 Hz, the value for V2/V1 is given by

V2/V1 = 1/{1 + (100X2

X60X100

)
2
}
0.5

= 0.256

Since the harmonics are attenuated even more, the variation in v
oltage across C
F

will be even less
then 0.256 times the variation in voltage across C
I
. Again this is a question best answered using
PSPICE.

11.

The Zener diode also causes a reduction in the ac voltage between the input at node 2 and
the output at node 3
. . The reduction in the ratio V3/V2 is given by

V3/V2 = {R
L

r
d
}

{R
S
+ R
L

r
d
}

where r
d
is the dynamic resistance of the Zener Diode. In E6 Dynamic Impedance a value r
d
= 5.6
ohms at I
Z

= 10 mA was measured. Using R
S

= 100 ohms and R
L

r
d

= 511

5.6 = 5.
5, the value for
V3/V2 is given by

V3/V2 = 5.5

{100

+ 5.5} = 0.052

Thus the ac voltage at V3 should be more than an order of magnitude less than that at V2. Usually,
voltage at V3 is a few mV. Averaging and manual adjustments of the cursors are the b
est
experimental techniques to use to measure voltage at V3.

12. The ac voltage at V3 can be now be estimated. Multiple the results obtained for the droop in V1
and the voltage ratios V3/V2 & V2/V1. The results are

V3
pp

= 5.1V X 0.256 X 0.052 = 0.067 V =

67 mV

V3
pp

= 3.6V X 0.256 X 0.052 = 0.048 V = 48 mV

These results are much larger than that observed. . Again this is a question best answered using
PSPICE.

13. The division of the sum of the resistors (R
F
+ R
S
) should have as its goal making the over
all
reduction in the ac voltage large by making the product of the two voltage expressions small. That
is make V3/V1 small where V3/V1 is given by

V3/V1 = 1/{1 + (R
F

2

fC
F
)
2
}
0.5

X {R
L

r
d
}

{R
S
+ R
L

r
d
}

The values for R
F
& R
S

are limited to what is avail
able on the Heathkit Resistance Substitution
Boxes. There are not too many combinations to test. It is recommended that neither R
F
nor R
S
be
set equal to zero. Remember that R
F
+ R
S
must exceed the minimum value calculated previously to
prevent zapping th
e Zener Diode. The first set of values recommended for E7 may be summarized
as follows:

R
F
= R
S

= 0.100 kohm = 100 ohm

R
L

= 511 ohm

C
I

= 250

F

C
F

= 100

F

TABLE 2 DATA FOR DC POWER SUPPLY WITH FULL
-
WAVE RECTIFIER

The Fluke 8000A DMM was used to me
asure V
T
. The HP54600B CRO was used to
measure V1(avg), V1(pp), V2(avg), V2(pp), and V3(pp). The Fluke 8010A DMM was used to
L

was calculated using I
L

= V3(avg)/R
L.
The
current I
L
+ I
Z

was calculated u
sing I
L
+ I
Z

= {V3(avg)
-

V1(avg)}

{R
F

+ R
S
}. The Zener Diode
current was calculated using I
Z

= I
L
+ I
Z

I
L
.

TABLE 2 DATA FOR DC POWER SUPPLY WITH FULL
-
WAVE RECTIFIER

C = R
L
CONNECTED & NC = R
L
NOT CONNECTED

X DENOTES INITIALS REQUIRED

%

V
T

R
L

V1

V1

V
2

V2

V3

V3

I
L

I
L
+I
Z

I
Z

VAC

rms

avg

pp

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pp

avg

pp

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PSPICE (or ELECTRONIC WORKBENCH) SIMULATION: Week 2

During the second week a PSPICE EXAM on the dc power supply circuit will be
conducted. PSPICE will be used to simulate the

half
-
wave rectifier dc power supply circuit and to
determine values for all the voltages and currents measured.

The 50

F capacitor was connected in parallel with the 200

F. However, it may be that a
better dc power supply results by connecting the 50

F capacitor in parallel with the 100

F
capacitor and using the parallel combination as the filter capacitor C
F
. This is one of the alternative
designs that can be checked using PSPICE. Other alternative designs compatible with the worst
-
case design proce
dure such as using 150 ohms & 47 ohms for R
F &

R
S

can also be checked.

References:

1.

Sedra/Smith, “Microelectronics Circuits,” 4
th

ed. New York: Oxford University Press, 1998,
Section 3.7 Rectifier Circuits, pp. 179
-
191.

2.

EE 352 Fall 1999 Lectu
re Slides on Experiment No. 8 DC Power Supply

3.

EE 312 Fall 2000 Lecture Slides on Experiment No. 7 DC Power Supply

Appendix: Resistance Values Available

There are 29 Heathkit boxes that are available. The Heathkit resistance substitution box resis
ter
choices are:

Low value:

15, 22, 33, 47, 68, 100, 150, 220, 330, 470, 680, 1k, 1.5k,

2.2k, 3.3k, 4.7k 6.8k 10k

A different resistance box is used to augment the 29 Heathkit boxes (20 boxes are available).

Low Values:

10, 47, 100, 220, 470, 1k
, 2.2, 3.3k, 4.7k 6.8k 10k